Help understanding changes of basis

Question

Help: I don't understand how we can claim that

$$[L(v_1)]_B = P^{-1} [L(v_1)]_c$$

Can someone explain why that is true?

Answer 1

This is because $P$ is the matrix of the identity map from $(\mathbf R^n,\mathcal B)$ to $(\mathbf R^n,\mathcal C)$. Therefore, if you take any vector $w\in(\mathbf R^n, \mathcal B)$, it will be represented in $(\mathbf R^n, \mathcal C)$ by the column-vector $(w_{_\mathcal C})=P(w_{_\mathcal B})$.

Answer 2

We know that for any vector $[x]_{\mathcal{B}}$, we can find its representation under the basis $\mathcal{C}$ by the change of coordinates matrix $P$, which means by definition that $P[x]_{\mathcal{B}}=[x]_{\mathcal{C}}$. If you multiply by $P^{-1}$ on the right of both sides, we get that $[x]_{\mathcal{B}} = P^{-1}[x]_{\mathcal{C}}$.

Since this is true for all $x\in\mathbb{R}^n$, it must certainly be true for $L(v_i)$!

Answer 3

The actual proof is using this: given a vector with respect to one basis $B$, if we have a change-of-basis matrix $P$ from $B$ to $C$, can be written in terms of the other basis $C$ by left-multiplication by $P$. Then, to reverse this process, we use left-multiplication by $P^{-1}$. However to understand what's going on I think an example is helpful.

Suppose $C$ is the standard basis $\vec v_1=\begin{bmatrix}1\\0\\0\end{bmatrix}, \vec v_2 =\begin{bmatrix}0\\1\\0\end{bmatrix}, \vec v_3 = \begin{bmatrix}0\\0\\1\end{bmatrix}$. And let's take $B$ to be another basis $\vec w_1=\begin{bmatrix}2\\1\\1\end{bmatrix}, \vec w_2 = \begin{bmatrix}-1\\0\\3\end{bmatrix}, \vec w_3= \begin{bmatrix}0\\1\\1\end{bmatrix}$.

Now, by definition of the entries in vectors, or by writing it out, you can see that $\vec w_1 = 2\vec v_1+\vec v_2+\vec v_3$. And $\vec w_2 = -\vec v_1 + 3\vec v_3$, and $\vec w_3 = \vec v_2+\vec v_3$. So: if we write the columns of $C$ into a matrix $P$, we get $P = \begin{bmatrix}2&-1&0\\ 1&0&1\\ 1&3&1.\end{bmatrix}$, and we have that $\vec w_1 = P\vec v_1$, $\vec w_2 = P\vec v_2, \vec w_3 = P\vec v_3$.

So, note that $[\vec w_1]_B = \begin{bmatrix}1\\0\\0\end{bmatrix}$, but $[\vec w_1]_C = \begin{bmatrix} 2\\1\\1\end{bmatrix}$.

Now suppose we have a linear transformation, $L$, such that $L(\vec v_1) = 2\vec v_1-5\vec v_2, L(\vec v_2)=\vec 0$, and $L(\vec v_3) = \vec v_1+5\vec v_2$.

If you want to find $L(\vec v_1)$ with respect to the basis $B$, you will have to solve for $2\vec v_1-5\vec v_2 = \begin{bmatrix}2\\-5\\0\end{bmatrix}$ as a linear combination of $\vec w_1, \vec w_2, \vec w_3$. To do this, you can use $P^{-1}$.