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)
Let
\[A=\begin{bmatrix}
1 & 3 & 3 \\
-3 &-5 &-3 \\
3 & 3 & 1
\end{bmatrix} \text{ and } B=\begin{bmatrix}
2 & 4 & 3 \\
-4 &-6 &-3 \\
3 & 3 & 1
\end{bmatrix}.\]
For this problem, you may use the fact that both matrices have the same characteristic polynomial:
\[p_A(\lambda)=p_B(\lambda)=-(\lambda-1)(\lambda+2)^2.\]
(a) Find all eigenvectors of $A$.
(b) Find all eigenvectors of $B$.
(c) Which matrix $A$ or $B$ is diagonalizable?
(d) Diagonalize the matrix stated in (c), i.e., find an invertible matrix $P$ and a diagonal matrix $D$ such that $A=PDP^{-1}$ or $B=PDP^{-1}$.
(Stanford University Linear Algebra Final Exam Problem)
Suppose that $\begin{bmatrix}
1 \\
1
\end{bmatrix}$ is an eigenvector of a matrix $A$ corresponding to the eigenvalue $3$ and that $\begin{bmatrix}
2 \\
1
\end{bmatrix}$ is an eigenvector of $A$ corresponding to the eigenvalue $-2$.
Compute $A^2\begin{bmatrix}
4 \\
3
\end{bmatrix}$.
Let $A$ be an $m \times n$ real matrix.
Then the kernel of $A$ is defined as $\ker(A)=\{ x\in \R^n \mid Ax=0 \}$.
The kernel is also called the null space of $A$.
Suppose that $A$ is an $m \times n$ real matrix such that $\ker(A)=0$. Prove that $A^{\trans}A$ is invertible.
Suppose that $A$ is a diagonalizable $n\times n$ matrix and has only $1$ and $-1$ as eigenvalues.
Show that $A^2=I_n$, where $I_n$ is the $n\times n$ identity matrix.