[ALGEBRA] Seen Questions

Worked solutions on linear maps, projections, kernels, images, rank-nullity and linear operators.

These past exam questions move between the abstract and coordinate views of a linear transformation. They run from basis data and coordinates, through operators on function spaces and geometric projections, to rank and dimension arguments.

Some questions use eigenvalue or diagonalisation ideas from the eigenvalues chapter. They are marked Later-topic preview, so you can skip them on a first reading and return after studying that chapter. For the geometric picture, review geometric examples.

In each solution, first identify the structural fact being tested, such as preservation of a linear combination or a kernel inclusion and only then carry out the calculation.

[THEOREM]

Values on a basis determine a linear map

Let $B=(u_1,\ldots,u_n)$ be a basis for $V$ and let $T:V\to W$ be linear. If

v=\omega_1u_1+\cdots+\omega_nu_n,

then

T(v)=\omega_1T(u_1)+\cdots+\omega_nT(u_n).

Therefore the values of $T$ on the basis vectors determine $T(v)$ for every $v\in V$ .

[EXAMPLE]

A linear transformation from a quadrilateral

A mapping $T:\mathbb R^2\to\mathbb R^2$ is a linear transformation. A quadrilateral $ABCD$ and its image under $T$ have vertices

A=(-3,1),\qquad B=(-1,1),\qquad C=(1,0),\qquad D=(1,-1),

and

T(A)=(-1,-3),\qquad T(B)=(-1,-1),\qquad T(C)=(0,1),\qquad T(D)=(1,1).

(a) Find the coefficients $\alpha_1,\beta_1,\alpha_2,\beta_2$ such that

\begin{pmatrix}1\\0\end{pmatrix} =\alpha_1 \begin{pmatrix}-3\\1\end{pmatrix} +\beta_1 \begin{pmatrix}-1\\1\end{pmatrix}; \qquad \begin{pmatrix}0\\1\end{pmatrix} =\alpha_2 \begin{pmatrix}-3\\1\end{pmatrix} +\beta_2 \begin{pmatrix}-1\\1\end{pmatrix}.

(b) Find a matrix $P$ such that $T(\mathbf{x})=P\mathbf{x}$ .

Solution

The vectors on the right in part (a) are $A$ and $B$ , so part (a) expresses the standard basis vectors $\mathbf e_1=(1,0)^{\mathsf T}$ and $\mathbf e_2=(0,1)^{\mathsf T}$ in terms of $A$ and $B$ . This is possible because $A$ and $B$ are not parallel: the determinant of the matrix with columns $A$ and $B$ is $(-3)(1)-(-1)(1)=-2\ne0$ .

For $\mathbf e_1$ , comparing coordinates gives

-3\alpha_1-\beta_1=1, \qquad \alpha_1+\beta_1=0.

The second equation gives $\beta_1=-\alpha_1$ . Substitution into the first equation gives

-3\alpha_1+\alpha_1=1 \quad\Longrightarrow\quad -2\alpha_1=1 \quad\Longrightarrow\quad \alpha_1=-\frac12, \qquad \beta_1=\frac12.

For $\mathbf e_2$ , the coordinate equations are

-3\alpha_2-\beta_2=0, \qquad \alpha_2+\beta_2=1.

Using $\beta_2=1-\alpha_2$ in the first equation gives

-3\alpha_2-(1-\alpha_2)=0 \quad\Longrightarrow\quad -2\alpha_2-1=0 \quad\Longrightarrow\quad \alpha_2=-\frac12, \qquad \beta_2=\frac32.

Hence

{ \alpha_1=-\frac12,\quad \beta_1=\frac12,\quad \alpha_2=-\frac12,\quad \beta_2=\frac32 }.

Because $T$ is linear, apply $T$ to these coordinate identities:

\begin{aligned} T(\mathbf e_1) &=-\frac12T(A)+\frac12T(B)\\ &=-\frac12 \begin{pmatrix}-1\\-3\end{pmatrix} +\frac12 \begin{pmatrix}-1\\-1\end{pmatrix}\\ &=\begin{pmatrix}0\\1\end{pmatrix}, \end{aligned}

and

\begin{aligned} T(\mathbf e_2) &=-\frac12T(A)+\frac32T(B)\\ &=-\frac12 \begin{pmatrix}-1\\-3\end{pmatrix} +\frac32 \begin{pmatrix}-1\\-1\end{pmatrix}\\ &=\begin{pmatrix}-1\\0\end{pmatrix}. \end{aligned}

The columns of the standard-coordinate matrix are the images of $\mathbf e_1$ and $\mathbf e_2$ . Therefore

{ P= \begin{pmatrix} 0&-1\\ 1&0 \end{pmatrix} }.

For a check, this matrix sends the four given vertices to

P \begin{pmatrix}-3\\1\end{pmatrix} =\begin{pmatrix}-1\\-3\end{pmatrix}, \quad P \begin{pmatrix}-1\\1\end{pmatrix} =\begin{pmatrix}-1\\-1\end{pmatrix},

P \begin{pmatrix}1\\0\end{pmatrix} =\begin{pmatrix}0\\1\end{pmatrix}, \quad P \begin{pmatrix}1\\-1\end{pmatrix} =\begin{pmatrix}1\\1\end{pmatrix}.

Thus the matrix agrees with every given image vertex. Part (a) recovers the standard-basis images from $T(A)$ and $T(B)$ and those images become the columns of $P$ . This is the coordinate form of the fact that a linear map is determined by its values on a basis.

[exam]

[THEOREM]

Values on a basis determine a linear map

Let $B=(u_1,\ldots,u_n)$ be a basis for $V$ and let $T:V\to W$ be linear. If

v=\omega_1u_1+\cdots+\omega_nu_n,

then

T(v)=\omega_1T(u_1)+\cdots+\omega_nT(u_n).

Therefore the values of $T$ on the basis vectors determine $T(v)$ for every $v\in V$ .

[EXAMPLE]

Values of a partially specified linear transformation

Let $T:\mathbb R^3\to\mathbb R^2$ be a linear transformation such that

T(\mathbf e_1)= \begin{pmatrix} -9\\ 8 \end{pmatrix} \qquad\text{and}\qquad T\begin{pmatrix} 0\\ 1\\ 1 \end{pmatrix} = \begin{pmatrix} 2\\ 8 \end{pmatrix},

where $\mathbf e_1=(1,0,0)^T$ .

Using the information given, compute the following, or state that there is not enough information by entering the word undetermined.

(a)

T\begin{pmatrix} 0\\ 0\\ 0 \end{pmatrix} =\ ?

(b)

T\begin{pmatrix} 0\\ 1\\ 0 \end{pmatrix} =\ ?

(c)

T\begin{pmatrix} 1\\ 1\\ 1 \end{pmatrix} =\ ?

Now suppose that you also know

T(\mathbf e_3)= \begin{pmatrix} 4\\ -9 \end{pmatrix}.

(d)

Compute all of the vectors that you could not compute above:

T\begin{pmatrix} 0\\ 0\\ 0 \end{pmatrix}, \qquad T\begin{pmatrix} 0\\ 1\\ 0 \end{pmatrix}, \qquad T\begin{pmatrix} 1\\ 1\\ 1 \end{pmatrix}.

Solution

Rewrite each input as a linear combination of the vectors whose images are known. Linearity applies the same coefficients to the corresponding output vectors.

For (a), every linear transformation sends the zero vector to the zero vector:

{ T\begin{pmatrix}0\\0\\0\end{pmatrix} = \begin{pmatrix}0\\0\end{pmatrix} }.

For (b), the given information does not determine $T(\mathbf e_2)$ . Let $\mathbf r=T(\mathbf e_3)$ . Since

\mathbf e_2 = \begin{pmatrix}0\\1\\1\end{pmatrix} -\mathbf e_3,

linearity gives

T(\mathbf e_2) = T\begin{pmatrix}0\\1\\1\end{pmatrix} -T(\mathbf e_3) = \begin{pmatrix}2\\8\end{pmatrix} -\mathbf r.

The vector $\mathbf r$ is not known at this stage, so the answer to (b) is

{\text{undetermined}}.

For (c),

\begin{pmatrix}1\\1\\1\end{pmatrix} =\mathbf e_1+ \begin{pmatrix}0\\1\\1\end{pmatrix}.

Therefore

\begin{aligned} T\begin{pmatrix}1\\1\\1\end{pmatrix} &=T(\mathbf e_1) +T\begin{pmatrix}0\\1\\1\end{pmatrix}\\ &=\begin{pmatrix}-9\\8\end{pmatrix} +\begin{pmatrix}2\\8\end{pmatrix}\\ &={\begin{pmatrix}-7\\16\end{pmatrix}}. \end{aligned}

For (d), the additional information is

T(\mathbf e_3)= \begin{pmatrix}4\\-9\end{pmatrix}.

For the previously undetermined vector, use

\mathbf e_2+\mathbf e_3 =\begin{pmatrix}0\\1\\1\end{pmatrix}.

Thus

\begin{aligned} T(\mathbf e_2) &=T\begin{pmatrix}0\\1\\1\end{pmatrix}-T(\mathbf e_3)\\ &=\begin{pmatrix}2\\8\end{pmatrix} -\begin{pmatrix}4\\-9\end{pmatrix}\\ &={\begin{pmatrix}-2\\17\end{pmatrix}}. \end{aligned}

The three requested entries in part (d) are therefore

\begin{aligned} T\begin{pmatrix}0\\0\\0\end{pmatrix} &=\begin{pmatrix}0\\0\end{pmatrix},\\[2mm] T\begin{pmatrix}0\\1\\0\end{pmatrix} &=\begin{pmatrix}-2\\17\end{pmatrix},\\[2mm] T\begin{pmatrix}1\\1\\1\end{pmatrix} &=\begin{pmatrix}-7\\16\end{pmatrix}. \end{aligned}

The first data set fixes $T(\mathbf e_1)$ and the sum $T(\mathbf e_2)+T(\mathbf e_3)$ , but not the two summands separately. The extra value of $T(\mathbf e_3)$ removes exactly that remaining freedom. Once the images of all three standard basis vectors are known, every value of the linear map is forced by its coordinates.

[exam]

[DEFINITION]

Trigonometric polynomial

A trigonometric polynomial is an expression of the form

a_0+\sum_{k=1}^N\bigl(a_k\cos(k\theta)+b_k\sin(k\theta)\bigr), \qquad \theta\in\mathbb R,

where the coefficients $a_k,b_k$ are complex numbers.

[1]:Trigonometric polynomials and a differential operator

[EXAMPLE]

Trigonometric polynomials and a differential operator

Later-topic preview: the interpretation in terms of eigenvalues uses ideas from the later eigenvalues chapter.

A real trigonometric polynomial of degree $N$ , where $N$ is a positive integer, is a function $f:\mathbb R\to\mathbb R$ of the form

f(x)=a_0+\sum_{n=1}^{N}a_n\cos(nx)+\sum_{n=1}^{N}b_n\sin(nx), \tag{1}

where $a_n,b_n\in\mathbb R$ and the highest-degree coefficients $a_N,b_N$ are not both zero. For $N=0$ , the polynomial has degree zero and $f(x)=a_0$ for all $x\in\mathbb R$ .

(a)

Using Euler's formula, write the function in (1) as

f(x)=\sum_{n=-N}^{N}c_ne^{inx}.

Find $c_0$ and $c_n$ for $n\ne0$ .

(b)

The set of all real trigonometric polynomials of degree at most $45$ , denoted by $\operatorname{Trig}_{45}(\mathbb R)$ , is a vector space over $\mathbb R$ with the usual addition and scalar multiplication.

For the following statements, let $d$ be an integer satisfying $1\le d\le45$ .

Consider the linear transformation

T:\operatorname{Trig}_{45}(\mathbb R)\longrightarrow \operatorname{Trig}_{45}(\mathbb R), \qquad T(f)=f''+16f.

(i)

Choose all correct statements:

For all such $d$ , $\cos(3x)+\sin(dx)$ is in $\ker(T)$ .
For all such $d$ , $\cos(3x)+\sin(dx)$ is not in $\ker(T)$ .
For some but not all such $d$ , $\cos(3x)+\sin(dx)$ is in $\ker(T)$ .
For all such $d$ , $\cos(3x)+\sin(dx)$ is in $\operatorname{im}(T)$ .
For all such $d$ , $\cos(3x)+\sin(dx)$ is not in $\operatorname{im}(T)$ .
For some but not all such $d$ , $\cos(3x)+\sin(dx)$ is in $\operatorname{im}(T)$ .

(ii)

Find a basis for $\ker(T)$ .

(iii)

Find $\operatorname{im}(T)$ and explain your answer.

Solution

For part (a), use

\cos(nx)=\frac{e^{inx}+e^{-inx}}2, \qquad \sin(nx)=\frac{e^{inx}-e^{-inx}}{2i}.

Therefore

\begin{aligned} a_n\cos(nx)+b_n\sin(nx) &=\frac{a_n-ib_n}{2}e^{inx} +\frac{a_n+ib_n}{2}e^{-inx}. \end{aligned}

The coefficient of the zero-frequency term is unchanged, so

{c_0=a_0}.

For $k=1,\ldots,N$ , the positive-frequency coefficient is

{c_k=\frac{a_k-ib_k}{2}.}

The corresponding negative-frequency coefficient is

{c_{-k}=\frac{a_k+ib_k}{2}.}

As a check, because the original polynomial is real-valued,

c_{-n}=\overline{c_n}.

For part (b), the restriction $1\le d\le45$ ensures that $\cos(3x)+\sin(dx)$ belongs to $\operatorname{Trig}_{45}(\mathbb R)$ . The operator $T=D^2+16I$ is linear because differentiation and addition preserve linear combinations.

For $k\ge1$ ,

\frac{d^2}{dx^2}\cos(kx)=-k^2\cos(kx), \qquad \frac{d^2}{dx^2}\sin(kx)=-k^2\sin(kx).

Consequently,

T(\cos(kx))=(16-k^2)\cos(kx), \qquad T(\sin(kx))=(16-k^2)\sin(kx),

and $T(1)=16$ .

(b)(i)

For the proposed polynomial,

T\bigl(\cos(3x)+\sin(dx)\bigr) =7\cos(3x)+(16-d^2)\sin(dx).

The coefficient of $\cos(3x)$ is always $7$ , so this expression can never be zero. Hence $\cos(3x)+\sin(dx)$ is in the kernel for no integer $d$ with $1\le d\le45$ .

The only frequency missing from the image is $4$ , because

16-4^2=0.

If $d=4$ , the term $\sin(4x)$ cannot be produced by $T$ , so

\cos(3x)+\sin(4x)\notin\operatorname{im}(T).

If $d\ne4$ , both frequencies $3$ and $d$ have non-zero multiplier $16-k^2$ , so the polynomial is in the image. For example, when $d\ne4$ ,

T\left(\frac17\cos(3x)+\frac1{16-d^2}\sin(dx)\right) =\cos(3x)+\sin(dx).

The correct selections are therefore

{\text{for all integers }d\text{ with }1\le d\le45,\ \cos(3x)+\sin(dx)\notin\ker(T)}

and

{\text{for some but not all integers }d\text{ with }1\le d\le45,\ \cos(3x)+\sin(dx)\in\operatorname{im}(T)}.

(b)(ii)

Write

f(x)=a_0+\sum_{k=1}^{45} \bigl(a_k\cos(kx)+b_k\sin(kx)\bigr).

Then

T(f)=16a_0+ \sum_{k=1}^{45}(16-k^2) \bigl(a_k\cos(kx)+b_k\sin(kx)\bigr).

The functions $1,\cos(kx),\sin(kx)$ at distinct frequencies are linearly independent. One way to see this is to use part (a), multiply a zero exponential sum by $e^{45ix}$ and obtain a complex polynomial in $e^{ix}$ that vanishes at infinitely many points of the unit circle; a non-zero polynomial cannot do that.

Therefore $T(f)=0$ forces every displayed coefficient to vanish, so the constant coefficient and all frequencies except $k=4$ must be zero:

a_0=0,\qquad a_k=b_k=0\quad(k\ne4).

The coefficients $a_4,b_4$ are unrestricted because $16-4^2=0$ . Hence

\ker(T)=\operatorname{span}\{\cos(4x),\sin(4x)\},

so one basis is

{\{\cos(4x),\sin(4x)\}}.

(b)(iii)

Every output of $T$ has zero $\cos(4x)$ and $\sin(4x)$ coefficients. Conversely, if

g(x)=\gamma_0+ \sum_{\substack{1\le k\le45\\k\ne4}} \bigl(\gamma_k\cos(kx)+\delta_k\sin(kx)\bigr),

then the polynomial

h(x)=\frac{\gamma_0}{16} +\sum_{\substack{1\le k\le45\\k\ne4}} \left( \frac{\gamma_k}{16-k^2}\cos(kx) +\frac{\delta_k}{16-k^2}\sin(kx) \right)

belongs to $\operatorname{Trig}_{45}(\mathbb R)$ and satisfies $T(h)=g$ . Therefore

{ \operatorname{im}(T) =\operatorname{span} \left( \{1\}\cup \{\cos(kx),\sin(kx):1\le k\le45,\ k\ne4\} \right) }.

Equivalently, the image is the set of trigonometric polynomials in $\operatorname{Trig}_{45}(\mathbb R)$ whose $\cos(4x)$ and $\sin(4x)$ coefficients are both zero. Its dimension is $91-2=89$ , in agreement with rank-nullity.

The operator $D^2+16I$ acts diagonally, multiplying each frequency $k$ by $16-k^2$ . Frequency $4$ has eigenvalue zero, so its two real directions form the kernel and removing exactly those two directions gives the image.

[exam]

[1]:Trigonometric polynomial
[2]:Linear map
[3]:Exponential forms of sine and cosine

[DEFINITION]

Kernel

If $T:V\to W$ is a linear map, its kernel is

\ker(T)=\{u\in V:T(u)=0\}.

The same set is also called the null space. If $A\in\mathbb F^{m\times n}$ is regarded as the map $\mathbb F^n\to\mathbb F^m$ , then

\ker(A)=\{x\in\mathbb F^n:Ax=0\}.

[DEFINITION]

Nullity

The nullity of a linear map $T$ is

\operatorname{nullity}(T)=\dim\ker(T).

For a matrix $A$ , the nullity is

\operatorname{nullity}(A)=\dim\ker(A).

[1]:Kernel
[2]:Dimension
[3]:Column independence and nullity
[4]:Rank-nullity theorem
[5]:Rank-nullity theorem for vector spaces
[6]:A projection, nullity and a composition

[THEOREM]

Rank-nullity theorem for vector spaces

If $T:V\to W$ is a linear map over a scalar field $\mathbb F$ and $V$ is finite-dimensional, then

\operatorname{rank}(T)+\operatorname{nullity}(T)=\dim(V).

[1]:Nullity
[2]:Rank
[3]:Properties read from a matrix representation
[4]:Integration gives an injective map
[5]:Bijectivity in equal dimensions
[6]:Proof
14 more

[EXAMPLE]

A projection, nullity and a composition

Suppose

\mathbf b= \begin{pmatrix} -2\\ 6\\ -2 \end{pmatrix},

and let $T:\mathbb R^3\to\mathbb R^3$ be the linear transformation defined by

T(\mathbf x)=\operatorname{proj}_{\mathbf b}(\mathbf x) =\left(\frac{\mathbf x\cdot\mathbf b}{\mathbf b\cdot\mathbf b}\right)\mathbf b, \qquad \mathbf x\in\mathbb R^3.

Additionally, suppose

A= \begin{pmatrix} -7&-9&-4\\ 5&-1&9\\ 7&-2&8 \end{pmatrix},

and let $S:\mathbb R^3\to\mathbb R^3$ be the linear transformation defined by

S(\mathbf x)=A\mathbf x, \qquad \mathbf x\in\mathbb R^3.

You are also given that $\det(A)\ne0$ .

(a) Nullity of $T$

(i)

The nullity of $T$ is $\operatorname{nullity}(T)=$

(ii)

Give a brief explanation of how this value is obtained.

(b) Kernel of $S\circ T$

The kernel of $S\circ T$ is the set of all $\mathbf x=(x,y,z)^T\in\mathbb R^3$ satisfying a Cartesian equation.

Briefly explain how the formula for this equation is obtained.

Solution

Part (a) needs only the projection, whose kernel can be found directly. For part (b), $\det(A)\ne0$ makes $S$ invertible, so $S$ sends no non-zero vector to zero and composing with $S$ does not change the kernel of $T$ .

First,

\mathbf b\cdot\mathbf b=(-2)^2+6^2+(-2)^2=4+36+4=44.

Every value of $T$ is a scalar multiple of $\mathbf b$ , so $\operatorname{im}(T)\subseteq\operatorname{span}\{\mathbf b\}$ . Also,

T(\mathbf b) =\left(\frac{\mathbf b\cdot\mathbf b}{\mathbf b\cdot\mathbf b}\right)\mathbf b =\mathbf b,

so $\operatorname{im}(T)=\operatorname{span}\{\mathbf b\}$ and $\operatorname{rank}(T)=1$ . Since the domain has dimension $3$ , the rank-nullity theorem gives

\operatorname{nullity}(T) =\dim(\mathbb R^3)-\operatorname{rank}(T) =3-1 ={2}.

This can also be checked by finding the kernel. For $\mathbf x=(x,y,z)^T$ ,

\begin{aligned} T(\mathbf x)=0 &\Longleftrightarrow \left(\frac{\mathbf x\cdot\mathbf b}{44}\right)\mathbf b=0\\ &\Longleftrightarrow \mathbf x\cdot\mathbf b=0, \end{aligned}

because $\mathbf b\ne0$ . Therefore

\ker(T) =\left\{ \begin{pmatrix}x\\y\\z\end{pmatrix} :-2x+6y-2z=0 \right\}.

Dividing the equation by $-2$ gives $x-3y+z=0$ , or $x=3y-z$ . Hence

\ker(T) =\left\{ s\begin{pmatrix}3\\1\\0\end{pmatrix} +t\begin{pmatrix}-1\\0\\1\end{pmatrix} :s,t\in\mathbb R \right\},

which is a two-dimensional plane, agreeing with the nullity calculation.

For part (b), $\det(A)\ne0$ means that $A$ is invertible. Thus

\ker(S)=\{0\}.

For any $\mathbf x\in\mathbb R^3$ ,

\begin{aligned} \mathbf x\in\ker(S\circ T) &\Longleftrightarrow S(T(\mathbf x))=0\\ &\Longleftrightarrow A\,T(\mathbf x)=0\\ &\Longleftrightarrow T(\mathbf x)=0\\ &\Longleftrightarrow \mathbf x\in\ker(T). \end{aligned}

Consequently,

\ker(S\circ T)=\ker(T) =\left\{ \begin{pmatrix}x\\y\\z\end{pmatrix} :-2x+6y-2z=0 \right\}.

The requested Cartesian equation is therefore

{-2x+6y-2z=0},

equivalently $x-3y+z=0$ .

In summary, the kernel of the projection is the orthogonal complement of $\mathbf b$ and rank-nullity turns the one-dimensional image into nullity $2$ . The determinant condition is used only to show that $S$ is injective, which gives $\ker(S\circ T)=\ker(T)$ .

[exam]

[1]:Kernel
[2]:Nullity
[3]:Rank-nullity theorem for vector spaces

[DEFINITION]

Subspace

Let $V$ be a vector space over $F$ . A subset $S\subseteq V$ is a subspace of $V$ when it is a vector space over the same field $F$ and uses the addition and scalar multiplication inherited from $V$ . A proper subspace is a subspace that is not equal to $V$ .

[DEFINITION]

Eigenvalue

Let $T:V\to V$ be a linear map over a field $F$ . A scalar $\lambda\in F$ is an eigenvalue of $T$ if there is a non-zero vector $v\in V$ such that

T(v)=\lambda v.

[1]:Linear map
[2]:Eigenvectors for distinct eigenvalues
[3]:A real matrix with complex eigenvalues
[4]:Invertibility and eigenvalues
[5]:Diagonalisation
15 more

[DEFINITION]

Eigenvector

If $T(v)=\lambda v$ for a non-zero vector $v$ , then $v$ is an eigenvector of $T$ corresponding to the eigenvalue $\lambda$ .

[1]:Eigenvalue
[2]:Eigenvectors for distinct eigenvalues
[3]:Eigenbasis
[4]:A real matrix with complex eigenvalues
[5]:Diagonalisation
7 more

[EXAMPLE]

A projection onto a subspace

Later-topic preview: parts (b) and (c) use eigenvalues and eigenvectors.

Suppose that $H$ is a subspace of $\mathbb R^3$ defined as follows:

H=\operatorname{span}\left\{ \begin{pmatrix}1\\0\\-1\end{pmatrix}, \begin{pmatrix}1\\-1\\0\end{pmatrix}, \begin{pmatrix}-5\\2\\3\end{pmatrix} \right\}.

(a) Choose the correct statement about $H$

The subspace $H$ is a plane through the origin.
The subspace $H$ is a line which does not go through the origin.
The subspace $H$ is a plane which does not go through the origin.
The subspace $H$ is a line through the origin.
The subspace $H$ is the entire space $\mathbb R^3$ .

The mapping $T:\mathbb R^3\to\mathbb R^3$ is the projection onto $H$ , defined as follows: for any vector $\mathbf x\in\mathbb R^3$ , $T(\mathbf x)$ is the unique vector in $H$ such that $\mathbf x-T(\mathbf x)$ is perpendicular to every vector in $H$ .

It can be proved that $T$ is a linear transformation. It is also known that $T(\mathbf n)=0$ for any normal $\mathbf n$ to $H$ and $T(\mathbf x)=\mathbf x$ for any vector $\mathbf x\in H$ . These facts do not need to be proved here.

(b) Select all that apply

Any nonzero vector in $H$ is an eigenvector of $T$ .
There exists a nonzero vector in $H$ that is not an eigenvector of $T$ .
Any nonzero normal vector to $H$ is an eigenvector of $T$ .
The linear transformation $T$ has at least one zero eigenvalue.
All eigenvalues of $T$ are zero.
All eigenvalues of $T$ are nonzero.
The linear transformation $T$ has at least one eigenvalue that equals $1$ .
The rank of $T$ is $1$ .
The rank of $T$ is $2$ .
The nullity of $T$ is $1$ .
The nullity of $T$ is $2$ .

(c) Select all that apply

The vector $\begin{pmatrix}-1\\-1\\-1\end{pmatrix}$ is an eigenvector of $T$ .
The vector $\begin{pmatrix}1\\0\\-1\end{pmatrix}$ is an eigenvector of $T$ .
The vector $\begin{pmatrix}-2\\-1\\0\end{pmatrix}$ is an eigenvector of $T$ .

Solution

Let

\mathbf u_1=\begin{pmatrix}1\\0\\-1\end{pmatrix}, \qquad \mathbf u_2=\begin{pmatrix}1\\-1\\0\end{pmatrix}, \qquad \mathbf u_3=\begin{pmatrix}-5\\2\\3\end{pmatrix}.

The third vector is redundant because

-3\mathbf u_1-2\mathbf u_2 =-3\begin{pmatrix}1\\0\\-1\end{pmatrix} -2\begin{pmatrix}1\\-1\\0\end{pmatrix} =\begin{pmatrix}-5\\2\\3\end{pmatrix} =\mathbf u_3.

The first two vectors are not scalar multiples, so they are linearly independent. Therefore

H=\operatorname{span}\{\mathbf u_1,\mathbf u_2\}, \qquad \dim(H)=2.

Since a two-dimensional subspace of $\mathbb R^3$ is a plane and every subspace contains the zero vector, the correct statement in part (a) is

{\text{The subspace $H$ is a plane through the origin.}}

A normal vector is

\mathbf n=\begin{pmatrix}-1\\-1\\-1\end{pmatrix},

because

\mathbf n\cdot\mathbf u_1=(-1)(1)+(-1)(0)+(-1)(-1)=0

and

\mathbf n\cdot\mathbf u_2=(-1)(1)+(-1)(-1)+(-1)(0)=0.

Thus

H=\left\{ \begin{pmatrix}x\\y\\z\end{pmatrix}\in\mathbb R^3:x+y+z=0 \right\},

and $\mathbb R^3$ decomposes as

\mathbb R^3=H\oplus\operatorname{span}\{\mathbf n\}.

The projection is the identity on $H$ and is zero on $\operatorname{span}\{\mathbf n\}$ . Relative to the basis $(\mathbf u_1,\mathbf u_2,\mathbf n)$ , its matrix is therefore

\begin{pmatrix} 1&0&0\\ 0&1&0\\ 0&0&0 \end{pmatrix}.

The matrix shows the eigenvalues and the dimensions of the corresponding eigenspaces:

\lambda=1 \text{ on }H, \qquad \lambda=0 \text{ on }\operatorname{span}\{\mathbf n\}.

Consequently, the statements selected in part (b) are

{ \begin{gathered} \text{Any nonzero vector in $H$ is an eigenvector of $T$},\\ \text{any nonzero normal vector to $H$ is an eigenvector of $T$},\\ \text{$T$ has at least one zero eigenvalue},\\ \text{$T$ has at least one eigenvalue equal to $1$},\\ \operatorname{rank}(T)=2,\qquad \operatorname{nullity}(T)=1. \end{gathered} }

Indeed, $\operatorname{im}(T)=H$ , so $\operatorname{rank}(T)=\dim(H)=2$ , while $\ker(T)=\operatorname{span}\{\mathbf n\}$ , so $\operatorname{nullity}(T)=1$ .

For part (c), the first vector is $\mathbf n$ , so

T\begin{pmatrix}-1\\-1\\-1\end{pmatrix} =\mathbf 0 =0\begin{pmatrix}-1\\-1\\-1\end{pmatrix}.

It is an eigenvector for $\lambda=0$ . The second vector is $\mathbf u_1\in H$ , so

T\begin{pmatrix}1\\0\\-1\end{pmatrix} =\begin{pmatrix}1\\0\\-1\end{pmatrix},

and it is an eigenvector for $\lambda=1$ .

For the third vector, let

\mathbf w=\begin{pmatrix}-2\\-1\\0\end{pmatrix}.

The orthogonal projection onto the plane perpendicular to $\mathbf n$ is

T(\mathbf w) =\mathbf w-\frac{\mathbf w\cdot\mathbf n}{\mathbf n\cdot\mathbf n}\mathbf n.

Here $\mathbf w\cdot\mathbf n=3$ and $\mathbf n\cdot\mathbf n=3$ , so

T(\mathbf w) =\begin{pmatrix}-2\\-1\\0\end{pmatrix} -\begin{pmatrix}-1\\-1\\-1\end{pmatrix} =\begin{pmatrix}-1\\0\\1\end{pmatrix}.

This is not a scalar multiple of $\mathbf w$ : comparing second coordinates forces the scalar to be $0$ , but then the first coordinates $-1$ and $0$ disagree. Therefore the third vector is not an eigenvector. The selected vectors in part (c) are

{ \begin{pmatrix}-1\\-1\\-1\end{pmatrix} \quad\text{and}\quad \begin{pmatrix}1\\0\\-1\end{pmatrix}. }

Geometrically, vectors in the plane are fixed, with eigenvalue $1$ and vectors in the normal direction are sent to zero, with eigenvalue $0$ . The same split into a two-dimensional image and a one-dimensional kernel gives the rank and nullity.

[exam]

[DEFINITION]

Kernel

If $T:V\to W$ is a linear map, its kernel is

\ker(T)=\{u\in V:T(u)=0\}.

The same set is also called the null space. If $A\in\mathbb F^{m\times n}$ is regarded as the map $\mathbb F^n\to\mathbb F^m$ , then

\ker(A)=\{x\in\mathbb F^n:Ax=0\}.

[DEFINITION]

Rank

The rank of a linear map $T$ is

\operatorname{rank}(T)=\dim\operatorname{im}(T).

For a matrix $A$ , the rank is

\operatorname{rank}(A)=\dim\operatorname{im}(A).

[1]:Image
[2]:Dimension
[3]:Rank-nullity theorem
[4]:Rank-nullity theorem for vector spaces
[5]:Kernels, ranks and powers of a matrix
[6]:Ranks of products involving a matrix and its transpose

[THEOREM]

Rank-nullity theorem for vector spaces

If $T:V\to W$ is a linear map over a scalar field $\mathbb F$ and $V$ is finite-dimensional, then

\operatorname{rank}(T)+\operatorname{nullity}(T)=\dim(V).

[1]:Nullity
[2]:Rank
[3]:Properties read from a matrix representation
[4]:Integration gives an injective map
[5]:Bijectivity in equal dimensions
[6]:Proof
14 more

[EXAMPLE]

Kernels, ranks and powers of a matrix

Let $A$ and $B$ be square matrices of the same size $n\times n$ . It is generally true that

\ker(B)\subseteq\ker(AB).

(a)

(i)

Prove this result by completing the following argument:

If $v\in\ker(B)$ and so $Bv={\phantom{0}}$ , then $ABv={\phantom{0}}$ and hence $v\in\ker(AB)$ .

(ii)

Complete the following statement. Some interesting special cases to consider are when $A=O$ (the zero matrix) and $A=I$ (the identity matrix). In these cases, $\ker(OB)={\phantom{\mathbb R^n}}$ and $\ker(IB)={\phantom{\ker(B)}}$ .

(b)

Choose all the correct statements, which are true for all $A$ and $B$ :

$\operatorname{nullity}(B)=\operatorname{nullity}(AB)$ ;
$\operatorname{rank}(B)=\operatorname{rank}(AB)$ ;
$\operatorname{rank}(B)\ge\operatorname{rank}(AB)$ ;
$\operatorname{rank}(B)\le\operatorname{rank}(AB)$ ;
$\operatorname{nullity}(B)\le\operatorname{nullity}(AB)$ ;
$\operatorname{nullity}(B)\ge\operatorname{nullity}(AB)$ .

(c)

Let $A$ be a square matrix and let

r_k=\operatorname{rank}(A^k), \qquad k=1,2,3,\ldots.

Choose all the correct statements, which are true for all square matrices:

The sequence $\{r_k\}$ is non-decreasing.
The sequence $\{r_k\}$ is constant.
The sequence $\{r_k\}$ is non-increasing.
The sequence $\{r_k\}$ is bounded.
The sequence $\{r_k\}$ is unbounded.
$\displaystyle\lim_{k\to\infty}r_k$ exists.
$\displaystyle\lim_{k\to\infty}r_k$ does not exist.

(d)

The following matrix computation and output is available:

>> A = [5, 7, -6; 10, 12, -11; 20, 24, -22]

A =

     5     7    -6
    10    12   -11
    20    24   -22

>> A^2

ans =

   -25   -25    25
   -50   -50    50
  -100  -100   100

>> A^3 + 5*A^2

ans =

     0     0     0
     0     0     0
     0     0     0

Find the smallest integer $N$ such that

\operatorname{rank}(A^k)=\operatorname{rank}(A^{k+1}) \qquad\text{for all }k\ge N.

Justify the answer using the displayed computation.

Solution

The kernel inclusion in part (a) gives a nullity inequality. Because $B$ and $AB$ have the same number of columns, rank-nullity turns it into a rank inequality, which part (c) applies to consecutive powers of $A$ .

(a) Kernel inclusion and the special cases

By definition,

v\in\ker(B)\Longrightarrow Bv=0.

Therefore

ABv=A(Bv)=A0=0.

This is exactly the condition $v\in\ker(AB)$ . The completed argument is

{Bv=0}, \qquad {ABv=0}.

If $A=O$ , then $OB=O$ , so every vector in $\mathbb R^n$ is sent to zero:

{\ker(OB)=\mathbb R^n}.

If $A=I$ , then $IB=B$ , so

{\ker(IB)=\ker(B)}.

The identity case shows that the inclusion can be an equality, while the zero-matrix case shows that $\ker(AB)$ can be much larger.

(b) Comparing rank and nullity

The inclusion from part (a) implies

\dim\ker(B)\le\dim\ker(AB),

{\operatorname{nullity}(B) \le\operatorname{nullity}(AB)}.

Both $B$ and $AB$ are $n\times n$ , so rank-nullity gives

\operatorname{rank}(B) =n-\operatorname{nullity}(B), \qquad \operatorname{rank}(AB) =n-\operatorname{nullity}(AB).

Reversing the inequality after subtracting from $n$ gives

{\operatorname{rank}(B)\ge\operatorname{rank}(AB)}.

Thus the correct selections in part (b) are exactly

{ \operatorname{rank}(B)\ge\operatorname{rank}(AB), \qquad \operatorname{nullity}(B)\le\operatorname{nullity}(AB). }

The equality statements are not always true: taking $A=O$ gives $AB=O$ , so both inequalities are strict whenever $B$ is non-zero.

(c) The rank sequence

Apply the rank inequality from part (b) with the left factor $A$ and the right factor $A^k$ :

\operatorname{rank}(A^{k+1}) =\operatorname{rank}(AA^k) \le\operatorname{rank}(A^k).

Therefore $\{r_k\}$ is non-increasing. Since an $n\times n$ matrix has rank between $0$ and $n$ ,

0\le r_k\le n \qquad\text{for every }k,

so the sequence is bounded. A non-increasing sequence of integers bounded below must eventually be constant and therefore its limit exists.

The sequence need not be constant from the start. For example, a non-zero nilpotent matrix can have $\operatorname{rank}(A)>\operatorname{rank}(A^2)$ . The correct selections are

{ \text{the sequence is non-increasing, it is bounded and } \lim_{k\to\infty}r_k\text{ exists}. }

(d) Finding the smallest stabilisation index

First determine $\operatorname{rank}(A)$ explicitly. Row operations give

\begin{aligned} A &= \begin{pmatrix} 5&7&-6\\ 10&12&-11\\ 20&24&-22 \end{pmatrix}\\ &\xrightarrow{\substack{R_2\leftarrow R_2-2R_1\\ R_3\leftarrow R_3-4R_1}} \begin{pmatrix} 5&7&-6\\ 0&-2&1\\ 0&-4&2 \end{pmatrix} \xrightarrow{R_3\leftarrow R_3-2R_2} \begin{pmatrix} 5&7&-6\\ 0&-2&1\\ 0&0&0 \end{pmatrix}. \end{aligned}

There are two pivots, so

\operatorname{rank}(A)=2.

The displayed value of $A^2$ is non-zero and its rows are multiples of the first row:

\begin{pmatrix} -25&-25&25\\ -50&-50&50\\ -100&-100&100 \end{pmatrix} \sim \begin{pmatrix} -25&-25&25\\ 0&0&0\\ 0&0&0 \end{pmatrix}.

Thus

\operatorname{rank}(A^2)=1.

The second MATLAB output says

A^3+5A^2=O, \qquad\text{so}\qquad A^3=-5A^2.

Multiplying by $A^{k-2}$ for any $k\ge2$ gives

A^{k+1}=A^{k-2}A^3 =-5A^{k-2}A^2 =-5A^k.

Because multiplying a matrix by the non-zero scalar $-5$ does not change its rank,

\operatorname{rank}(A^{k+1}) =\operatorname{rank}(A^k) \qquad(k\ge2).

However,

\operatorname{rank}(A)=2\ne1=\operatorname{rank}(A^2),

so the equality does not hold at $k=1$ . Therefore the smallest possible integer is

{N=2}.

[exam]

[1]:Kernel
[2]:Rank
[3]:Rank-nullity theorem for vector spaces

[THEOREM]

Rank-nullity theorem for vector spaces

If $T:V\to W$ is a linear map over a scalar field $\mathbb F$ and $V$ is finite-dimensional, then

\operatorname{rank}(T)+\operatorname{nullity}(T)=\dim(V).

[1]:Nullity
[2]:Rank
[3]:Properties read from a matrix representation
[4]:Integration gives an injective map
[5]:Bijectivity in equal dimensions
[6]:Proof
14 more

[EXAMPLE]

Ranks of products involving a matrix and its transpose

Wenbo is given the matrix

A= \begin{pmatrix} 1&-1&2&12\\ -1&1&-2&-8\\ -1&-1&-8&-2\\ -5&-4&-37&-13 \end{pmatrix}.

Jess has found the eigenvalues of $A$ to be approximately

0,\qquad -1.72981886,\qquad -18.03929570,\qquad 0.76911457,

where any integer eigenvalues are exact, not approximations.

The eigenvalue list is supplied as data; the rank calculation below is verified directly from $A$ by row reduction.

Calculate:

(a) $\operatorname{rank}(A^{\mathsf T}A)$ ;

(b) $\operatorname{rank}(AA^{\mathsf T})$ .

(c) If $B$ is another $4\times4$ matrix with $\operatorname{rank}(B)=1$ , what are the possible values of $\operatorname{rank}(BA^{\mathsf T})$ ?

Solution

Find $\operatorname{rank}(A)$ first, then compare kernels. For any real matrix, the quadratic forms $x^{\mathsf T}A^{\mathsf T}Ax$ and $y^{\mathsf T}AA^{\mathsf T}y$ are squared Euclidean lengths, so the two products have the same kernels as $A$ and $A^{\mathsf T}$ . Rank-nullity then gives their ranks.

Row reduction gives

\begin{aligned} A &= \begin{pmatrix} 1&-1&2&12\\ -1&1&-2&-8\\ -1&-1&-8&-2\\ -5&-4&-37&-13 \end{pmatrix}\\ &\xrightarrow[\ R_3\leftarrow R_3+R_1\ ]{ \substack{R_2\leftarrow R_2+R_1\\R_4\leftarrow R_4+5R_1}} \begin{pmatrix} 1&-1&2&12\\ 0&0&0&4\\ 0&-2&-6&10\\ 0&-9&-27&47 \end{pmatrix}\\ &\xrightarrow{R_2\leftrightarrow R_3} \begin{pmatrix} 1&-1&2&12\\ 0&-2&-6&10\\ 0&0&0&4\\ 0&-9&-27&47 \end{pmatrix}\\ &\xrightarrow{R_2\leftarrow-\frac12R_2} \begin{pmatrix} 1&-1&2&12\\ 0&1&3&-5\\ 0&0&0&4\\ 0&-9&-27&47 \end{pmatrix}\\ &\xrightarrow{R_4\leftarrow R_4+9R_2} \begin{pmatrix} 1&-1&2&12\\ 0&1&3&-5\\ 0&0&0&4\\ 0&0&0&2 \end{pmatrix}\\ &\xrightarrow{\substack{R_3\leftarrow R_3-2R_4\\R_4\leftarrow\frac12R_4}} \begin{pmatrix} 1&-1&2&12\\ 0&1&3&-5\\ 0&0&0&0\\ 0&0&0&1 \end{pmatrix} \xrightarrow{R_3\leftrightarrow R_4} \begin{pmatrix} 1&-1&2&12\\ 0&1&3&-5\\ 0&0&0&1\\ 0&0&0&0 \end{pmatrix}. \end{aligned}

There are pivots in columns $1$ , $2$ and $4$ , so

\operatorname{rank}(A)=3.

The exact zero eigenvalue agrees with this singularity, but it is the row reduction that shows the nullity is exactly one; the rounded eigenvalues alone would not.

For part (a), if $x\in\ker(A)$ then $A^{\mathsf T}Ax=0$ . Conversely, if $A^{\mathsf T}Ax=0$ , then

0=x^{\mathsf T}A^{\mathsf T}Ax =(Ax)^{\mathsf T}(Ax) =\|Ax\|^2,

so $Ax=0$ . Hence

\ker(A^{\mathsf T}A)=\ker(A).

Both matrices have four columns. Since $\operatorname{nullity}(A)=4-3=1$ , rank-nullity gives

{\operatorname{rank}(A^{\mathsf T}A)=4-1=3}.

Similarly,

\ker(AA^{\mathsf T})=\ker(A^{\mathsf T}),

because

y^{\mathsf T}AA^{\mathsf T}y =(A^{\mathsf T}y)^{\mathsf T}(A^{\mathsf T}y) =\|A^{\mathsf T}y\|^2.

Transposition preserves rank, so $\operatorname{rank}(A^{\mathsf T})=3$ and $\operatorname{nullity}(A^{\mathsf T})=1$ . Therefore

{\operatorname{rank}(AA^{\mathsf T})=4-1=3}.

For part (c), write a rank-one matrix as

B=uv^{\mathsf T}, \qquad u\ne0,\quad v\ne0.

Then

BA^{\mathsf T} =uv^{\mathsf T}A^{\mathsf T} =u(Av)^{\mathsf T}.

This product has rank at most $1$ . It has rank $0$ when $Av=0$ . Such a non-zero $v$ exists because $\operatorname{nullity}(A)=1$ . It has rank $1$ when $Av\ne0$ , which can be achieved by choosing $v\notin\ker(A)$ . Consequently, the possible values are

{\operatorname{rank}(BA^{\mathsf T})\in\{0,1\}.}

The kernel identities give the product ranks without multiplying the matrices and the rank-one factorisation $B=uv^{\mathsf T}$ reduces part (c) to whether $Av$ is zero.

[exam]

[1]:Rank-nullity theorem for vector spaces
[2]:Rank

[THEOREM]

Rank-nullity theorem for vector spaces

If $T:V\to W$ is a linear map over a scalar field $\mathbb F$ and $V$ is finite-dimensional, then

\operatorname{rank}(T)+\operatorname{nullity}(T)=\dim(V).

[1]:Nullity
[2]:Rank
[3]:Properties read from a matrix representation
[4]:Integration gives an injective map
[5]:Bijectivity in equal dimensions
[6]:Proof
14 more

[DEFINITION]

Dimension

If a vector space $V$ has a finite basis, then $V$ is finite-dimensional and its dimension, denoted $\dim(V)$ , is the common number of vectors in all bases of $V$ .

[1]:Basis
[2]:All bases have the same size
[3]:A system with zero initial derivative
[4]:Nullity
[5]:Rank
[6]:Rank-nullity
7 more

[DEFINITION]

Surjective map

A map $T:V\to W$ is surjective, or onto, when every $w\in W$ equals $T(v)$ for some $v\in V$ . Equivalently, $\operatorname{im}(T)=W$ .

[EXAMPLE]

Possible nullities of a linear map

Let

T:M_{9\times7}(\mathbb R)\longrightarrow P_{25}(\mathbb R)

be a linear map. Suppose the kernel of $T$ has dimension $k$ . Then:

(a) the maximum possible value of $k$ is ___;

(b) the minimum possible value of $k$ is ___; and

Solution

The domain consists of $9\times7$ matrices, so one basis is given by the matrices with a single $1$ and all other entries $0$ . Hence

\dim\bigl(M_{9\times7}(\mathbb R)\bigr)=9\cdot7=63.

The space $P_{25}(\mathbb R)$ has basis $\{1,x,\ldots,x^{25}\}$ , so

\dim\bigl(P_{25}(\mathbb R)\bigr)=26.

Let $r=\operatorname{rank}(T)$ . Rank-nullity gives

r+k=\dim\bigl(M_{9\times7}(\mathbb R)\bigr)=63,

k=63-r.

The rank is at least $0$ and cannot exceed the dimension of the codomain:

0\le r\le26.

Therefore

37=63-26\le k\le63=63-0.

The upper bound is attained by the zero map, whose kernel is the whole 63-dimensional domain. The lower bound is attainable because a 63-dimensional domain can map onto the 26-dimensional codomain by sending 26 domain basis vectors to a basis of $P_{25}(\mathbb R)$ and the remaining 37 basis vectors to zero. Thus

{\text{(a) the maximum possible value of $k$ is }63}

and

{\text{(b) the minimum possible value of $k$ is }37}.

Finally, $T$ is surjective exactly when its image is the whole codomain. This is equivalent to

\operatorname{rank}(T)=\dim\bigl(P_{25}(\mathbb R)\bigr)=26.

Using $k=63-r$ ,

T\text{ is surjective} \Longleftrightarrow r=26 \Longleftrightarrow k=63-26=37.

Hence

{\text{(c) $T$ is surjective if and only if $k=37$.}}

[routine]

[1]:Rank-nullity theorem for vector spaces
[2]:Dimension
[3]:Surjective map

[DEFINITION]

Dimension

If a vector space $V$ has a finite basis, then $V$ is finite-dimensional and its dimension, denoted $\dim(V)$ , is the common number of vectors in all bases of $V$ .

[1]:Basis
[2]:All bases have the same size
[3]:A system with zero initial derivative
[4]:Nullity
[5]:Rank
[6]:Rank-nullity
7 more

[THEOREM]

Rank-nullity theorem for vector spaces

If $T:V\to W$ is a linear map over a scalar field $\mathbb F$ and $V$ is finite-dimensional, then

\operatorname{rank}(T)+\operatorname{nullity}(T)=\dim(V).

[1]:Nullity
[2]:Rank
[3]:Properties read from a matrix representation
[4]:Integration gives an injective map
[5]:Bijectivity in equal dimensions
[6]:Proof
14 more

[DEFINITION]

Eigenvalue

Let $T:V\to V$ be a linear map over a field $F$ . A scalar $\lambda\in F$ is an eigenvalue of $T$ if there is a non-zero vector $v\in V$ such that

T(v)=\lambda v.

[1]:Linear map
[2]:Eigenvectors for distinct eigenvalues
[3]:A real matrix with complex eigenvalues
[4]:Invertibility and eigenvalues
[5]:Diagonalisation
15 more

[EXAMPLE]

Classifying mathematical objects

Later-topic preview: several classifications use eigenvalues, eigenspaces or diagonalisation.

Consider the following mathematical objects that we encountered in MATH1251:

$\displaystyle \begin{pmatrix} 1&0&0\\ 0&1&0\\ 0&0&1 \end{pmatrix}$ .
$T:V\to\mathbb R$ defined by
$T(f)=\int_0^1f(x)\,dx,$
where $V$ is the vector space of all integrable functions on the interval $[0,1]$ .
$\displaystyle \begin{pmatrix} 11&12&0&3&-5\\ 1&-1&1&4&2\\ 0&-1&3&-1&-2\\ 0&0&0&7&1 \end{pmatrix}$ .
$T:\mathbb R^3\to\mathbb R$ defined by
$T(\mathbf v) =\mathbf v\cdot \begin{pmatrix}-5\\11\\1\end{pmatrix}.$
$\displaystyle \begin{pmatrix} 1&-1\\ 5&-5 \end{pmatrix}$ .
$P_3$ .
$\displaystyle \operatorname{span}\left\{ \begin{pmatrix}1\\0\\0\\0\end{pmatrix}, \begin{pmatrix}-1\\0\\0\\0\end{pmatrix}, \begin{pmatrix}0\\-5\\-1\\0\end{pmatrix}, \begin{pmatrix}1\\1\\1\\1\end{pmatrix} \right\}$ .
$T:P_1\to P_1$ defined by
$T(p)=\frac{dp}{dx}.$
$\displaystyle \begin{pmatrix} 0&1\\ 1&0 \end{pmatrix}$ .
$P_4$ .
$\displaystyle\operatorname{span}\{1,x^2,x^4,x^6\}$ .

(a) Select all examples on the list above for which the following requirements are well-defined and are satisfied.

(i) ___ has at least 3 linearly independent columns.

(ii) ___ is rank 1.

(iii) ___ has a 3-dimensional eigenspace.

(iv) ___ is a 4-dimensional vector space.

(v) ___ has a positive eigenvalue.

(vi) ___ has positive nullity.

(vii) ___ is a subspace of $P_{11}$ .

(viii) ___ is diagonalisable.

(ix) ___ has linearly dependent columns.

Solution

A requirement can be satisfied only where it is well defined for the object. Rank and nullity apply to linear maps and matrices. Eigenvalues, eigenspaces and diagonalisation apply to endomorphisms, such as square matrices or maps from a vector space to itself. Statements about columns apply to matrices, and the vector-space and subspace statements apply to the listed sets or spaces.

For object 3, denote the matrix by $A_3$ . Row reduction gives

\begin{aligned} A_3 &= \begin{pmatrix} 11&12&0&3&-5\\ 1&-1&1&4&2\\ 0&-1&3&-1&-2\\ 0&0&0&7&1 \end{pmatrix}\\ &\xrightarrow{R_1\leftrightarrow R_2} \begin{pmatrix} 1&-1&1&4&2\\ 11&12&0&3&-5\\ 0&-1&3&-1&-2\\ 0&0&0&7&1 \end{pmatrix}\\ &\xrightarrow{R_2\leftarrow R_2-11R_1} \begin{pmatrix} 1&-1&1&4&2\\ 0&23&-11&-41&-27\\ 0&-1&3&-1&-2\\ 0&0&0&7&1 \end{pmatrix}\\ &\xrightarrow{R_2\leftrightarrow R_3} \begin{pmatrix} 1&-1&1&4&2\\ 0&-1&3&-1&-2\\ 0&23&-11&-41&-27\\ 0&0&0&7&1 \end{pmatrix}\\ &\xrightarrow{R_3\leftarrow R_3+23R_2} \begin{pmatrix} 1&-1&1&4&2\\ 0&-1&3&-1&-2\\ 0&0&58&-64&-73\\ 0&0&0&7&1 \end{pmatrix}. \end{aligned}

There are pivots in columns $1,2,3,4$ , so

\operatorname{rank}(A_3)=4.

Consequently, object 3 has at least three linearly independent columns. It also has five columns in $\mathbb R^4$ , so its columns are linearly dependent and rank-nullity gives

\operatorname{nullity}(A_3)=5-4=1.

The matrix in object 1 is the $3\times3$ identity matrix. Its columns are independent, its rank is $3$ , its nullity is $0$ and its only eigenvalue is $1$ , with a three-dimensional eigenspace. It is already diagonal, so it is diagonalisable.

For object 5, write

A_5= \begin{pmatrix} 1&-1\\ 5&-5 \end{pmatrix}.

Its second column is the negative of its first column, while the first column is nonzero. Therefore $\operatorname{rank}(A_5)=1$ and its columns are linearly dependent. Its characteristic polynomial is

\det(A_5-\lambda I) =\det\begin{pmatrix}1-\lambda&-1\\5&-5-\lambda\end{pmatrix} =\lambda(\lambda+4).

Thus its eigenvalues are $0$ and $-4$ . They are distinct, so $A_5$ is diagonalisable, but it has no positive eigenvalue.

For object 9,

A_9= \begin{pmatrix}0&1\\1&0\end{pmatrix}, \qquad A_9\begin{pmatrix}1\\1\end{pmatrix} =\begin{pmatrix}1\\1\end{pmatrix}, \qquad A_9\begin{pmatrix}1\\-1\end{pmatrix} =-\begin{pmatrix}1\\-1\end{pmatrix}.

Its eigenvalues are $1$ and $-1$ , so it has a positive eigenvalue and is diagonalisable. Its columns are independent and its nullity is $0$ .

For object 2, linearity of integration makes this a linear map. It is nonzero because

T(1)=\int_0^1 1\,dx=1.

Its codomain is one-dimensional, so its nonzero image is all of $\mathbb R$ :

\operatorname{rank}(T)=1.

Its kernel is nontrivial, since, for example,

\int_0^1\left(x-\frac12\right)\,dx =\left[\frac{x^2}{2}-\frac{x}{2}\right]_0^1=0.

Thus object 2 has positive nullity.

For object 4, the vector $(-5,11,1)^{\mathsf T}$ is nonzero. The dot-product map therefore has image $\mathbb R$ and rank $1$ . Its kernel is the plane perpendicular to this vector, so its nullity is $3-1=2$ , which is positive.

For object 8, write $p=a+bx$ . Then

T(p)=b.

The image consists of the constant polynomials, so

\operatorname{rank}(T)=1, \qquad \ker(T)=\operatorname{span}\{1\}, \qquad \operatorname{nullity}(T)=1.

With respect to the basis $(1,x)$ , the matrix of $T$ is

\begin{pmatrix}0&1\\0&0\end{pmatrix}.

Its only eigenvalue is $0$ and its eigenspace is the one-dimensional space of constant polynomials. Since $P_1$ has dimension $2$ , object 8 is not diagonalisable and has no positive eigenvalue.

The remaining vector-space dimensions are

\dim(P_3)=4, \qquad \dim(P_4)=5.

Object 7 is not four-dimensional. Its second displayed vector is the negative of the first and the other three vectors are independent: if

a\begin{pmatrix}1\\0\\0\\0\end{pmatrix} +b\begin{pmatrix}0\\-5\\-1\\0\end{pmatrix} +c\begin{pmatrix}1\\1\\1\\1\end{pmatrix} =\begin{pmatrix}0\\0\\0\\0\end{pmatrix},

then the fourth coordinate gives $c=0$ , the first gives $a=0$ and the second gives $b=0$ . Hence object 7 has dimension $3$ .

The four polynomials $1,x^2,x^4,x^6$ in object 11 are linearly independent because a linear combination equal to the zero polynomial has all four coefficients equal to zero. Therefore object 11 has dimension $4$ .

Each of $P_3$ , $P_4$ and object 11 is a subspace of $P_{11}$ : every polynomial in these spaces has degree at most $11$ and the span in object 11 is closed under addition and scalar multiplication. Object 7 is a subspace of $\mathbb R^4$ , not of $P_{11}$ .

Collecting the results, the selected object numbers are:

{ \begin{aligned} (\mathrm{i})&\quad 1,3,\\ (\mathrm{ii})&\quad 2,4,5,8,\\ (\mathrm{iii})&\quad 1,\\ (\mathrm{iv})&\quad 6,11,\\ (\mathrm{v})&\quad 1,9,\\ (\mathrm{vi})&\quad 2,3,4,5,8,\\ (\mathrm{vii})&\quad 6,10,11,\\ (\mathrm{viii})&\quad 1,5,9,\\ (\mathrm{ix})&\quad 3,5. \end{aligned} }

Different objects need different tests: row reduction counts pivots and free variables, coefficient comparison gives polynomial dimensions and eigenvalue statements require a square map.

[exam]

[1]:Dimension
[2]:Rank-nullity theorem for vector spaces
[3]:Eigenvalue

[THEOREM]

The kernel is a subspace

If $T:V\to W$ is a linear map, then $\ker(T)$ is a subspace of $V$ .

[1]:Kernel
[2]:Subspace test
[3]:Proof
[4]:Consequences of prescribed kernel and image vectors

[THEOREM]

The image is a subspace

If $T:V\to W$ is a linear map, then $\operatorname{im}(T)$ is a subspace of $W$ .

[1]:Image
[2]:Subspace test
[3]:Proof
[4]:Consequences of prescribed kernel and image vectors

[THEOREM]

Rank-nullity theorem for vector spaces

If $T:V\to W$ is a linear map over a scalar field $\mathbb F$ and $V$ is finite-dimensional, then

\operatorname{rank}(T)+\operatorname{nullity}(T)=\dim(V).

[1]:Nullity
[2]:Rank
[3]:Properties read from a matrix representation
[4]:Integration gives an injective map
[5]:Bijectivity in equal dimensions
[6]:Proof
14 more

[EXAMPLE]

Consequences of prescribed kernel and image vectors

Let $V$ and $W$ be vector spaces and let

T:V\to W

be a linear transformation. Let $v_1,v_2,v_3,v_4$ be vectors in $V$ and let $p,q$ be vectors in $W$ .

It is given that $v_1,v_2$ are linearly independent and also $p,q$ are linearly independent. It is also given that

T(v_1)=0,\qquad T(v_2)=0,\qquad T(v_3)=p,\qquad T(v_4)=q.

(a)

Select the statements which are definitely true. If it is not possible to be certain whether a statement is true or false, do not select it.

$\ker(T)$ is a subspace of $V$ .
$\ker(T)$ is a subspace of $W$ .
$\operatorname{nullity}(T)\le2$ .
$\operatorname{nullity}(T)=2$ .
$\operatorname{nullity}(T)\ge2$ .
$-8v_2+5v_3$ is not an element of $\ker(T)$ .

(b)

Select the statements which are definitely true. If it is not possible to be certain whether a statement is true or false, do not select it.

$\operatorname{im}(T)$ is a subspace of $V$ .
$\operatorname{im}(T)$ is a subspace of $W$ .
$\operatorname{rank}(T)\le2$ .
$\operatorname{rank}(T)=2$ .
$\operatorname{rank}(T)\ge2$ .
$4p+3q$ is not an element of $\operatorname{im}(T)$ .

(c)

Now suppose that you are told that the vector space $V$ is in fact $\mathbb R^5$ . Select the correct statement from the following list, then give a brief explanation:

$\operatorname{rank}(T)=3$ and $\operatorname{nullity}(T)\ne3$ .
$\operatorname{nullity}(T)=5$ and $\operatorname{rank}(T)=5$ .
Either $\operatorname{nullity}(T)=3$ or $\operatorname{rank}(T)=3$ , but we do not have enough information to tell which.
$\operatorname{nullity}(T)=3$ and $\operatorname{rank}(T)=3$ .
$\operatorname{nullity}(T)=3$ and $\operatorname{rank}(T)\ne3$ .
$\operatorname{nullity}(T)=2$ and $\operatorname{rank}(T)=2$ .

(d)

By selecting a statement from each list, complete a carefully explained proof that $\{v_1,v_2,v_3,v_4\}$ is a linearly independent set.

Solution

Write

K=\ker(T) \qquad\text{and}\qquad I=\operatorname{im}(T).

For (a), the kernel is a subspace of the domain $V$ , so the first statement is definitely true and the second is false.

Since $T(v_1)=T(v_2)=0$ , both $v_1$ and $v_2$ belong to $K$ . Being linearly independent, they give $K$ at least two independent directions, so

\operatorname{nullity}(T)=\dim K\ge2.

There is no information ruling out further independent vectors in $K$ , so neither $\operatorname{nullity}(T)\le2$ nor $\operatorname{nullity}(T)=2$ is forced.

Finally,

\begin{aligned} T(-8v_2+5v_3) &=-8T(v_2)+5T(v_3)\\ &=-8(0)+5p\\ &=5p. \end{aligned}

The independence of $p,q$ implies $p\ne0$ , so $5p\ne0$ . Hence $-8v_2+5v_3\notin\ker(T)$ . The definitely true choices in (a) are

{\ker(T)\text{ is a subspace of }V,\qquad \operatorname{nullity}(T)\ge2,\qquad -8v_2+5v_3\notin\ker(T).}

For (b), the image is a subspace of the codomain $W$ , not necessarily a subspace of $V$ . Because

T(v_3)=p,\qquad T(v_4)=q,

both $p$ and $q$ belong to $I$ . Their independence gives

\operatorname{rank}(T)=\dim I\ge2.

There may be further independent vectors in the image, so rank at most $2$ and rank equal to $2$ are not forced. Also,

4p+3q =4T(v_3)+3T(v_4) =T(4v_3+3v_4),

so $4p+3q\in\operatorname{im}(T)$ . Therefore the definitely true choices in (b) are

{ \operatorname{im}(T)\text{ is a subspace of }W, \qquad \operatorname{rank}(T)\ge2. }

For (c), the rank-nullity theorem applies to the domain $V=\mathbb R^5$ :

\operatorname{rank}(T)+\operatorname{nullity}(T)=5.

The conclusions from (a) and (b) give

\operatorname{nullity}(T)\ge2, \qquad \operatorname{rank}(T)\ge2.

Thus

2\le\operatorname{rank}(T)\le3, \qquad 2\le\operatorname{nullity}(T)\le3.

The only possible pairs are

(\operatorname{rank}(T),\operatorname{nullity}(T))=(2,3) \quad\text{or}\quad (3,2).

The given information does not determine whether an additional independent image direction occurs. Therefore the correct listed statement is

{ \text{either }\operatorname{nullity}(T)=3\text{ or }\operatorname{rank}(T)=3, \text{ but there is not enough information to tell which}. }

For (d), suppose that scalars $\lambda_1,\lambda_2,\lambda_3,\lambda_4$ satisfy

\lambda_1v_1+\lambda_2v_2+\lambda_3v_3+\lambda_4v_4=0.

Apply $T$ to both sides. Since $T$ is linear and $T(0)=0$ ,

\begin{aligned} 0 &=T\left(\lambda_1v_1+\lambda_2v_2+\lambda_3v_3+\lambda_4v_4\right)\\ &=\lambda_1T(v_1)+\lambda_2T(v_2) +\lambda_3T(v_3)+\lambda_4T(v_4)\\ &=\lambda_3p+\lambda_4q. \end{aligned}

Because $p,q$ are linearly independent, the last equation implies

\lambda_3=\lambda_4=0.

Substituting these values into the original relation gives

\lambda_1v_1+\lambda_2v_2=0.

Since $v_1,v_2$ are linearly independent,

\lambda_1=\lambda_2=0.

Therefore

\lambda_1=\lambda_2=\lambda_3=\lambda_4=0,

which is precisely the homogeneous criterion for $\{v_1,v_2,v_3,v_4\}$ to be linearly independent.

[exam]

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