# Diagonal matrix of a Basis

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!

## 1 Answer

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)