Characteristic polynomial of a $3\times 3$ matrix Let $$
 M = \begin{bmatrix}
       1 &1 & 3 \\[0.3em]
       1 & 5 & 1 \\[0.3em]
       3  & 1 & 1
     \end{bmatrix}
$$ a matrix. I am stuck on solving this question from my book. Find eigenvalues, characteristic polynomial and diagonalize this matrix. 
 A: You have that the characteristic polynomial is given by: $\det(M-\lambda I)$, so we have:
$$M = \begin{bmatrix}
       1-\lambda &1 & 3 \\[0.3em]
       1 & 5-\lambda & 1 \\[0.3em]
       3  & 1 & 1-\lambda
     \end{bmatrix}$$
Find the determinant.
$$\begin{align*}
\det(M-\lambda I) &=(1-\lambda)(5-\lambda)(1-\lambda)+3+3-9(5-\lambda)-(1-\lambda)-(1-\lambda) \\
&=(1-\lambda)(5-\lambda)(1-\lambda)-2(1-\lambda)+6.
\end{align*}$$
This is the characteristic polynomial. To find the eingenvalues, make it equal to zero then find the roots of such polynomial.Then you're done.
Now, for the diagonal matrix :$M$ is called diagonalizable if it is similar to a diagonal matrix, i.e., if there exists an invertible matrix $P$ such that $P^{−1}MP$ is a diagonal matrix. You have to find the eigenvector related to the eigenvalues you found.Suppose they are $\lambda _1,\lambda _2, \lambda _3.$ They will appear in the diagonal of the diagonal matrix you look for.The proccess to find such matrix is the following:


*

*You find the eigenvectors related to the eigenvalues you found;

*Suppose they are:
$$v_1=[a_1, a_2, a_3],\ v_2=[b_1, b_2, b_3],\ v_3=[c_1, c_2,c_3].$$
Now let $P$ be the matrix of these eigenvectors as its columns:
$$\begin{bmatrix}
           a_1 &b_1 & c_1 \\[0.3em]
           a_2 & b_2 & c_2 \\[0.3em]
           a_3  & b_3 & c_3
         \end{bmatrix}$$ 
Now you will have to find $P^{-1}$.Once you found it you have to
compute:
$P^{-1}MP$, which is equal to your diagonal matrix with the
eigenvalues in the main diagonal.It's a good exercise to check this
way.
