Diagonalization of matrix and eigenvalues of $A+A^2+A^3$

Question

Let $$A= \begin{pmatrix} 6 & 4 & 1 \\ -6 & -1 & 3 \\ 8 & 8 & 4 \\ \end{pmatrix} $$

Find a non-sigular matrix P and a diagonal matrix D such that $A+A^2+A^3=PDP^{-1}$

No idea what to do

Answer 1

You do not have to calculate $A+A^2+A^3$.

Suppose that you know how to diagonalize the matrix $A$, i.e., you can find an invertible matrix $P$ and a diagonal matrix $D$ such that $A=PDP^{-1}$. (I will leave the computations to you. You can have a look at other posts tagged diagonalization or on Wikipedia. I guess you can find there something to get you started. Then you can check your result on WolframAlpha or using some other tools.)

If you already have $A=PDP^{-1}$, then it is easy to see that $$ \begin{align*} A^2&=(PDP^{-1})(PDP^{-1}) = PD^2P^{-1}\\ A^3&=(PD^2P^{-1})PDP^{-1} = PD^3P^{-1} \end{align*} $$ and we get $$A+A^2+A^3=PDP^{-1}+PD^2P^{-1}+PD^3P^{-1} = P(D+D^2+D^3)P^{-1}.$$ Clearly, the matrix $D+D^2+D^3$ is a diagonal matrix. If $d_1$, $d_2$, $d_3$ are the diagonal elements of the matrix the, then $D+D^2+D^3$ has $$d'_i=d_i+d_i^2+d_i^3$$ on the diagonal.