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