There are 2 bases, $B$ and $C$ that are bases for $\Bbb R^2$. Let $B=\{b_1, b_2\}$ and $C=\{c_1,c_2\}$. Find the change of coordinates matrix from $B$ to $C$.

$b_1= \left[ \begin{matrix} 7 \\ 5 \end{matrix} \right]$ $b_2= \left[ \begin{matrix} -3 \\ -1 \end{matrix} \right]$

$c_1= \left[ \begin{matrix} 1 \\ -5 \end{matrix} \right]$ $c_2= \left[ \begin{matrix} -2 \\ 2 \end{matrix} \right]$

Let $ [b_1]_C = \left[ \begin{matrix} x_1 \\ x_2 \end{matrix} \right]$ and $ [b_2]_C = \left[ \begin{matrix} y_1 \\ y_2 \end{matrix} \right]$

$\left[ \begin{matrix} 1 && -2\\ -5 && 2 \end{matrix} \right] \left[ \begin{matrix} x_1 \\ x_2 \end{matrix} \right] = \left[ \begin{matrix} 7 \\ 5 \end{matrix} \right]$

$\left[ \begin{matrix} 1 && -2\\ -5 && 2 \end{matrix} \right] \left[ \begin{matrix} y_1 \\ y_2 \end{matrix} \right] = \left[ \begin{matrix} -3 \\ -1 \end{matrix} \right]$

Then by using row operations you can find $[b_1]_C$ and $[b_2]_C$.

Thus the change of coordinates matrix from B to C would be $ \left[ \; [b_1]_C \, \, [b_2]_C \; \right]$

Why are you able to find $[b_1]_C$ and $[b_2]_C$ by doing this?

Why is it that you can augment the matrix with both $[b_1]_C$ and $[b_2]_C$ and it works as well for solving both of these?


I just realized why $[b_1]_C$ and $[b_2]_C$ can be obtained from the calculations from above by using this equation learned from an earlier lesson.

$ x = P_B [x]_B $

$x$ can be considered to be $b_1$ while $P_B$ can be seen as $P_C$ which would simply be $C$ as it's the change of coordinates matrix from C to the standard basis in $\Bbb R^n$. And of course $ [x]_B $ can be considered as $[b_1]_C$.


1 Answer 1


Forget about the augmented matrices and let's just focus on why is the change of basis matrix from $B$ to $C$ given by $$\begin{bmatrix} [b_1]_C & [b_2]_C\end{bmatrix}$$

The change of basis transformation is a type of linear transformation. That means that it can be completely determined by its action on a basis. So let's see if this matrix transforms the basis vectors the correct way.

First you need to remember that $[b_1]_B = \begin{bmatrix} 1 \\ 0 \end{bmatrix}$ and $[b_2]_B = \begin{bmatrix} 0 \\ 1 \end{bmatrix}$ (why?).

So what we need is a matrix $A$ such that $$A[b_1]_B = [b_1]_C \\ A[b_2]_B = [b_2]_C$$

That is, we want $A$ to transform this basis from its representation wrt to $B$ to its representation wrt to $C$.

But $A\begin{bmatrix} 1 \\ 0 \end{bmatrix}$ is just the first column of $A$. And likewise $A\begin{bmatrix} 0 \\ 1 \end{bmatrix}$ is just the second column of $A$ (why?). So clearly we just want the two columns of $A$ to be $[b_1]_C$ and $[b_2]_C$.

Do you see why this is the change of basis matrix now?

EDIT: Or are you asking why solving the matrix equation

$$\begin{bmatrix}[c_1]_{\mathcal E} & [c_2]_{\mathcal E}\end{bmatrix}\begin{bmatrix} x_1 \\ x_2 \end{bmatrix} = [b_1]_{\mathcal E}$$

gets you $[b_1]_C$ (where $\mathcal E$ is the standard basis)?

If so then look over the above explanation until you convince yourself that $\begin{bmatrix}[c_1]_\varepsilon & [c_2]_\varepsilon\end{bmatrix}$ is the change of basis matrix from $C$ to $\mathcal E$. Then because change of basis matrices are always invertible, all you're doing is solving

$$\begin{bmatrix} x_1 \\ x_2\end{bmatrix} = \begin{bmatrix}[c_1]_\varepsilon & [c_2]_\varepsilon\end{bmatrix}^{-1}[b_1]_{\mathcal E}$$

where $\begin{bmatrix}[c_1]_\varepsilon & [c_2]_\varepsilon\end{bmatrix}^{-1}$ is the change of basis matrix from $\mathcal E$ to $C$. So you're just changing that basis of $b_1$ from $\mathcal E$ to $C$.

  • $\begingroup$ Thanks for the response. I also just realized something. $\endgroup$ Oct 17, 2015 at 0:01

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