Examples from Seen Questions

selecting true statements

Later-topic preview: statements 2, 5, 6 and 7 use linear transformations, rank-nullity or eigenvalues.

Choose all of the true statements below:

1. The columns of the matrix $A$ are linearly independent if and only if $Ax=\mathbf0$ has only the trivial solution.
2. If $A$ is an $m\times n$ matrix, then $T(\mathbf{x})=A\mathbf{x}$ defines a linear transformation from $\mathbb R^n$ to $\mathbb R^m$.
3. Let $A$ be an $n\times n$ matrix with $n\ge2$. If $\mathbf{x}$ is a nontrivial solution of $A\mathbf{x}=\mathbf0$, then every entry in $\mathbf{x}$ is non-zero.
4. Let $\mathbf{u},\mathbf{v}$ be vectors in $\mathbb R^5$. The set $\operatorname{span}\{\mathbf{u},\mathbf{v}\}$ is always a plane through the origin.
5. If two rows of an $n\times n$ matrix $A$, where $n\ge2$, are the same, then $\operatorname{rank}(A)>0$.
6. If two rows of an $n\times n$ matrix $A$, where $n\ge2$, are the same, then $\operatorname{nullity}(A)>0$.
7. If $A$ is an $n\times n$ matrix with the property that the sum of the entries in each row is $74$, then $A$ has a non-zero eigenvalue.

Solution

The true statements are 1, 2, 6 and 7. Each statement is checked against its defining equation or refuted by a counterexample.

Statement 1 is true. If the columns of $A$ are $\mathbf a_1,\ldots,\mathbf a_n$, then

Thus $A\mathbf{x}=\mathbf0$ has only the trivial solution exactly when the only linear combination of the columns that gives $\mathbf0$ is the trivial combination. This is precisely linear independence.

Statement 2 is true. For $\mathbf u,\mathbf v\in\mathbb R^n$ and $c\in\mathbb R$, matrix multiplication satisfies

and

These are the two linearity conditions and the dimensions of the product show that the codomain is $\mathbb R^m$.

Statement 3 is false. A nontrivial vector is only required to be non-zero; it may still have zero entries. For example, take $A$ to be the zero matrix and $\mathbf{x}=(1,0,\ldots,0)^{\mathsf T}$. Then $A\mathbf{x}=\mathbf0$, $\mathbf{x}\ne\mathbf0$ and most entries of $\mathbf{x}$ are zero.

Statement 4 is false because two vectors need not be independent. If $\mathbf u=\mathbf v\ne\mathbf0$, then

which is a line through the origin. If both vectors are zero, the span is just $\{\mathbf0\}$.

Statement 5 is false. The zero square matrix has two equal rows, but

so its rank is not greater than zero.

Statement 6 is true. Equal rows are linearly dependent, so an $n\times n$ matrix with two equal rows has rank at most $n-1$. Rank-nullity then gives

Finally, statement 7 is true. Let $\mathbf1=(1,\ldots,1)^{\mathsf T}\in\mathbb R^n$ be the all-ones vector. The row-sum condition says exactly that

Since $\mathbf1\ne\mathbf0$, this shows that $74$ is an eigenvalue and it is non-zero.

linear independence of real-valued functions

Kaiwen is given $f_1,f_2,f_3\in\mathcal R[\mathbb R]$, where $\mathcal R[\mathbb R]$ is the vector space of all real-valued functions on $\mathbb R$ and where $f_1,f_2,f_3$ are defined by the rules

Are these functions linearly dependent?

• No.
• Yes.

Solution

The correct choice is No. Suppose that $a,b,c\in\mathbb R$ satisfy

Equality of functions means equality at every $x\in\mathbb R$, so

The trigonometric part is bounded:

Therefore the assumed relation would give

If $a\ne0$, the left-hand side becomes arbitrarily large as $x\to\infty$, which is impossible. Hence $a=0$. The relation reduces to

At $x=0$ this gives

With $c=0$, substitute $x=\frac{\pi}{14}$:

Thus $a=b=c=0$, so the functions are linearly independent.

This is the homogeneous independence test in a function space, where a relation must hold at every point of the domain. Boundedness separates the affine function from the trigonometric ones and two evaluations settle the remaining coefficients.

span, independence and possible bases

Lance and Tamzid were given four sets of vectors in $\mathbb R^3$ or $\mathbb R^5$:

In each set, the vectors are distinct.

For each of the first two sets, Lance and Tamzid are asked to determine whether it forms a spanning set for $\mathbb R^3$ and to prove the result. For each of the third and fourth sets, they are asked to determine whether it is linearly dependent or linearly independent.

(a) They think that $\{u_1,u_2,u_3,u_4\}$ forms a spanning set for $\mathbb R^3$.

(b) They think that $\{v_1,v_2,v_3,v_4\}$ is not a spanning set for $\mathbb R^3$.

(c) They think that $x_1,x_2,x_3$ are linearly independent.

(d) They think that $y_1,y_2,y_3$ are linearly dependent.

(e) Is it possible for $\{u_1,u_2,u_3,u_4\}$ to be a basis for $\mathbb R^3$ and for $\{x_1,x_2,x_3\}$ to be a basis for $\mathbb R^5$?

Solution

The vectors have no coordinates, so there is nothing to row-reduce. The task is to state what would have to be proved for each claim. With the named vectors placed as columns, the vector equations are the matrix equations $Uc=\mathbf x$, $Vc=\mathbf y$, $X\alpha=\mathbf v$ and $Y\beta=\mathbf w$.

(a) Proving that the $u$-set spans $\mathbb R^3$

By definition, a set spans $\mathbb R^3$ when every vector in $\mathbb R^3$ is a linear combination of the set. Therefore the required statement is

The coefficients may depend on $\mathbf x$ and one choice for each target vector is enough; uniqueness is not part of spanning.

(b) Proving that the $v$-set does not span $\mathbb R^3$

To disprove spanning, one target vector that cannot be produced is enough. The required statement is

Equivalently, the column space of the matrix $V=[\,v_1\ v_2\ v_3\ v_4\,]$ is a proper subset of $\mathbb R^3$. A contradictory row in the augmented matrix $[V\mid\mathbf y]$ would exhibit such a vector $\mathbf y$.

(c) Proving that the $x$-set is linearly independent

Independence is tested with the homogeneous equation, so the right-hand side is $\mathbf 0$. The required statement is

In matrix language, if $X=[\,x_1\ x_2\ x_3\,]$, then $X\alpha=\mathbf0$ must have only the zero solution. With actual coordinates, row reduction would show a pivot in every column of $X$.

(d) Proving that the $y$-set is linearly dependent

Dependence is the failure of the homogeneous test: some non-zero coefficient vector must produce the zero vector. The required statement is

The right-hand side is again $\mathbf0$, but now a non-trivial solution is required.

(e) Testing the two possible bases

The answer choices are:

1. It is possible for both $\{u_1,u_2,u_3,u_4\}$ and $\{x_1,x_2,x_3\}$ to be bases for their respective vector spaces.
2. $\{u_1,u_2,u_3,u_4\}$ cannot form a basis for $\mathbb R^3$, but $\{x_1,x_2,x_3\}$ could possibly form a basis for $\mathbb R^5$.
3. $\{x_1,x_2,x_3\}$ cannot form a basis for $\mathbb R^5$, but $\{u_1,u_2,u_3,u_4\}$ could possibly form a basis for $\mathbb R^3$.
4. Neither $\{u_1,u_2,u_3,u_4\}$ nor $\{x_1,x_2,x_3\}$ can possibly be bases for their respective vector spaces.

The correct choice is 4. A basis for $\mathbb R^3$ must contain exactly $\dim(\mathbb R^3)=3$ vectors. Although the $u$-set is assumed to span $\mathbb R^3$, it has four vectors, so it must be linearly dependent and cannot be a basis. A basis for $\mathbb R^5$ must contain five vectors, while the $x$-set contains only three. It may be independent, but it cannot span $\mathbb R^5$.

polynomials, dependence and dimension

Recall that $P_2(\mathbb R)$, the set of real polynomials whose degree is at most two, is a vector space over $\mathbb R$.

(a)

Consider

You are given the following matrices and their corresponding row-echelon forms:

and

(i)

Find scalars $\alpha,\beta,\gamma$, not all zero, such that

(ii)

Find a basis of $P_2(\mathbb R)$ that contains as many elements of $\{A,B,C\}$ as possible.

(b)

Let $p,q\in P_2(\mathbb R)$, where

for some $a_1,b_1,c_1,a_2,b_2,c_2\in\mathbb R$ with

Define

(i)

Choose all correct statements:

• The set $\{p,q\}$ is linearly dependent.
• The set $\{p,q,R\}$ is linearly independent.
• There is not enough information to determine whether the set $\{p,q\}$ is linearly independent.
• The set $\{p,q,R\}$ is linearly dependent.
• The set $\{p,q\}$ is linearly independent.
• There is not enough information to determine whether the set $\{p,q,R\}$ is linearly independent.

(ii)

Let

Find $\dim(\mathcal A)$.

Solution

The first matrix has the coefficient vectors of $A,B,C$ as its rows and the appended identity records which row combinations row reduction produces. The second matrix has those coefficient vectors as columns, so its pivot columns identify a largest linearly independent subset of $\{A,B,C\}$.

(a)(i)

Starting with the first augmented matrix, the displayed row-echelon form is obtained by

where the last operation uses the updated second row. The last row of the resulting augmented matrix is

Its first three entries say that a row combination of the coefficient vectors is zero and its last three entries give the coefficients of that combination. Therefore

Checking directly,

Thus one valid answer is

(a)(ii)

For the transpose coefficient matrix, the row operations

give, in the first three columns,

There are pivots in columns one and two, so $A$ and $B$ are linearly independent. The relation from part (i) shows that no independent set contains all three of $A,B,C$.

Since $\dim P_2(\mathbb R)=3$, one more polynomial is needed; try $1$. To check independence, suppose

The coefficients of $t^2$ and $t$ give

Adding these equations gives $\mu=0$ and then $\lambda=0$. The constant coefficient then gives $\nu=0$. Hence

is a basis of $P_2(\mathbb R)$ containing two of $A,B,C$, the most possible.

(b)(i)

Suppose that

Comparing the coefficients of $t^2$ and $t$ gives

The determinant of this coefficient matrix is

Therefore the homogeneous system has only the zero solution, so $\{p,q\}$ is linearly independent.

On the other hand,

so

is a non-trivial dependence relation. The correct statements are therefore

and

(b)(ii)

The relations defining $P$ and $Q$ can be inverted:

Thus

Hence $\operatorname{span}\{P,Q\}=\operatorname{span}\{p,q\}$. Since $p,q$ are independent, $P,Q$ are also independent: if $\lambda P+\mu Q=0$, then

so independence of $p,q$ gives $\lambda+3\mu=0$ and $\lambda-3\mu=0$, hence $\lambda=\mu=0$.

Finally, $R=9p-3q$ is already in $\operatorname{span}\{P,Q\}$. Therefore

In part (a), row reduction finds the dependence and the dimension count says one extra polynomial completes the basis. In part (b), the determinant condition gives two independent directions, the invertible change from $(p,q)$ to $(P,Q)$ keeps them and $R$ adds no new one.

a set of polynomial divisors

Let

Is $S$ a subspace of $P_3(\mathbb R)$? Justify the answer.

Solution

Let

Every subspace must contain the zero polynomial. Here $0\notin S$: if $0$ divided the non-zero polynomial $q$, then $q=0r=0$ for some polynomial $r$, a contradiction. Therefore

Closure under addition also fails. Both

belong to $S$, because

However,

does not belong to $S$. If $q(x)+1$ divided $q(x)$, there would be a polynomial $r(x)$ such that

Both $q(x)$ and $q(x)+1$ have degree $3$, so $r(x)$ would have to be a constant. Comparing leading coefficients would force $r(x)=1$, which would give $q(x)=q(x)+1$, a contradiction. Thus $S$ is not closed under addition.

Divisibility is a multiplicative condition, so it has no reason to survive the addition inherited from $P_3(\mathbb R)$.

subspaces of $M_{2\times2}(\mathbb R)$

Later-topic preview: parts of this question use eigenvalues, diagonalisation, kernels and rank-nullity.

Let

be the vector space of $2\times2$ matrices with real entries over the scalar field $\mathbb R$.

(a)

Choose from the following list the subsets of $V$ that are subspaces of $V$:

1. $\{A\in V:A^{\mathsf T}=4A\}$;
2. $\{A\in V:0\text{ is an eigenvalue of }A\}$;
3. $\left\{A\in V: v=\begin{pmatrix}-4\\4\end{pmatrix} \text{ is an eigenvector of }A\text{ for the eigenvalue }\lambda=3 \right\}$;
4. $\left\{A\in V: v=\begin{pmatrix}-4\\4\end{pmatrix} \text{ is an eigenvector of }A \right\}$;
5. $\{A\in V:A\text{ is diagonalisable}\}$;
6. $\{A\in V:A^2=I\}$, where $I$ is the identity matrix in $V$;
7. $\{A\in V:A\text{ is a diagonal matrix}\}$.

(b)

For a fixed matrix $B\in V$, define

where $O$ is the zero matrix in $V$. It is given that $S_B$ is a subspace of $V$.

(i)

Under simple conditions imposed on $B$, achieve $\dim(S_B)=0$ and $\dim(S_B)=4$. If $C$ is a basis of $S_B$, what can be said about the number of elements in $C$?

(ii)

Find a matrix $C$ such that, for

the equation $AB=O$ can be written as

Use this result to prove that if

then $\dim(S_B)=2$.

Solution

Each set is checked for the zero matrix and for closure under addition and scalar multiplication. Homogeneous linear conditions on the entries of $A$ usually pass. Conditions involving a fixed non-zero eigenvalue, invertibility or the nonlinear equation $A^2=I$ need separate checking.

(a) Testing the listed sets

For the first set, if $A^{\mathsf T}=4A$, transpose both sides:

Hence $15A=0$, so $A=O$. The set is therefore $\{O\}$, which is a subspace.

The second set is the set of singular matrices, since $0$ is an eigenvalue exactly when the matrix has a non-zero vector in its kernel. It is not closed under addition. For example,

both have $0$ as an eigenvalue, but

does not.

For the third set, the condition is

The zero matrix does not satisfy this condition, because $Ov=0\ne3v$, so this set is not a subspace. It is an affine set of matrices rather than a homogeneous solution set.

For the fourth set, saying that $v$ is an eigenvector for some eigenvalue means that

If $A_1v=\alpha v$ and $A_2v=\beta v$, then

The zero matrix also satisfies the condition, with eigenvalue $0$. Thus this set is a subspace.

The fifth set is not a subspace because diagonalisability is not preserved under addition. For example,

are both diagonalisable: each has two distinct eigenvalues. Their sum is

which has only one independent eigenvector and is not diagonalisable.

The sixth set does not contain the zero matrix, since $O^2=O\ne I$. Therefore it is not a subspace.

Finally, the diagonal matrices form a subspace: the zero matrix is diagonal, and sums and scalar multiples of diagonal matrices are diagonal.

The correct selections are therefore

(b)(i) Extreme choices of $B$

If $B$ is invertible and $AB=O$, multiply on the right by $B^{-1}$:

Thus $S_B=\{O\}$ and

In particular, this happens for $B=I$.

At the other extreme, if $B=O$, then $AB=O$ for every $A\in V$. Hence $S_B=V$. Since a $2\times2$ matrix has four independent entries,

If $C$ is a basis of $S_B$, the defining property of dimension gives

(b)(ii) Converting $AB=O$ into a homogeneous system

In this part, $C$ denotes the coefficient matrix requested in the question, not the basis called $C$ in part (i).

Multiplying the two matrices gives

Therefore $AB=O$ is equivalent to the four scalar equations

and

With the variable order $(x_1,x_2,x_3,x_4)^{\mathsf T}$, this is

For the specified matrix

the coefficient matrix becomes

Row reduction gives

Thus $\operatorname{rank}(C)=2$. The system has four variables, so rank-nullity gives

The entries of $A$ identify $V$ with $\mathbb R^4$ and the solutions of $C\mathbf x=0$ are precisely the matrices in $S_B$. Hence

More explicitly, $x_2$ and $x_4$ are free. Writing $x_2=s$ and $x_4=t$ gives

so

These two matrices form a basis of $S_B$, which confirms the dimension directly. The homogeneous condition $AB=O$ defines a kernel, the coordinate matrix $C$ describes it and rank-nullity gives its dimension.

parameter-dependent vectors and possible bases

Hilda and Jonathan have picked three vectors $\vec v_1,\vec v_2$ and $\vec v_3$ in $\mathbb R^3$. These vectors depend on a parameter $q$. Hilda and Jonathan use these vectors to define a matrix $A$: the first column is $\vec v_1$, the second column is $\vec v_2$ and the third column is $\vec v_3$. They then add

as the fourth column. The full matrix is not needed; the elimination output below is sufficient. The only additional assumption is that $q$ is a real number.

The supplied elimination output is

Let $d$ be the dimension of the span of

(a)

Select all statements about $S$ and $d$ that are true:

• There is no value of $q$ for which $S$ is linearly independent.
• There is exactly one $q$ for which $S$ is linearly independent.
• There are infinitely many values of $q$ for which $S$ is linearly independent.
• There is no value of $q$ for which $d=2$.
• There is exactly one $q$ for which $d=2$.
• There are infinitely many values of $q$ for which $d=2$.
• There is no value of $q$ for which $d=1$.
• There is exactly one $q$ for which $d=1$.
• There are infinitely many values of $q$ for which $d=1$.
• There is no value of $q$ for which $S$ spans $\mathbb R^3$.
• There is exactly one $q$ for which $S$ spans $\mathbb R^3$.
• There are infinitely many values of $q$ for which $S$ spans $\mathbb R^3$.

Multiple selection advice: In a multiple selection question, one point is awarded for each correct selection and one point is deducted for each incorrect selection, but you cannot get a negative mark for the question in the end.

Case 1: $q\ne0$

(b)

If possible, find an expression of $\vec v_3$ as a linear combination of $\vec v_1$ and $\vec v_2$.

(c)

Determine the span of $S$ when $q\ne0$. The answer may depend on $q$.

(d)

Recall that $\vec e_1,\vec e_2,\vec e_3$ denote the standard basis vectors of $\mathbb R^3$. The supplied computations show that

is a basis of $\mathbb R^3$ for every value of $q$. Consider all bases of $\mathbb R^3$ obtained by selecting vectors from

When $q\ne0$ and $q\ne-15$, how many different bases can be formed that contain as many elements of $S$ as possible? When $q=-15$, how many can be formed?

Case 2: $q=0$

(e)

Determine the span of $S$ when $q=0$.

(f)

When $q=0$, how many different bases of $\mathbb R^3$ can be formed from

that contain as many elements of $S$ as possible?

Solution

The first three columns of the row-reduced matrix are

If $q\ne0$, there are pivots in columns $1$ and $2$, so

The third column relation in the row-reduced matrix is

The displayed matrix comes from row operations, which preserve linear relations among columns, although it need not be in echelon form at special values such as $q=0$. Therefore, when $q\ne0$,

The first two columns have pivots, so $\vec v_1,\vec v_2$ are independent. Consequently,

is a two-dimensional plane in $\mathbb R^3$.

If $q=0$, the first three columns reduce to

so

and $d=1$. The last column of the displayed row reduction shows that a vector $\vec b$ lies in this line exactly when

Thus

when $q=0$.

These dimensions settle part (a). The set $S$ is never linearly independent and never spans $\mathbb R^3$. It has dimension $2$ for every non-zero real $q$ and dimension $1$ for the single value $q=0$. Therefore the true selections are