How to find a matrix that transforms vectors from basis $B$ into vectors from basis $C$?

Question

$$B=\left\{\begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix}, \begin{bmatrix} 3 \\ 2 \\ 1 \end{bmatrix}, \begin{bmatrix} 1 \\ 1 \\ 0 \end{bmatrix}\right\} \quad C=\left\{\begin{bmatrix} 4 \\ 5 \\ 6 \end{bmatrix}, \begin{bmatrix} 2 \\ 4 \\ 7 \end{bmatrix}, \begin{bmatrix} 6 \\ 7 \\ 8 \end{bmatrix}\right\} $$

I need to find such a matrix $A=\begin{bmatrix} a & b & c \\ d & e & f \\ g & h & i \\ \end{bmatrix}$ that transforms vectors from $B$ into vectors from $C$, so for example:

$$\begin{bmatrix} a & b & c \\ d & e & f \\ g & h & i \\ \end{bmatrix} \begin{bmatrix} 1 \\ 2 \\ 3 \\ \end{bmatrix}=\begin{bmatrix} 4 \\ 5 \\ 6 \\ \end{bmatrix}\text{ etc...}$$ I want to understand the general idea of solving such tasks, so please don't give away the answer!

Answer 1

Consider the following so called commutative diagram: $$\require{AMScd}\begin{CD} \mathbb R^3 @>A>> \mathbb R^3\\ @A(b_1 b_2 b_3)AA @AA(c_1 c_2 c_3)A \\ B @>>I> C \end{CD}$$ If you have a vector with respect to basis $B$, such as (1,0,0), this vector is actually $\mathbf b_1$, the first vector in the basis of $B$. More generally, (x,y,z) with respect to $B$ is $x\mathbf b_1+y\mathbf b_2+z\mathbf b_3$. It means that we can convert a vector with respect to $B$ to the standard basis of $\mathbb R^3$ by simply putting the basis vectors of $B$ as columns in a matrix.

In your case, you're interested in the conversion from $B$ to $C$, which is effectively the identity matrix $I$. That is, for instance (1,0,0) with respect to $B$, which is $\mathbf b_1$, must be mapped to (1,0,0) with respect to $C$, which is $\mathbf c_1$.

To find the corresponding matrix $A$ with respect to the standard basis, we need to follow the arrows, and in reverse where necessary: $$A = \begin{pmatrix}\mathbf c_1&\mathbf c_2&\mathbf c_3\end{pmatrix} \cdot I \cdot\begin{pmatrix}\mathbf b_1&\mathbf b_2&\mathbf b_3\end{pmatrix}^{-1} $$ Reading from right to left, we start with a vector in the standard basis, convert it to $B$ by the corresponding inverse, multiply by $I$, and convert from $C$ to the standard basis.

Answer 2

Here's a good way to start: let $$ E = \Bigg\{\begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix}, \begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix}, \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix}\Bigg\} $$ and ask "Can I find a transformation taking the $E$ vectors to the $B$ vectors?" The answer is "sure, just put the $B$ vectors into the columns of a 3x3 matrix."

You should do this and convince yourself that this matrix, which I'll call $M_{E\to B}$does in fact take the $E$ vectors to the corresponding $B$ vectors.

Now...what matrix would do the opposite, would take the $B$ vectors to the $E$ vectors?

It's $M_{E\to B}^{-1}$, of course! So we now know that $$ M_{B \to E} = M_{E \to B}^{-1}. $$

Now you can do the same thing and find a matrix that takes the $E$ vectors to the $C$ vectors, right?

Now what happens if you multiply the $E$ vectors by the matrix $$ Q = M_{E\to B}M_{C \to E}? $$

If you take it one step at a time, you'll see that $c_1$ gets send to $e_1$, which then (by the first matrix) gets sent to $b_1$.

Try this out on a $2 \times 2$ example to convince yourself it works --- it's a lot easier to invert a $2 \times 2$ matrix!