Change of basis/R^2/Standard and 12,-23/Example

We consider in ${ℝ}^2$ the standard basis,

$𝔲=\left(\begin{array}{c}1\\ 0\end{array}\right),\left(\begin{array}{c}0\\ 1\end{array}\right),$

and the basis

$𝔳=\left(\begin{array}{c}1\\ 2\end{array}\right),\left(\begin{array}{c}-2\\ 3\end{array}\right).$

The basis vectors of $𝔳$ can be expressed directly with the standard basis, namely

$v_1=\left(\begin{array}{c}1\\ 2\end{array}\right)=1\left(\begin{array}{c}1\\ 0\end{array}\right)+2\left(\begin{array}{c}0\\ 1\end{array}\right)\text{ and }{v}_2=\left(\begin{array}{c}-2\\ 3\end{array}\right)=-2\left(\begin{array}{c}1\\ 0\end{array}\right)+3\left(\begin{array}{c}0\\ 1\end{array}\right).$

Therefore, we get immediately

$M_{𝔲}^{𝔳}=\left(\begin{array}{cc}1& -2\\ 2& 3\end{array}\right).$

For example, the vector that has the coordinates $(4,-3)$ with respect to $𝔳$, has the coordinates

$M_{𝔲}^{𝔳}\left(\begin{array}{c}4\\ -3\end{array}\right)=\left(\begin{array}{cc}1& -2\\ 2& 3\end{array}\right)\left(\begin{array}{c}4\\ -3\end{array}\right)=\left(\begin{array}{c}10\\ -1\end{array}\right)$

with respect to the standard basis $𝔲$. The transformation matrix $M_{𝔳}^{𝔲}$ is more difficult to compute. We have to write the standard vectors as linear combinations of $v_1$ and $v_2$. A direct computation (solving two linear systems) yields

$\left(\begin{array}{c}1\\ 0\end{array}\right)=\frac{3}{7}\left(\begin{array}{c}1\\ 2\end{array}\right)-\frac{2}{7}\left(\begin{array}{c}-2\\ 3\end{array}\right)$

and

$\left(\begin{array}{c}0\\ 1\end{array}\right)=\frac{2}{7}\left(\begin{array}{c}1\\ 2\end{array}\right)+\frac{1}{7}\left(\begin{array}{c}-2\\ 3\end{array}\right).$

Hence,

$M_{𝔳}^{𝔲}=\left(\begin{array}{cc}\frac{3}{7}& \frac{2}{7}\\ -\frac{2}{7}& \frac{1}{7}\end{array}\right).$