Find new bases with respect to which the matrix of the linear map is diagonal Let $U=\mathbb{R}^3$ and $V=\mathbb{R}^3$. Let $T$ be the linear map $U\rightarrow V$ defined by the matrix
 $$A= \left[ \begin{matrix}
 6 &3&9\\
3&4&2\\
3&6&0
\end{matrix} \right]$$
with respect to the standard bases of $U$ and $V$. Find new bases for $U$ and $V$ with respect to which the matrix of $T$ is diagonal. 
Could someone please provide me with a method to answering this question as I have no idea where to start. 
 A: To find a basis for $U=V=\mathbb{R}^3$ such that $A$ is diagonal with respect to this basis, we need to find the eigenvalues and corresponding eigenvectors. The characteristic polynomial of this matrix is $-\lambda^3+10\lambda^2+24\lambda$, so the set of eigenvalues is $\{12,-2,0\}$. Then $\lambda_1=12$ corresponds to the eigenvector $v_1=(2,1,1)$, $\lambda_2=-2$ corresponds to the eigenvector $v_2=(-48,11,39)$, and $\lambda_3=0$ corresponds to the eigenvector $v_3=(-2,1,1)$. Now let $P$ be the matrix with rows $v_1,v_2$ and $v_3$, then $P$ is invertible and $P^{-1}AP=D$, where $D$ is the diagonal matrix consisting of the eigenvalues of $A$ (the order of the eigenvalues corresponds to the order of the corresponding eigenvectors, so in this case $D=\mathrm{diag}(12,-1,0)$. Note that $P$ is just a base change matrix, from $\{e_1,e_2,e_3\}$ to $\{v_1,v_2,v_3\}$.
For an example of a matrix which is not diagonalisable, there are many, but for example one can take
\begin{align}
\begin{pmatrix}
1 & 0 \\
1 & 1
\end{pmatrix},
\end{align}
or any nilpotent matrix.
