Given two basis, find the transformation matrix from one to another

Question

I have these two basis of $M^R_{2x2}$:

$C= (\begin{pmatrix}1 & 0 \\ 0 & 0\end{pmatrix}, \begin{pmatrix}0 & 1 \\ 0 & 0\end{pmatrix}, \begin{pmatrix}0 & 0 \\ 1 & 0\end{pmatrix}, \begin{pmatrix}0 & 0 \\ 0 & 1\end{pmatrix} )$

$B= (\begin{pmatrix}1 & 1 \\ 0 & 0\end{pmatrix}, \begin{pmatrix}0 & 1 \\ 1 & 0\end{pmatrix}, \begin{pmatrix}0 & 0 \\ 1 & 1\end{pmatrix}, \begin{pmatrix}0 & 0 \\ 0 & 1\end{pmatrix} )$

And I need to find: $$M_C^B, M_B^C$$

Now, I'm pretty now sure what is the process of finding the transformation matrix from one vector to another.

Can someone please help me understand it?

Answer 1

Hint:

Let $C_1,C_2,C_3,C_4$ be the elements of the canonical basis $C$, than you can represent any matrix $X$ in $M(2,\mathbb{R})$ as a linear combination $$ X=x_1C_1+x_2C_2+x_3C_3+x_4C_4 $$ that is a vector of components $(x_1,x_2,x_3,x_4)^T$ with respect to the basis $C$.

For the elements of the new basis $B$ you have: $$ B_1=C_1+C_2=(1,1,0,0)^T \quad B_2=C_2+C_3=(0,1,1,0)^T $$ $$ B_3=C_3+C_4=(0,0,1,1)^T \quad B_4=C_4=(0,0,0,1)^T $$

So the matrix:

$$ M= \begin{bmatrix} 1&0&0&0\\ 1&1&0&0\\ 0&1&1&0\\ 0&0&1&1 \end{bmatrix} $$ and its inverse $M^{-1}$ represent the transformations between the basis $B$ and $C$.

Answer 2

Consider the coefficient vector $$ \begin{pmatrix}1 & 0 & 0 & 0\end{pmatrix}, $$ in $B$, i.e., 1 times the first element of $B$, plus 0 times the second, and so on. That represents the matrix $$ \begin{pmatrix}1 & 1 \\ 0 & 0\end{pmatrix}, $$ right?

That's easily expressed in the $C$ basis as $1$ times the first element plus one times the second, i.e., in vector form, it's $$ \begin{pmatrix}1 \\ 1 \\ 0 \\ 0\end{pmatrix}. $$

So to find $M^B_C$, you need a $4 \times 4$ matrix that sends $$ \begin{pmatrix}1 \\ 0 \\ 0 \\ 0\end{pmatrix} \to \begin{pmatrix}1 \\ 1 \\ 0 \\ 0\end{pmatrix}. $$ That means that the first column of $M^B_C$ must be $$ \begin{pmatrix}1 \\ 1 \\ 0 \\ 0\end{pmatrix}. $$ Repeat with the other three items in the $B$ basis to find the other three columns.

Going the other way might seem harder, but it's not much harder; alternatively, you can invert the matrix $M^B_C$ to get $M^C_B$, but since I'm rotten at matrix inversion, I'd probably get that wrong. :)