# True of False Problems on Determinants and Invertible Matrices

## Problem 438

Determine whether each of the following statements is True or False.

**(a)** If $A$ and $B$ are $n \times n$ matrices, and $P$ is an invertible $n \times n$ matrix such that $A=PBP^{-1}$, then $\det(A)=\det(B)$.

**(b)** If the characteristic polynomial of an $n \times n$ matrix $A$ is

\[p(\lambda)=(\lambda-1)^n+2,\]
then $A$ is invertible.

**(c)** If $A^2$ is an invertible $n\times n$ matrix, then $A^3$ is also invertible.

**(d)** If $A$ is a $3\times 3$ matrix such that $\det(A)=7$, then $\det(2A^{\trans}A^{-1})=2$.

**(e)** If $\mathbf{v}$ is an eigenvector of an $n \times n$ matrix $A$ with corresponding eigenvalue $\lambda_1$, and if $\mathbf{w}$ is an eigenvector of $A$ with corresponding eigenvalue $\lambda_2$, then $\mathbf{v}+\mathbf{w}$ is an eigenvector of $A$ with corresponding eigenvalue $\lambda_1+\lambda_2$.

(Stanford University, Linear Algebra Exam Problem)

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Contents

- Problem 438
- Solution.
- (a) True or False: If $A=PBP^{-1}$, then $\det(A)=\det(B)$.
- (b) True or False: If $p(\lambda)=(\lambda-1)^n+2$, then $A$ is invertible.
- (c) True or False: If $A^2$ is an invertible $n\times n$ matrix, then $A^3$ is also invertible.
- (d) True or False: If $A$ is a $3\times 3$ matrix such that $\det(A)=7$, then $\det(2A^{\trans}A^{-1})=2$.
- (e) True or False: $\mathbf{v}+\mathbf{w}$ is an eigenvector of $A$ with corresponding eigenvalue $\lambda_1+\lambda_2$.

## Solution.

We use the following properties of the determinant.

- For $n\times n$ matrices $A, B$, we have

\[\det(AB)=\det(A)\det(B).\] - If $A$ is invertible, we have

\[\det(A^{-1})=\det(A)^{-1}.\]

Also recall that a matrix $A$ is invertible if and only if $\det(A)\neq 0$.

### (a) True or False: If $A=PBP^{-1}$, then $\det(A)=\det(B)$.

True. We have

\begin{align*}

&\det(A)=\det(PBP^{-1})\\

&=\det(P)\det(B)\det(P)^{-1} && \text{by properties 1, 2}\\

&=\det(P)\det(P)^{-1}\det(B) && \text{since determinants are just numbers}\\

&=\det(B).

\end{align*}

### (b) True or False: If $p(\lambda)=(\lambda-1)^n+2$, then $A$ is invertible.

True. Note that we have $p(0)=(-1)^n+2\neq 0$.

Thus $0$ is not an eigenvalue of $A$.

Recall that a matrix is invertible if and only if it does not have $0$ as an eigenvalue.

Thus, $A$ is invertible.

### (c) True or False: If $A^2$ is an invertible $n\times n$ matrix, then $A^3$ is also invertible.

True. If $A^2$ is invertible, then we have $\det(A^2)\neq 0$.

Then by property 1, we have

\[\det(A)^2=\det(A^2)\neq 0,\]
and hence $\det(A)\neq 0$.

It follows from property 1 that we have

\begin{align*}

&\det(A^3)=\det(A)^3\neq 0.

\end{align*}

Thus the matrix $A^3$ is invertible.

### (d) True or False: If $A$ is a $3\times 3$ matrix such that $\det(A)=7$, then $\det(2A^{\trans}A^{-1})=2$.

False. Recall that we have $\det(A^{\trans})=\det(A)$.

Together with properties 1, 2, we have

\begin{align*}

&\det(2A^{\trans}A^{-1})=\det(2I)\det(A^{\trans})\det(A^{-1})\\

&=\det(2I)\det(A)\det(A)^{-1}\\

&=\det(2I)=8.

\end{align*}

Here $I$ is the $3\times 3$ identity matrix.

Remark that in general for an $n\times n$ matrix and a scalar $r$, we have

\[\det(rA)=r^n\det(A).\]

### (e) True or False: $\mathbf{v}+\mathbf{w}$ is an eigenvector of $A$ with corresponding eigenvalue $\lambda_1+\lambda_2$.

False. For example, let $A$ be a $2\times 2$ identity matrix.

The only eigenvalue of $A$ is $1$.

Note that $\mathbf{v}=\begin{bmatrix}

1 \\

0

\end{bmatrix}$ and $\mathbf{w}=\begin{bmatrix}

0 \\

1

\end{bmatrix}$ are both eigenvector corresponding to the eigenvalue $\lambda_1=\lambda_2=1$.

However, $\lambda_1+\lambda_2=2$ is not an eigenvalue of $A$, and thus $\mathbf{v}+\mathbf{w}$ cannot be an eigenvector corresponding to $\lambda_1+\lambda_2=2$.

Note that the set of eigenvectors corresponding to an eigenvalue $\lambda$ together with the zero vector form a vector space, called the eigenspace of $\lambda$. Thus, if $\mathbf{v}$ and $\mathbf{w}$ are both eigenvectors of $\lambda$, then the sum $\mathbf{v}+\mathbf{w}$ is also an eigenvector of $\lambda$ or the zero vector.

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