Finding a basis with respect to a matrix.

Question

Let $v$ be an endomorphism of $\mathbb{R^3}$ with the following matrix in the standard basis:

$M=\begin{pmatrix}0&1&0\\0&0&1\\1&-3&3\end{pmatrix}$.

Find a basis of $\mathbb{R^3}$ in which the matrix of $v$ is : $M'=\begin{pmatrix}1&1&0\\0&1&1\\0&0&0\end{pmatrix}$.

What i have so far:

Let $B$ be the standard basis , thus $B=${$(1,0,0),(0,1,0),(0,0,1)$}

We have :

$v(e_1)=e_3$.

$v(e_2)=e_1-3e_3$.

$v(e_3)=e_2+3e_3$.

Let $B'$ be the basis we're looking for , $B'=${$v1,v2,v3$}

We have :

$v(v_1)=v_1$

$v(v_2)=v_1+v_2$

$v(v_3)=v_2$

I'm stuck here and I have no idea on how to find B'. Could anyone give me hints ?

Thanks in advance.

Answer 1

Note that det(M)=$1$ while det(M')=$0$. Is it even possible to find a basis for which a one-to-one linear transformation changes to a linear transformation which is not one-to-one?