<s>[INST] Linear transformation, transformation in base1 I have the following problem, I know how to solve it (knowing the answer) however I do not understand what I'm doing
We have the following base in $\mathbb{R}^4$, called $b_1$,$b_2$,$b_3$,$b_4$
$$B = \left\{
                  \vphantom{\begin{array}{c}1\\1\\1\\1\\1\end{array}}
                  \smash{\underbrace{
                      \begin{array}{ccccc}
                             1 \\0 \\0 \\ 0 \\
                      \end{array}
                      }_{\text{ $b_1$}}}
              ,
\vphantom{\begin{array}{c}1\\1\\1\\1\\1\end{array}}                   \smash{\underbrace{
                      \begin{array}{ccccc}
                             1 \\1 \\0 \\ 0 \\
                      \end{array}
                      }_{\text{ $b_2$}}},
\vphantom{\begin{array}{c}1\\1\\1\\1\\1\end{array}}                   \smash{\underbrace{
                      \begin{array}{ccccc}
                             0 \\1 \\1 \\ 0 \\
                      \end{array}
                      }_{\text{ $b_3$}}},
\vphantom{\begin{array}{c}1\\1\\1\\1\\1\end{array}}                   \smash{\underbrace{
                      \begin{array}{ccccc}
                             0 \\0 \\1 \\ 1 \\
                      \end{array}
                      }_{\text{ $b_4$}}}
\right\}
              \
$$
And the following linear transformation from $\mathbb{R}^4 \rightarrow \mathbb{R}^4$:
$$
                  f\left( \vphantom{\begin{array}{c}1\\1\\1\\1\\1\end{array}}
                  \smash{\underbrace{
                      \begin{array}{ccccc}
                             1 \\0 \\0 \\ 0 \\
                      \end{array}
                      }} \right) =  \left( \vphantom{\begin{array}{c}1\\1\\1\\1\\1\end{array}}
                  \smash{\underbrace{
                      \begin{array}{ccccc}
                             0 \\0 \\1 \\ 1 \\
                      \end{array}
                      }} \right) \text{   and   }                   f\left( \vphantom{\begin{array}{c}1\\1\\1\\1\\1\end{array}}
                  \smash{\underbrace{
                      \begin{array}{ccccc}
                             0 \\1 \\1 \\ 0 \\
                      \end{array}
                      }} \right) =  \left( \vphantom{\begin{array}{c}1\\1\\1\\1\\1\end{array}}
                  \smash{\underbrace{
                      \begin{array}{ccccc}
                             1 \\1 \\0 \\ 0 \\
                      \end{array}
                      }} \right)
$$ 
Also, the rest of the transformations of the base remain the same, in other words: $f(b_2) = b_2$ and $f(b_4) = b_4$
What is the representing matrix of F in the base B?
What I did was the following:
I found the inverse of the matrix B, then multiplied this inverse by the left with the "transformed" matrix (using the values that f gives us). Doing this I get to the right answer, which is:
Answer = $\begin{bmatrix}0 & 0 &0&0\\0&1&1&0\\0&0&0&0\\1&0&0&1 \end{bmatrix}$
However I don't understand why the answer is this, I thought I was supposed to multiply the inverse by the right in order to get the transformation of the matrix in the other base. Am I missing something, or misunderstanding what's being asked?
Thanks in advance
 [/INST] You don't need to use inverse matrix (you need it if you change base). Let $f\colon V\to V$ be a linear transformation ($V$ for vector space). If $B=\{e_1, \ldots, e_n\}$ is a basis of $V$, and
$$
f(e_1) = a_{11}e_1 + a_{21}e_2 + \ldots + a_{n1}e_n\\
\cdots\\
f(e_n) = a_{n1}e_1 + a_{n1}e_2 + \ldots + a_{nn}e_n\\
$$
then matrix
$$
\begin{pmatrix}
a_{11} & \cdots & a_{1n}\\
\vdots & \vdots & \vdots\\
a_{n1} & \cdots & a_{nn}
\end{pmatrix}
$$
called transformation matrix in the base $B$ (see wiki, for example). Just look:
$$
f(b_1) = b_4\\
f(b_2) = b_2\\
f(b_3) = b_2\\
f(b_4) = b_4\\
$$
And matrix is ...?
</s>