Find a real number $t$ such that $\alpha$ is represented with respect to the basis Doing some self study as this topic really stumps me, every time I see it I cringe. I believe this question requires knowledge of change of basis and using $\phi_{BD}(\alpha)$. 
Question: Let $\alpha\in End(\Bbb R^3)$ be represented with respect to the canonical basis by the matrix $\begin{bmatrix}
        4 & 4 & 1 \\
        2 & 2 & 3 \\
        2 & 2 & 3 \\
        \end{bmatrix}$.
Find a real number $t$ such that $\alpha$ is represented with respect to the basis $B=\begin{Bmatrix}\begin{bmatrix} t \\-2\\0 \end{bmatrix} \begin{bmatrix} -1 \\t  \\2  \\\end{bmatrix}\begin{bmatrix} 2\\2 \\t \\ \end{bmatrix}\end{Bmatrix}$ by the matrix $\begin{bmatrix}
        0 & 0 & 0 \\
        0 & 1 & 0 \\
        0 & 0 & 4 \\
        \end{bmatrix}$.
Can anyone explain this question? I know what the canonical basis is but im not seeing how to connect everything in this problem. Do you work backwards from the first matrix to find $\alpha$ then use alpha to find t?
I greatly appreciate someone explaining this to me.
Edited to add correct matrices: (I am leaving the original so the comments and answer given still make sense)
Let $\alpha\in End(\Bbb R^3)$ be represented with respect to the canonical basis by the matrix $\begin{bmatrix}
        2 & 2 & 0 \\
        1 & 1 & 2 \\
        1 & 1 & 2 \\
        \end{bmatrix}$.
Find a real number $t$ such that $\alpha$ is represented with respect to the basis $B=\begin{Bmatrix}\begin{bmatrix} t \\-1\\0 \end{bmatrix} \begin{bmatrix} -2 \\t  \\1  \\\end{bmatrix}\begin{bmatrix} 1\\1 \\t \\ \end{bmatrix}\end{Bmatrix}$ by the matrix $\begin{bmatrix}
        0 & 0 & 0 \\
        0 & 1 & 0 \\
        0 & 0 & 4 \\
        \end{bmatrix}$.

Because the matrix is diagonal, the form is very simple. If $B=\{b_1,b_2,b_3\}$, the matrix $$\begin{bmatrix}
        0 & 0 & 0 \\
        0 & 1 & 0 \\
        0 & 0 & 4 \\
        \end{bmatrix}$$ means that $$\tag{*} \alpha b_1=0,\ \ \alpha b_2=b_2,\ \ \alpha b_3=4b_3.$$ So
$$
\begin{bmatrix}
        0  \\
        0  \\
        0 \\
        \end{bmatrix}
=\alpha b_1=\begin{bmatrix}
        4 & 4 & 1 \\
        2 & 2 & 3 \\
        2 & 2 & 3 \\
        \end{bmatrix}
\begin{bmatrix}
        t \\
        -2\\
        0\\
        \end{bmatrix}
=\begin{bmatrix}
        4t-8 \\
        2t-4 \\
        2t-4 \\
        \end{bmatrix}.
$$
So you need $t=2$. Now you still need to check that this $t$ also works for $b_2$ and $b_3$, which doesn't in this case. 
In general, that $A $ is the matrix for the basis $b_1,b_2,b_3$ means that
$$\alpha b_1=A_{11}b_1+A_{12}b_2+A_{13}b_3, $$ and so on.
Edit: with the new data, the equations in $(*)$ now are 
$$
\begin{bmatrix}
        2 & 2 & 0 \\
        1 & 1 & 2 \\
        1 & 1 & 2 \\
        \end{bmatrix}\begin{bmatrix} t \\-1\\0 \end{bmatrix}
=\begin{bmatrix} 2t-2 \\t-1\\t-1 \end{bmatrix},
$$
which forces $t=1$. Now, for the other two vectors, 
$$
\begin{bmatrix}
        2 & 2 & 0 \\
        1 & 1 & 2 \\
        1 & 1 & 2 \\
        \end{bmatrix}
\begin{bmatrix}
        -2  \\
        1   \\
        1   \\
        \end{bmatrix}
=
\begin{bmatrix}
        -2 \\
        1  \\
        1 \\
        \end{bmatrix}
$$
and
$$
\begin{bmatrix}
        2 & 2 & 0 \\
        1 & 1 & 2 \\
        1 & 1 & 2 \\
        \end{bmatrix}
\begin{bmatrix}
        1 \\
        1 \\
        1  \\
        \end{bmatrix}
=
\begin{bmatrix}
       4 \\
        4 \\
        4 \\
        \end{bmatrix}=4\begin{bmatrix}
        1 \\
        1   \\
        1   \\
        \end{bmatrix}
$$
