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!


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)


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