Diagonal matrix of a Basis

Question

Let $V$ be a vector space and $B=\{u, v, w\}$ be a basis of $V$. Let $T: V\to V$ be a linear map such that $T(u)=u$, $T(v)=4v-w$ and $T(w)=2v+w$. Is there a basis $B_1$ such that $[T]_{B_1}$ is diagonal?

So I got that the matrix representation for this linear map was:

$\begin{pmatrix}1&0&0\\ 0&4&2\\ 0&-1&1\end{pmatrix}$

And the eigenvalues were $1$, $2$, and $3$ with the eigenvectors:

$\begin{pmatrix}1\\ 0\\ 0\end{pmatrix}$, $\begin{pmatrix}0\\ 1\\ -1\end{pmatrix}$ and $\begin{pmatrix}0\\ 2\\ -1\end{pmatrix}$ respectively. So this led me to believe that this matrix was diagonalizable and a diagonal matrix would be:

$\begin{pmatrix}1&0&0\\ 0&2&0\\ 0&0&3\end{pmatrix}$

So although this matrix may be correct, I am confused by the idea of identifying a new basis $B1$ for this diagonal matrix. Am I actually supposed to change the initial basis or is having this diagonal matrix good enough?

Any help would be highly appreciated!

Answer 1

To find this matrix, you'll have to find the eigenvectors corresponding to these eigenvalues, and put them in the order of the eigenvalues; that is: $B_1=((1, 0, 0), (0, 1, -1), (0, 2, -1))$ is a basis in which $[T]_{B_1}$ is diagonal: $$ [T]_{B_1}=\begin{pmatrix}1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 3\end{pmatrix} $$

(Assuming the eigenvalues and eigenvectors you found are correct)