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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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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