2.8b Change of Basis

Change of Basis

Let $\{{\vec{v}}_1,{\vec{v}}_2,\dots ,{\vec{v}}_n\}$ be a V-basis (nonstandard basis) of ${\mathbb{R}}^n$. Let $\vec{x}$ be a vector given in the standard basis. The representation of $\vec{x}$ in the V-basis is
$\vec{z}=V^{-1}\vec{x},$
where $V:=\left[\begin{array}{cccc}{\vec{v}}_1& {\vec{v}}_2& \dots & {\vec{v}}_n\end{array}\right]$ is an invertible square basis matrix.

Intuition

Think of it this way: we can translate a vector $\vec{z}$ in V-basis to $\vec{x}$ in our basis using a matrix $V$ that contains the representation in our basis of what V-basis members see as their standard basis ($\left\{\left[\begin{array}{c}1\\ 0\end{array}\right],\left[\begin{array}{c}0\\ 1\end{array}\right]\right\}$). Suppose V-basis unit vectors ${\vec{v}}_1$ and ${\vec{v}}_2$ can be translated into standard basis using our unit vectors as ${\vec{v}}_1=2{\vec{e}}_1+3{\vec{e}}_2$ and ${\vec{v}}_2={\vec{e}}_1-2{\vec{e}}_2$. Therefore, the V-basis matrix representation in our (standard) basis is $V=\left[\begin{array}{cc}{\vec{v}}_1& {\vec{v}}_2\end{array}\right]=\left[\begin{array}{cc}2& 1\\ 3& -2\end{array}\right]$. Algebraically, we can convert a V-basis vector $\vec{z}=\left[\begin{array}{c}3\\ 2\end{array}\right]=3{\vec{v}}_1+2{\vec{v}}_2$ to our basis as $\vec{x}=3(2{\vec{e}}_1+3{\vec{e}}_2)+2({\vec{e}}_1-2{\vec{e}}_2)=8{\vec{e}}_1+5{\vec{e}}_2=\left[\begin{array}{c}8\\ 5\end{array}\right].$ We can quickly verify this by equivalently solving $\vec{x}=\left[\begin{array}{cc}2& 1\\ 3& -2\end{array}\right]\left[\begin{array}{c}3\\ 2\end{array}\right]=3\left[\begin{array}{c}2\\ 3\end{array}\right]+2\left[\begin{array}{c}1\\ -2\end{array}\right]=\left[\begin{array}{c}8\\ 5\end{array}\right]$. Then, in order to reverse the process, we redo it backwards (by taking the inverse).

Why do we take the inverse to translate to V-basis?

Think of it in the following way: the way people using V-basis see our basis is $V^{-1}=\left[\begin{array}{cc}\frac{2}{7}& \frac{1}{7}\\ \frac{3}{7}& -\frac{2}{7}\end{array}\right]$. For them, ${\vec{e}}_1=\frac{2}{7}{\vec{v}}_1+\frac{3}{7}{\vec{v}}_2=$ and ${\vec{e}}_2=\frac{1}{7}{\vec{v}}_1-\frac{2}{7}{\vec{v}}_2$. We can quickly verify this by running our representations of the V-basis vectors ${\vec{v}}_1$ and ${\vec{v}}_2$ to find out that ${\vec{e}}_1=\left[\begin{array}{c}1\\ 0\end{array}\right]$ and ${\vec{e}}_2=\left[\begin{array}{c}0\\ 1\end{array}\right]$. Therefore, our vector $\vec{x}=\left[\begin{array}{c}8\\ 5\end{array}\right]=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]\left[\begin{array}{c}8\\ 5\end{array}\right]=8{\vec{e}}_1+5{\vec{e}}_2$ can be written in V-basis as $\vec{z}=8\left(\frac{2}{7}{\vec{v}}_1+\frac{3}{7}{\vec{v}}_2\right)+5\left(\frac{1}{7}{\vec{v}}_1+-\frac{2}{7}{\vec{v}}_2\right)=\frac{16}{7}{\vec{v}}_1+\frac{24}{7}{\vec{v}}_2+\frac{5}{7}{\vec{v}}_1-\frac{10}{7}{\vec{v}}_2=3{\vec{v}}_1+2{\vec{v}}_2=\left[\begin{array}{c}3\\ 2\end{array}\right]$. This can also be verified by solving $\vec{z}=\left[\begin{array}{cc}\frac{2}{7}& \frac{1}{7}\\ \frac{3}{7}& -\frac{2}{7}\end{array}\right]\left[\begin{array}{c}8\\ 5\end{array}\right]=8\left[\begin{array}{c}\frac{2}{7}\\ \frac{3}{7}\end{array}\right]+5\left[\begin{array}{c}\frac{1}{7}\\ -\frac{2}{7}\end{array}\right]=\left[\begin{array}{c}3\\ 2\end{array}\right]$.

How can we find the inverse coefficients algebraically?

Consider our main example where ${\vec{v}}_1=2{\vec{e}}_1+3{\vec{e}}_2$ and ${\vec{v}}_2={\vec{e}}_1-2{\vec{e}}_2$. We can solve for the representations of ${\vec{e}}_1$ and ${\vec{e}}_2$ in V-basis such that

$$\{\begin{array}{l}{\vec{e}}_1={\vec{v}}_2+2{\vec{e}}_2\\ {\vec{e}}_1=\frac{1}{2}{\vec{v}}_1-\frac{3}{2}{\vec{e}}_2\end{array}⟺\frac{7}{2}{\vec{e}}_2=\frac{1}{2}{\vec{v}}_1-{\vec{v}}_2⟺\{\begin{array}{l}{\vec{e}}_1=\frac{2}{7}{\vec{v}}_1+\frac{3}{7}{\vec{v}}_2\\ {\vec{e}}_2=\frac{1}{7}{\vec{v}}_1-\frac{2}{7}{\vec{v}}_2\end{array}.$$

We can verify that this is true by substituting for the respective values of ${\vec{v}}_1$ and ${\vec{v}}_2$. As such, our standard basis vectors are represented in V-basis as $\left\{\left[\begin{array}{c}\frac{2}{7}\\ \frac{3}{7}\end{array}\right],\left[\begin{array}{c}\frac{1}{7}\\ -\frac{2}{7}\end{array}\right]\right\}$.

Why does this equal $V^{-1}$?