My method is:Suppose that A is a square matrix and $\displaystyle \lambda$ is an eigenvalue of A.

(i) Show that $\displaystyle \lambda ^n$ is an eigenvalue of $\displaystyle A^n$ for all positive integers n.

(ii) Suppose A is invertible. Show $\displaystyle \lambda$ is non-zero and that $\displaystyle \lambda ^{-1}$ is an eigenvalue of $\displaystyle A^{-1}$

Let $\displaystyle |A|=f(x)$

Since $\displaystyle \lambda$, $\displaystyle f(\lambda)=0$

I need to show that $\displaystyle f(\lambda ^{n})= 0 $

$\displaystyle f ( \lambda)=0$

$\displaystyle (f(\lambda))^n=0^n$

$\displaystyle (f(\lambda)^n)=f(\lambda ^n)=0$ which is what I needed to show.

Is this right?

For part 2:

If A is invertible then $\displaystyle A^{-1}$ exists.

Like before $\displaystyle f(\lambda)=0$.

$\displaystyle A=\begin{pmatrix}

{a_{11}}&{.....}&{a_{1n}}\\

{...}&{...}&{....}\\

{a_{n1}}&{....}&{a_{nn}}

\end{pmatrix}$

$\displaystyle Det(A- \lambda I)=0$

Hence:

$\displaystyle \begin{vmatrix}

{a_{11}}&{....}&{a_{1n}}\\

{....}&{....}&{....}\\

{a_{n1}}&{....}&{a_{nn}}

\end{vmatrix}-\lambda I=\begin{vmatrix}

{a_{11}-\lambda}&{....}&{a_{1n}}\\

{....}&{....}&{....}\\

{a_{n1}}&{....}&{a_{nn}-\lambda}

\end{vmatrix}=f(\lambda)=0$

This is where my method goes awry. $\displaystyle f(\lambda)$ cannot equal zero or the matrix A is not invertible.

What am I doing wrong??