Prove that for any diagonalizable matrix $A$, $A^n$ is diagonalizable and also $aA^m+bA^n$

Question

Suppose that A is a diagonalizable matrix.

1) Prove that $A^n$ is diagonalizable

2) Prove that $aA^n + b A^m$ is diagnalizable, for every $a,b\in\mathbb{K}$

I thank you any help or hint you can give me,

Regards

Answer 1

Hint. There exists an invertible matrix $P$ and a diagonal matrix $D$ such that: $$A=PDP^{-1}.$$ Prove by induction on $k$, that: $$\forall k\in\mathbb{N},(PDP^{-1})^k=PD^kP^{-1}.$$ Notice that for all integer $k$, $D^k$ is a diagonal matrix.

Answer 2

Let $A=PDP^{-1}$ for diagonal matrix $D$ and non-singular matrix $P$.

Now, $cA$ is diagonalizable for some scalar $c$ because we have $$cA=cPDP^{-1}=PcDP^{-1}=P(cD)P^{-1}$$ Here, $cD$ is the diagonal matrix and $P$ is the non-singular matrix.

Also, $A^n$ is diagonalizable because we know that: $$A^k=(PDP^{-1})^k=PD^kP^{-1}$$ Here, $D^k$ is the diagonal matrix and $P$ is the non-singular matrix.

Now, polynomials add expressions in the form of $cA^n$ together, so we get: $$c_1A^{n_1}+c_2A^{n_2}$$ We need to prove the above expression is diagonalizable. From the steps above, we know that this is the same as: $$P(c_2D^{n_1})P^{-1}+P(c_2D^{n_2})P^{-1}$$ Can you prove this is diagonalizable?

Answer 3

That $\bf A$ is diagonalizable means that we have $\bf T$ and $\bf D$ so that:

$${\bf A = TDT}^{-1}$$

and $\bf D$ is a diagonal matrix. Then a polynomial in $\bf A$ can be written like:

$$\sum_{k=0}^N c_k{\bf A}^k = \sum_{k=0}^N c_k({{\bf TDT}^{-1}})^k = \sum_{k=0}^N c_k({\bf{TD}}^k{\bf T}^{-1}) = {\bf T} \left(\sum_{k=0}^N c_k{{\bf D}^k}\right) {\bf T}^{-1}$$

Now what can be said about the central sum?