<s>[INST] Finding the matrix for a linear transformation on a vector space when the basis changes Let B={$u_1,u_2,u_3$} as basis of Vector Space V, and Let T: V→V be the linear operator defined by,
$$
        [T]_B=\begin{bmatrix}
        -3 & 4 & 7 \\
        1 & 0 & -2 \\
        0 & 1 & 0 \\
        \end{bmatrix}
$$
Find $[T]_{B'}$. B'={$v_1,v_2,v_3$} is basis of V defined by $v_1 = u_1, v_2 = u_1 + u_2, v_3 = u_1 + u_2 + u_3$.
I have solved questions by my effort, but I'm not sure for this answer. Please check and give me a hint for this question. Thank you.
My Answer:  
$v_1 =\begin{pmatrix}
        1\\
        0\\
        0\\
        \end{pmatrix}$, $v_2 =\begin{pmatrix}
        1\\
        1\\
        0\\
        \end{pmatrix}$, $v_3 =\begin{pmatrix}
        1\\
        1\\
        1\\
        \end{pmatrix}$
and, $T(v_1) =\begin{pmatrix}
        -3 & 4 & 7 \\
        1 & 0 & -2 \\
        0 & 1 & 0 \\
        \end{pmatrix}\begin{pmatrix}
        1\\
        0\\
        0\\
        \end{pmatrix}=\begin{pmatrix}
        -3\\
        1\\
        0\\
        \end{pmatrix}$, $T(v_2) =\begin{pmatrix}
        -3 & 4 & 7 \\
        1 & 0 & -2 \\
        0 & 1 & 0 \\
        \end{pmatrix}\begin{pmatrix}
        1\\
        1\\
        0\\
        \end{pmatrix}=\begin{pmatrix}
        1\\
        1\\
        1\\
        \end{pmatrix}$$, T(v_3) =\begin{pmatrix}
        -3 & 4 & 7 \\
        1 & 0 & -2 \\
        0 & 1 & 0 \\
        \end{pmatrix}\begin{pmatrix}
        1\\
        1\\
        1\\
        \end{pmatrix}=\begin{pmatrix}
        8\\
        -1\\
        1\\
        \end{pmatrix}$  
$[T(v_1)]_{B'}$ is consisted by $-4v_1+v_2$, $[T(v_2)]_{B'}$ is consisted by $v_3$, $[T(v_3)]_{B'}$ is consisted by $9v_1-2v_2+v_3$.
finally, vector representaion of $[T]_{B'} = \begin{pmatrix}
        -4 & 0 & 9 \\
        1 & 0 & -2 \\
        0 & 1 & 1 \\
        \end{pmatrix}$.
 [/INST] Here is a trick.
Let $B' = \begin{pmatrix} 1&1&1\\0&1&1\\0&0&1\end{pmatrix}$ This is the basis of B' in terms of the basis of B.
To change the basis of T.
$[T]_B = [B'^{-1}TB']_{B'}$ 
</s>