Examples from Seen Questions

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

and

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

(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

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

For $\mathbf e_2$, the coordinate equations are

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

Hence

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

and

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

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

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.

Values of a partially specified linear transformation

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

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)

(b)

(c)

Now suppose that you also know

(d)

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

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:

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

linearity gives

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

For (c),

Therefore

For (d), the additional information is

For the previously undetermined vector, use

Thus

The three requested entries in part (d) are therefore

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.

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

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

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

(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

Therefore

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

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

The corresponding negative-frequency coefficient is

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

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 $k\ge1$,

Consequently,

and $T(1)=16$.

(b)(i)

For the proposed polynomial,

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

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

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 $d\ne4$,

The correct selections are therefore

and

(b)(ii)

Write

Then

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:

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

so one basis is

(b)(iii)

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

then the polynomial

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

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.

A projection, nullity and a composition

Suppose

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

Additionally, suppose

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

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,

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

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

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

because $\mathbf b\ne0$. Therefore

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

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

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

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

Consequently,

The requested Cartesian equation is therefore

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 $\ker(S\circ T)=\ker(T)$.

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:

(a) Choose the correct statement about $H$

1. The subspace $H$ is a plane through the origin.
2. The subspace $H$ is a line which does not go through the origin.
3. The subspace $H$ is a plane which does not go through the origin.
4. The subspace $H$ is a line through the origin.
5. 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

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

(c) Select all that apply

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

Solution

Let

The third vector is redundant because

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

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

A normal vector is

because

and

Thus

and $\mathbb R^3$ decomposes as

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

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

Consequently, the statements selected in part (b) are

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

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

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

For the third vector, let

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

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

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

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.

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

(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

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

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,

Therefore

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

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

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

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

so

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

Reversing the inequality after subtracting from $n$ gives

Thus the correct selections in part (b) are exactly

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 $A^k$:

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