True of False Problems on Determinants and Invertible Matrices

Stanford University Linear Algebra Exam Problems and Solutions

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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Solution.

We use the following properties of the determinant.

  1. For $n\times n$ matrices $A, B$, we have
    \[\det(AB)=\det(A)\det(B).\]
  2. 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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