Examples

Matrix in basis B’ acting on vector in basis B’

Matrix written in ${\mathcal{B}}^{\prime }$ acting on vector written in ${\mathcal{B}}^{\prime }$

Problem: The set of vectors

$${\mathcal{B}}^{\prime }=\left\{{\mathbf{f}}_1=\left(\begin{array}{c}1\\ -1\end{array}\right),{\mathbf{f}}_2=\left(\begin{array}{c}1\\ 1\end{array}\right)\right\}$$

is a basis of ${\mathbb{R}}^2$. (These vectors are independent and there are two of them, so they span all of ${\mathbb{R}}^2$.)

Consider the matrix $A=\left(\begin{array}{cc}-2& 2\\ 0& 1\end{array}\right)$ and the vector $\mathbf{x}=\left(\begin{array}{c}2\\ 4\end{array}\right)$. Compute the array $[\mathbf{x}{]}_{{\mathcal{B}}^{\prime }}$ that represents $\mathbf{x}$ in the new basis ${\mathcal{B}}^{\prime }$, and then compute $A_{{\mathcal{B}}^{\prime }}$. Calculate the image $A\mathbf{x}$ and compare this to the image $A_{{\mathcal{B}}^{\prime }}[\mathbf{x}{]}_{{\mathcal{B}}^{\prime }}$. They should agree!

Solution: First we find $[\mathbf{x}{]}_{{\mathcal{B}}^{\prime }}$. We must solve the system

$${\nu }_1\left(\begin{array}{c}1\\ -1\end{array}\right)+{\nu }_2\left(\begin{array}{c}1\\ 1\end{array}\right)=\left(\begin{array}{c}2\\ 4\end{array}\right)\text{so:}\begin{array}{rl}{\nu }_1+{\nu }_2& =2,\\ -{\nu }_1+{\nu }_2& =4.\end{array}$$

The solution is ${\nu }_2=3,{\nu }_1=-1$. Therefore $[\mathbf{x}{]}_{{\mathcal{B}}^{\prime }}=\left(\begin{array}{c}-1\\ 3\end{array}\right)$.

Next we seek $A_{{\mathcal{B}}^{\prime }}$. The columns of $A_{{\mathcal{B}}^{\prime }}$ are given by the images under $A$ of the items ${\mathbf{f}}_1$ and ${\mathbf{f}}_2$. So we compute:

$$A{\mathbf{f}}_1=\left(\begin{array}{cc}-2& 2\\ 0& 1\end{array}\right)\left(\begin{array}{c}1\\ -1\end{array}\right)=\left(\begin{array}{c}-4\\ -1\end{array}\right),A{\mathbf{f}}_2=\left(\begin{array}{cc}-2& 2\\ 0& 1\end{array}\right)\left(\begin{array}{c}1\\ 1\end{array}\right)=\left(\begin{array}{c}0\\ 1\end{array}\right).$$

Therefore we have $A_{{\mathcal{B}}^{\prime }}=\left(\begin{array}{cc}-4& 0\\ -1& 1\end{array}\right)$. This matrix sends ${\mathbf{f}}_1={\left(\begin{array}{c}1\\ 0\end{array}\right)}_{{\mathcal{B}}^{\prime }}$ to the vector $\left(\begin{array}{c}-4\\ -1\end{array}\right)$ and it sends ${\mathbf{f}}_2={\left(\begin{array}{c}0\\ 1\end{array}\right)}_{{\mathcal{B}}^{\prime }}$ to the vector $\left(\begin{array}{c}0\\ 1\end{array}\right)$.

Finally, observe on the one hand that

$$A_{{\mathcal{B}}^{\prime }}[\mathbf{x}{]}_{{\mathcal{B}}^{\prime }}=\left(\begin{array}{cc}-4& 0\\ -1& 1\end{array}\right)\left(\begin{array}{c}-1\\ 3\end{array}\right)=\left(\begin{array}{c}4\\ 4\end{array}\right),$$

while on the other hand

$$A\mathbf{x}=\left(\begin{array}{cc}-2& 2\\ 0& 1\end{array}\right)\left(\begin{array}{c}2\\ 4\end{array}\right)=\left(\begin{array}{c}-4+8\\ 0+4\end{array}\right)=\left(\begin{array}{c}4\\ 4\end{array}\right).$$

Therefore $A_{{\mathcal{B}}^{\prime }}[\mathbf{x}{]}_{{\mathcal{B}}^{\prime }}=A\mathbf{x}$ as the problem statement had anticipated.

Matrix in basis B’ on input and output via transfer matrices

Finding ${}_{{\mathcal{B}}^{\prime }}{A}_{{\mathcal{B}}^{\prime }}$ using ‘change of basis’ transfer matrices

Problem: Find the transfer matrix of the example in the previous section and use it to calculate ${}_{{\mathcal{B}}^{\prime }}{A}_{{\mathcal{B}}^{\prime }}$.

Solution: We know that

$${}_{\mathcal{B}}{T}_{{\mathcal{B}}^{\prime }}=\left(\begin{array}{cc}{\mathbf{f}}_1& {\mathbf{f}}_2\end{array}\right)=\left(\begin{array}{cc}1& 1\\ -1& 1\end{array}\right).$$

Notice that the matrix $A_{{\mathcal{B}}^{\prime }}$ found in the example is quickly computed as the matrix product:

$$A_{{\mathcal{B}}^{\prime }}=A\cdot {}_{\mathcal{B}}{T}_{{\mathcal{B}}^{\prime }}=\left(\begin{array}{cc}-2& 2\\ 0& 1\end{array}\right)\left(\begin{array}{cc}1& 1\\ -1& 1\end{array}\right)=\left(\begin{array}{cc}-4& 0\\ -1& 1\end{array}\right).$$

To find the matrix ${}_{{\mathcal{B}}^{\prime }}{A}_{{\mathcal{B}}^{\prime }}$, we compute the inverse using the usual 2x2 formula for inverses:

$${}_{{\mathcal{B}}^{\prime }}{T}_{\mathcal{B}}={}_{\mathcal{B}}{T}_{{\mathcal{B}}^{\prime }}^{-1}={\left(\begin{array}{cc}1& 1\\ -1& 1\end{array}\right)}^{-1}=\left(\begin{array}{cc}1/2& -1/2\\ 1/2& 1/2\end{array}\right).$$

Then we can calculate ${}_{{\mathcal{B}}^{\prime }}{A}_{{\mathcal{B}}^{\prime }}$ by taking the product:

$${}_{{\mathcal{B}}^{\prime }}{A}_{{\mathcal{B}}^{\prime }}={}_{{\mathcal{B}}^{\prime }}{T}_{\mathcal{B}}\cdot A_{{\mathcal{B}}^{\prime }}=\left(\begin{array}{cc}1/2& -1/2\\ 1/2& 1/2\end{array}\right)\left(\begin{array}{cc}-4& 0\\ -1& 1\end{array}\right)=\left(\begin{array}{cc}\frac{-3}{2}& \frac{-1}{2}\\ \frac{-5}{2}& \frac{1}{2}\end{array}\right).$$

Directly computing change of basis matrix using row reduction

Directly computing change of basis matrix using row reduction

Consider the following two bases:

$$\mathcal{B}=\left\{{\mathbf{b}}_1=\left(\begin{array}{c}7\\ 5\end{array}\right),{\mathbf{b}}_2=\left(\begin{array}{c}-3\\ -1\end{array}\right)\right\},\mathcal{C}=\left\{{\mathbf{c}}_1=\left(\begin{array}{c}1\\ -5\end{array}\right),{\mathbf{c}}_2=\left(\begin{array}{c}-2\\ 2\end{array}\right)\right\}.$$

Problem: Find the transfer matrices ${}_{\mathcal{B}}{T}_{\mathcal{C}}$ and ${}_{\mathcal{C}}{T}_{\mathcal{B}}$ representing change of basis.

Solution: We only need to find one, since the other will be determined as the inverse of the first. Let us start with ${}_{\mathcal{B}}{T}_{\mathcal{C}}$.

Let $B=\left(\begin{array}{cc}{\mathbf{b}}_1& {\mathbf{b}}_2\end{array}\right)$ and $C=\left(\begin{array}{cc}{\mathbf{c}}_1& {\mathbf{c}}_2\end{array}\right)$. Now suppose we have found a matrix $X$ such that $BX=C$. Then $X$ is in fact the transfer matrix ${}_{\mathcal{B}}{T}_{\mathcal{C}}$. Why? Observe that by definition, ${}_{\mathcal{B}}{T}_{\mathcal{C}}\left(\begin{array}{c}1\\ 0\end{array}\right)=[{\mathbf{c}}_1{]}_{\mathcal{B}}=\left(\begin{array}{c}x\\ y\end{array}\right)$ where $x{\mathbf{b}}_1+y{\mathbf{b}}_2={\mathbf{c}}_1$, and similarly ${}_{\mathcal{B}}{T}_{\mathcal{C}}\left(\begin{array}{c}0\\ 1\end{array}\right)=[{\mathbf{c}}_2{]}_{\mathcal{B}}=\left(\begin{array}{c}z\\ w\end{array}\right)$ where $z{\mathbf{b}}_1+w{\mathbf{b}}_2={\mathbf{c}}_2$. But these two equations precisely combine to the matrix equation $BX=C$, where $X=\left(\begin{array}{cc}x& z\\ y& w\end{array}\right)$.

So, we solve the equation $BX=C$ using row reduction on the augmented matrix $\left(\begin{array}{c}B\mid C\end{array}\right)$:

$$\begin{array}{rl}\left(\begin{array}{rr}B& C\end{array}\right)& =\left(\begin{array}{rrrr}7& -3& 1& -2\\ 5& -1& -5& 2\end{array}\right)\\ & \sim \left(\begin{array}{rrrr}7& -3& 1& -2\\ 0& 8/7& -40/7& 24/7\end{array}\right)\\ & \sim \left(\begin{array}{rrrr}1& -3/7& 1/7& -2/7\\ 0& 1& -5& 3\end{array}\right)\\ & \sim \left(\begin{array}{rrrr}1& 0& -2& 1\\ 0& 1& -5& 3\end{array}\right).\end{array}$$

It follows that ${}_{\mathcal{B}}{T}_{\mathcal{C}}=X=\left(\begin{array}{cc}-2& 1\\ -5& 3\end{array}\right)$. By computing the inverse ${}_{\mathcal{C}}{T}_{\mathcal{B}}={}_{\mathcal{B}}{T}_{\mathcal{C}}^{-1}$ using the $2\times 2$ formula, we have ${}_{\mathcal{C}}{T}_{\mathcal{B}}=\left(\begin{array}{cc}-3& 1\\ -5& 2\end{array}\right)$.