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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!

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  • $\begingroup$ If $P^{-1}AP$ is diagonal if and only if the columns of $P$ are eigenvectors of $A$ $\endgroup$ – Omnomnomnom Jun 26 at 14:08
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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$$M=\begin{bmatrix}1&2&0&0\\2&1&0&1\\0&0&1&2\\3&0&1&-1\end{bmatrix}.$$

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  • $\begingroup$ 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. $\endgroup$ – Gatsby Jun 26 at 14:32
  • $\begingroup$ Your first piece of information is that$$M.(1,2,0,3)^T=(4,8,0,12)^T=4\times(1,2,0,3)^T.$$ $\endgroup$ – José Carlos Santos Jun 26 at 14:34
  • $\begingroup$ I see it now. Thank you! Now I can proceed with my studies... :) $\endgroup$ – Gatsby Jun 26 at 14:37
  • $\begingroup$ I'm glad I could help. $\endgroup$ – José Carlos Santos Jun 26 at 14:39

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