Computing matrices of linear transformation under different basis

Came across a problem in my textbook and I am having trouble with it, the book provides an answer but no workings. Looking to get an explanation of what is going on help I prepare for an exam. If we let $T : \mathbb R^2 \to \mathbb R^2$ be a linear transformation. Consdider two bases $V$ and $W$ of the vector space $\mathbb R^2$, where $V=\{v_1,v_2\}$ with $ v_1 =
\begin{bmatrix}
-2
\\1
\end{bmatrix}$ and $ v_2 =
\begin{bmatrix}
1
\\1
\end{bmatrix}$ and where $W = \{w_1,w_2\}$ with $ w_1 =
\begin{bmatrix}
-1
\\4
\end{bmatrix}$ and $ w_2 =
\begin{bmatrix}
0
\\1
\end{bmatrix}$. Suppose that $ A =
\begin{bmatrix}
1&2
\\-1 &4
\end{bmatrix}$ is the matrix of $T$ under basis $W$. I need to find the matrix $B$ the matrix of $T$ under basis $V$.

The solution is:

$ B = P^{-1}AP$ = $
\begin{bmatrix}
-30&22
\\-48 & 35
\end{bmatrix} $

But I don’t know why.

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Here’s the idea. In the basis $V$, we’re expressing every vector as a linear combination of $v_1$ and $v_2$. That is, for any vector $x$ we write as $x = c_1 v_1 + c_2 v_2$. For linear transformations,

$$ Mx = M(c_1 v_1 + c_2 v_2) = c_1 M v_1 + c_2 M v_2.$$

Thus, it makes sense to investigate where the transformation sends $v_1$ and $v_2$. In this case, your transformation is in the basis $W$.

  1. Convert $v_1$ into the basis $W$. Concretely, this means express $v_1$ as a linear combination of $w_1$ and $w_2$. Do the same for $v_2$. Arranged in matrix notation, this will be some matrix $P$.

  2. Apply the transformation $A$.

  3. Revert back to the basis $V$. In concrete terms, this will be expressing $w_1$ and $w_2$ as linear combinations of $v_1$ and $v_2$. Luckily, this is very closely related to what we did in the first step.

In order to find the matrix of $T$ w.r.t.$V$ we write each vector in $V$ as a linear combination of vectors in $W$.

So we get $(-2,1)=2(-1,4)+(-7)(0,1)$ and

$(1,1)=-1(-1,4)+5(0,1)$

Also $T(-1,4)=(-1,3)$ and $T(0,1)=(-2,12)$.(Obtained from $A$).

Hence $T(-2,1)=2T(-1,4)-7T(0,1)=(12,-78)$

And $T(1,1)=-1T(-1,4)+5T(0,1)=(-9,57)$.

Now the last step we have to write $(12,-78),(-9,57)$ as a linear combination of the vectors $(-2,1),(1,1)$

So $(12,-78)=c_1(-2,1)+c_2(1,1)\implies c_1=-30;c_2=-48$

Also $(-9,57)=d_1(-2,1)+d_2(1,1)\implies d_1=22;d_2=35$