A is $n\times n$ diagonalizable matrix and that A has only 2, ad 4 as its eigenvalues. Show that $A^2=6A-8I$

Question

suppose A is $n\times n$ diagonalizable matrix and that A has only 2, ad 4 as its eigenvalues. How would you show that $A^2=6A-8I$

I've tried using eigenvectors but Im really stuck on showing this

Answer 1

Let $k$ be the multiplicity of 2 as an eigenvalue, then $n-k$ is the multiplicity of 4.

Diagonalizing $A$, with $P$ as a transition matrix, yields:

$$D=\begin{bmatrix} 2 & & & & & \\ & \ddots & & & & \\ & & 2 & & & \\ & & & 4 & & \\ & & & & \ddots & \\ & & & & & 4 \end{bmatrix} $$

With $k$ twos and $n-k$ fours.

Then $A^2$ is similar to (with the same transition matrix): $$ D^2=\begin{bmatrix} 4 & & & & & \\ & \ddots & & & & \\ & & 4 & & & \\ & & & 16 & & \\ & & & & \ddots & \\ & & & & & 16 \end{bmatrix}$$

Since the identity matrix commutes with all matrices, it commutes with $P$ and we find that $6A-8I$ is also similar to $D^2$ (with the same transition matrix) because $2\times 6 -8=4$ and $4\times 6 -8 =16$.

In the end, we found, $PA^2P^{-1} = P(6A-8I)P^{-1}$. Since $P$ is invertible, $A^2 = 6A-8I$ holds.

Answer 2

Note that $=(X-2)(X-4)=X^2-6X+8$, but since these are the only eigenvalues, you know that the characteristic polynomial is something like$(X-2)^k(X-4)^n$. From here, try Cayley-hamilton Cayley-Hamilton and a little algebra.