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.

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