So I have a question below as well as my solution but I'm told my answer is wrong and would appreciate some help/guidance.
For the matrix below, provide a $P$ and a diagonal matrix $D$ such that $A=PDP^{−1}$
$$A= \begin{bmatrix}{l}9-10i&8i&-8i&-4+4i\\0&1+2i&0&0\\16-24i&16i&1-14i&-8+8i\\8-64i&16+32i&-16-32i&-7+26i\end{bmatrix}$$
This matrix has eigenvalues $λ = 1−2i, 1+2i$
My solution:
For $λ = 1−2i$, I subtracted this value from the four diagonal spaces in the matrix
$$\begin{bmatrix}{l}8-8i&8i&-8i&-4+4i\\0&4i&0&0\\16-24i&16i&-12i&-8+8i\\8-64i&16+32i&-16-32i&-8+28i\end{bmatrix}$$
RREF this matrix:
$$\begin{bmatrix}{l}1&0&0&-1/5 - i/10\\0&1&0&0\\0&0&1&-2/5 - i/5\\0&0&0&0\end{bmatrix}$$
This gave the values $x1 = 1/5 + i/10, x2 = 0, x3 = 2/5 + i/5, x4 = 1$
For $λ = 1+2i$, I subtracted this value from the four diagonal spaces in the matrix
$$\begin{bmatrix}{l}8-12i&8i&-8i&-4+4i\\0&0&0&0\\16-24i&16i&-16i&-8+8i\\8-64i&16+32i&-16-32i&-8+24i\end{bmatrix}$$
RREF this matrix:
$$\begin{bmatrix}{l}1&0&6/13 - 4i/13&-5/13 - i/13\\0&1&0&0\\0&0&1&0\\0&0&0&0\end{bmatrix}$$
I got two vectors for this one because there are two free variables in columns 3 and 4
The first vector: $x1 = -6/13 + 4i/13, x2 = 0, x3 = 1, x4 = 0$
The second vector: $x1 = 5/13 + i/13, x2 = 0, x3 = 0, x4 = 1$
Placing all these vectors in the columns, I got:
$$P = \begin{bmatrix}{l}1/5 + i/10&-6/13 + 4i/13&5/13 + i/13&0\\0&0&0&0\\2/5 + i/5&1&0&0\\1&0&1&0\end{bmatrix}$$
Since there was only 1 vector for $λ = 1−2i$, it only appears once in matrix $D$. Since there are 2 vectors for $λ = 1+2i$, it appears twice.
$$D = \begin{bmatrix}{l}1-2i&0&0&0\\0&1+2i&0&0\\0&0&1+2i&0\\0&0&0&0\end{bmatrix}$$
I'm told my answer is wrong. I'm not sure if it's for both $P$ and $D$ but I was wondering if someone could look over my solution and help me understand what I did wrong. I'm really frustrated as I've spent a while on this question but I can't seem to get it right. I'm not sure if it's a simple calculation error or if my whole logic is wrong.
This might help get you in the right direction: It's worth double checking to make sure you determined all the eigenvalues. With an $4 \times 4$ matrix, I believe you should have 4 eigenvalues, with possible repeats. That is one place to verify.
It's worth double checking the RREF for $\lambda=1+2i$. I did a separate numerical check, and I believe there should be a third free variable. In other words, you should have 3 vectors coming out of that RREF.
(Please note: I'm assuming the small l in the top left entry of the matrices in front of numbers is a typo and not a numerical one.)
I hope this helps.