# Determining $P$ such that $P^{-1}AP$ is a diagonal matrix without knowing $A$ but only its eigenvalues and two eigenvectors?

I'm studying for a linear algebra exam and there is an exercise (about diagonalizable matrices) that I don't know how to resolve.

$$M$$ is a diagonalizable matrix such that $$\det(M-\lambda I4) = (-2-\lambda)(-2-\lambda)(3-\lambda)(4-\lambda)$$

where

$$M \begin{bmatrix} 1\\ 2\\ 0\\ 3\\ \end{bmatrix} = \begin{bmatrix} 4\\ 8\\ 0\\ 12\\ \end{bmatrix}$$

$$VM(-2) = \begin{bmatrix} 2y\\ y\\ -w\\ w\\ \end{bmatrix} : (y,w) \neq (0,0)$$

$$VM(3) = \begin{bmatrix} 0\\ y\\ 2y\\ -y\\ \end{bmatrix} : (y) \neq 0$$

Build a matrix $$P$$ so that $$P^{-1}MP = \mathrm{diag}(4; -3; 3; -2)$$.

It’s my first question here so sorry for my pour formation. I really need to know how to do this. Thanks in advance for any help!

• If $P^{-1}AP$ is diagonal if and only if the columns of $P$ are eigenvectors of $A$ – Omnomnomnom Jun 26 at 14:08

## 1 Answer

Your first piece of information tells us that $$(1,2,0,3)^T$$ is an eigenvector of $$M$$ with eigenvalue $$4$$.

From your second piece of information, you can deduce that $$(2,1,0,0)^T$$ and $$(0,0,-1,1)^T$$ are eigenvectors of $$M$$ with eigenvalue $$2$$. They are clearly linearly independent.

From your third piece of information, it follows that $$(0,1,2,-1)^T$$ is an eigenvector of $$M$$ with eigenvalue $$3$$.

So, take$$M=\begin{bmatrix}1&2&0&0\\2&1&0&1\\0&0&1&2\\3&0&1&-1\end{bmatrix}.$$

• Oh, I knew about your second and third point, but I didn't know about the first one. Can you explain me how/why I can take that conclusion from my first piece of information? Thank you, your answer helped me a lot. – Gatsby Jun 26 at 14:32
• Your first piece of information is that$$M.(1,2,0,3)^T=(4,8,0,12)^T=4\times(1,2,0,3)^T.$$ – José Carlos Santos Jun 26 at 14:34
• I see it now. Thank you! Now I can proceed with my studies... :) – Gatsby Jun 26 at 14:37
• I'm glad I could help. – José Carlos Santos Jun 26 at 14:39