Use the defining relation $A\mathbf{v}=\lambda \mathbf{v}$.

Solution.

(a) If $A$ is invertible, is $\mathbf{v}$ an eigenvector of $A^{-1}$?

The answer is yes. First note that the eigenvalue $\lambda$ is not zero since $A$ is invertible.

By definition, we have $A\mathbf{v}=\lambda \mathbf{v}$. Multiplying it by $A^{-1}$ from the left, we have
\[\mathbf{v}=\lambda A^{-1}\mathbf{v}.\]

As noted above, $\lambda$ is not zero, so we divide this equality by $\lambda$ and obtain
\[A^{-1}\mathbf{v=}\frac{1}{\lambda}\mathbf{v}.\]
Since $\mathbf{v}$ is not a zero vector, this implies that $\mathbf{v}$ is an eigenvector corresponding to the eigenvalue $1/\lambda$ of $A$.

(b) Is $3\mathbf{v}$ an eigenvector of $A$?

The answer is yes. We calculate
\[A(3\mathbf{v})=3A\mathbf{v}=3\lambda \mathbf{v}=\lambda (3\mathbf{v}).\]

Thus we have $A(3\mathbf{v})=\lambda (3\mathbf{v})$.
Since $3\mathbf{v}\neq 0$, this implies that $3\mathbf{v}$ is an eigenvector corresponding to the eigenvalue $\lambda$.

Comment.

In part (a), you shouldn’t divide by $\lambda$ without stating it is nonzero.

To prove that if a matrix $B$ is invertible, then an eigenvalue of $B$ is nonzero, you might want to consider for example

The determinant of the matrix $B$ is the product of all eigenvalues of $B$, or

If $0$ is an eigenvalue of $B$ then $B\mathbf{x}=\mathbf{0}$ has a nonzero solution, but if $B$ is invertible, then it’s impossible.

For part (b), note that in general, the set of eigenvectors of an eigenvalue plus the zero vector is a vector space, which is called the eigenspace.

Thus, a scalar multiplication of an eigenvector is again an eigenvector of the same eigenvalue.
Part (b) is a special case of this fact.

More Eigenvalue and Eigenvector Problems

Problems about eigenvalues and eigenvectors are collected on the page:

Two Matrices with the Same Characteristic Polynomial. Diagonalize if Possible.
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 […]

True of False Problems on Determinants and Invertible Matrices
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 $A$ […]

Given Eigenvectors and Eigenvalues, Compute a Matrix Product (Stanford University Exam)
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 […]

If the Kernel of a Matrix $A$ is Trivial, then $A^T A$ is Invertible
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 […]

Diagonalizable Matrix with Eigenvalue 1, -1
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.
(Stanford University Linear Algebra Exam)
See below for a generalized problem.
Hint.
Diagonalize the […]

Maximize the Dimension of the Null Space of $A-aI$
Let
\[ A=\begin{bmatrix}
5 & 2 & -1 \\
2 &2 &2 \\
-1 & 2 & 5
\end{bmatrix}.\]
Pick your favorite number $a$. Find the dimension of the null space of the matrix $A-aI$, where $I$ is the $3\times 3$ identity matrix.
Your score of this problem is equal to that […]

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