Given $T:\mathbb{R^3}\rightarrow \mathbb{R^3}$ defined by: $$ [T]^E_E=\begin{pmatrix} -2 & 4 & 5 \\ -8 & 12 & 12 \\ 8 & -11 & -10 \\ \end{pmatrix} $$
I need to find a basis $B=(\vec b_1,\vec b_2,\vec b_3)$ such that: $$ [T]^B_B=\begin{pmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \\ \end{pmatrix} $$
I tried using the relation $[I]^B_E[T]^B_B[I]^E_B=[T]^E_E$ but it lead me nowhere since I couldn't find $[I]^B_E, [I]^E_B$. Any help would be appretiated, thank you!
The first column of $[T]_B^B$ is the zero vector. This means that $$T(\vec{b}_1) = 0\vec{b}_1 + 0\vec{b}_2 + 0\vec{b}_3 = 0.$$ The second column of $[T]_B^B$ is $(1, 0, 0)^\top$. This means that $$T(\vec{b}_2) = 1\vec{b}_1 + 0\vec{b}_2 + 0\vec{b}_3 = \vec{b}_1.$$ The third column of $[T]_B^B$ is $(0, 1, 0)^\top$. This means that $$T(\vec{b}_3) = 0\vec{b}_1 + 1\vec{b}_2 + 0\vec{b}_3 = \vec{b}_2.$$ So, I would start by finding a non-zero solution to $T(\vec{x}) = 0$, and let it be $\vec{b}_1$. Then, take this solution, find a solution to $T(\vec{x}) = \vec{b}_1$, and let $\vec{b}_2$ be this solution. Finally, solve $T(\vec{x}) = \vec{b}_2$, and let $\vec{b}_3$ be one of the solutions.