Showing that if $A$ is diagonalizable then $A^2-4A+8I$ is diagonalizable

Question

Let $A_{n\times n}$ be a real matrix then:

1. if $A^4 = 8A$ then $A$ is not invertible.
2. if $A$ is diagonalizable over $\mathbb R$ then $A^2-4A+8I$ is diagonalizable over $\mathbb R$.

1. From $A^4 = 8A$ we get that $x(x^3-8)=0$ zero $A$ so $0$ can be one of its eigenvalues but it doesn't have to be, so that's false.

2. we know that $A=P^{-1}DP$, where $D$ is a diagonal matrix, $P$ is an invertible matrix, plug that into $A^2-4A+8I$ and we get $P^{-1}(D^2 -4D+8I)P$, so we get that $(D^2 -4D+8I)$ is a diagonal matrix, so $A^2-4A+8I = P^{-1}(D^2 -4D+8I)P$ which means that its diagonalizable?

I wasn't sure how to conclude that $(D^2 -4D+8I)$ is similar to $A^2-4A+8I$...

Answer 1

1. For the record, a counterexample to 1 (which you correctly disproved) is $A=2I$: $A^4=16I=8A$.

2. If $A=P^{-1}DP$, substitute that into $A^2-4A+8I$, you get:

   \begin{align*} A^2-4A+8I={}&(P^{-1}DP)^2-4P^{-1}DP+8I=(P^{-1}DP)(P^{-1}DP)-P^{-1}\cdot4D\cdot P+8I={} \\ {}={}&P^{-1}D(PP^{-1})DP-P^{-1}4DP+P^{-1}8IP=P^{-1}(D^2P-4DP+8P)={} \\ {}={}&P^{-1}(D^2-4D+8I)P. \end{align*}

   This proves $A^2-4A+8I$ is similar to $D^2-4D+8I$ via the same matrix as $A$, which implies it is diagonalizable since $D^2-4D+8I$ is obviously diagonal: it is, after all, a sum of diagonal terms.

Answer 2

In general, if $p(x)$ is a polynomial an $A$ a diagonalisable matrix, then so is $p(A)$.

First, note that if $A=U^{-1}DU$, then $A^n=U^{-1}D^nU$, and hence $$ p(A)=\sum_{k=0}^n c_kA^k=\sum_{k=0}^n c_kU^{-1}D^kU= U^{-1}\left(\sum_{k=0}^n c_k D^k\right)U. $$ Clearly, $\sum_{k=0}^n c_k D^k$ is also a diagonal matrix.