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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 $[T]_{B'} = \begin{pmatrix} -4 & 0 & 9 \\ 1 & 0 & -2 \\ 0 & 1 & 1 \\ \end{pmatrix}$.

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  • $\begingroup$ It's better to distinguish the difference between a vector and its coordinates. For instance, $(v_1)_{\mathcal{B}}=[1,0,0]^T$. $\endgroup$ – Jack May 25 '16 at 16:47
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I would like to do it in a systematic way. It is instructive to see how your question fits in the following steps.

Step 1

Let $V$ be a $n$-dimensional vector space over $\mathbb{F}$ (say, $\mathbb{R}$). Consider the following two ordered bases of $V$ $$ \alpha=(v_1,\cdots,v_n),\quad \beta=(w_1,\cdots,w_n). $$ For $x\in V$, denote respectively the coordinate of $x$ with respect to the ordered bases $\alpha$ and $\beta$ as $$ [x]_\alpha=(x_1,\cdots,x_n)^T,\\ [x]_\beta=(y_1,\cdots,y_n)^T. $$ If we define the $n\times n$ matrix as $$ B=[[w_1]_\alpha,\cdots,[w_n]_\alpha], $$ one can formally write $\beta=\alpha B$ and also, $$ x=\alpha[x]_\alpha=\beta[x]_\beta=\alpha B[x]_\beta $$ which implies that $$ [x]_\beta=B^{-1}[x]_\alpha\tag{*} $$

Step 2

Let $T:V\to V$ be a linear transformation. By definition, the matrix of $T$ w.r.t. $\alpha$ is $$ [T]_\alpha=[[Tv_1]_\alpha,\cdots,[Tv_n]_\alpha]. $$ Your want to find $$ [T]_\beta=[[Tw_1]_\beta,\cdots,[Tw_n]_\beta]. $$ By $(*)$, for each $1\leq k\leq n$ $$ [Tw_k]_\beta=B^{-1}[Tw_k]_\alpha=B^{-1}[T]_\alpha[w_k]_\alpha\tag{**} $$ where I use a simple exercise that $[Tx]_\alpha=[T]_\alpha[x]_\alpha$ for any $x\in V$. Now by the definition of $B$ and $(**)$, we can see that $$ [T]_\beta=B^{-1}[T]_\alpha B. $$

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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.

$[T]_B = [B'^{-1}TB']_{B'}$

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So first things first, your answer is correct. Since you're confused, it means that you don't totally understand the process that you used to get to your answer, so let's break it down step by step.

The first thing you did is put each new basis vector in $B' = \{v_1, v_2, v_3\}$ in terms of your original basis $B = \{ u_1, u_2, u_3 \} $. From there you have the equations:

$$ [T]_B[v_i]_B = [T(v_i)]_B $$

For $i$ in $\{1,2,3\}$. For $i = 3$ (the other two are very obvious so I'm going to use the third as an example), it takes this form:

$$ \begin{bmatrix} -3 & 4 & 7 \\ 1 & 0 & -2 \\ 0 & 1 & 0 \end{bmatrix} \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix} = \begin{bmatrix} 8 \\ -1 \\ 1 \end{bmatrix} $$

Or,

$$ T(v_3) = T(u_1 + u_2 + u_3) = 8u_1 - u_2 + u_3 $$ $$= 9(u_1) - 2(u_1+u_2) + (u_1+u_2+u_3) $$ $$ = 9v_1 -2v_2 + v_3 $$

This is the same as saying that, if you take $B' $ to be the basis for a matrix representation of $T$,

$$ [T]_{B'}[v_3]_{B'} = [T]_{B'} \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix} = \begin{bmatrix} 9 \\ -2 \\ 1 \end{bmatrix} $$

Which determines the third column of $[T]_{B'}$. By this same logic, this process also determines the other two columns of $[T]_{B'} $, which you have already done.

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