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4.7 Change of Basis: Class Notes

Coordinate Maps

Note: Recall that $(𝐯{)}_B$ is the coordinate vector for a vector $𝐯$ relative to a basis $B$. When considered as a column vector, this author uses the notation $[𝐯{]}_B$.

Definition

If $S=\{{𝐯}_1,{𝐯}_2,\dots ,{𝐯}_n\}$ is a basis for a finite-dimensional vector space $V$, and if $(𝐯{)}_S=(c_1,c_2,\dots ,c_n)$ is the coordinate vector of $𝐯$ relative to $S$, then the mapping $𝐯\to (𝐯{)}_S$ is the coordinate map relative to $𝐒$ from $V$ to $R^n$.

The standard basis for $M_{22}$ is

$$S=\{\left[\begin{array}{cc}1& 0\\ 0& 0\end{array}\right],\left[\begin{array}{cc}0& 1\\ 0& 0\end{array}\right],\left[\begin{array}{cc}0& 0\\ 1& 0\end{array}\right],\left[\begin{array}{cc}0& 0\\ 0& 1\end{array}\right]\}.$$

Relative to this basis, the coordinate vector for $A=\left[\begin{array}{cc}6& 2\\ 5& 3\end{array}\right]$ is $(A{)}_S=(6,2,5,3)$. But as we saw in 4.5, the coordinate vector relative to

$$B=\{\left[\begin{array}{cc}1& 0\\ 1& 0\end{array}\right],\left[\begin{array}{cc}1& 1\\ 0& 0\end{array}\right],\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right],\left[\begin{array}{cc}0& 0\\ 1& 0\end{array}\right]\}$$

is $(A{)}_B=(1,2,3,4)$. The coefficients change because the basis is different.

This produces a mapping from $M_{22}$ into $R^4$:

The matrix $A=\left[\begin{array}{cc}6& 2\\ 5& 3\end{array}\right]$ in $M_{22}$ is sent to two different points of $R^4$: using $S$ it goes to $(A{)}_S=(6,2,5,3)$, and using $B$ it goes to $(A{)}_B=(1,2,3,4)$.

The goal of this section is to provide a link between $(A{)}_S$ and $(A{)}_B$, or more generally, between $(\overrightarrow{v}{)}_B$ and $(\overrightarrow{v}{)}_{B\prime }$.

Transition Matrices

To transition from one basis $B$ to another basis $B\prime$ for the same vector space, we use a transition matrix, denoted by $P_{B\to B\prime }$.

In 4.5, we found $(𝐯{)}_B$ for $𝐯=(-1,7,2)$ with respect to the basis $B=\{{𝐯}_1,{𝐯}_2,{𝐯}_3\}$, where ${𝐯}_1=(2,-1,-1)$, ${𝐯}_2=(-2,1,2)$, and ${𝐯}_3=(3,5,4)$ by row reducing an augmented matrix. Here we achieve the same result by using a transition matrix from $S$, the standard basis for $R^3$, to $B$.

In what follows, $B$ will take the place of $S$, and $B\prime$ will take the place of $B$, to match the definition below.

To find a relationship between $B$ and $B\prime$, we start by finding a linear combination of vectors in $B\prime$ that map to vectors in $B$:

$$a\left[\begin{array}{c}2\\ -1\\ -1\end{array}\right]+b\left[\begin{array}{c}-2\\ 1\\ 2\end{array}\right]+c\left[\begin{array}{c}3\\ 5\\ 4\end{array}\right]=\left[\begin{array}{c}1\\ 0\\ 0\end{array}\right](\text{and likewise for }\left[\begin{array}{c}0\\ 1\\ 0\end{array}\right],\left[\begin{array}{c}0\\ 0\\ 1\end{array}\right])$$

This is 3 systems that we can solve simultaneously. The columns on the left are the vectors in $B\prime$, and the columns on the right are the vectors in $B$:

$${\displaystyle \left[\begin{array}{cccccc}\hfill {\displaystyle 2}\hfill & \hfill {\displaystyle -2}\hfill & \hfill {\displaystyle 3}\hfill & \hfill {\displaystyle 1}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 0}\hfill \\ \hfill {\displaystyle -1}\hfill & \hfill {\displaystyle 1}\hfill & \hfill {\displaystyle 5}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 1}\hfill & \hfill {\displaystyle 0}\hfill \\ \hfill {\displaystyle -1}\hfill & \hfill {\displaystyle 2}\hfill & \hfill {\displaystyle 4}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 1}\hfill \end{array}\right]⟶\left[\begin{array}{cccccc}\hfill {\displaystyle 1}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle \frac{6}{13}}\hfill & \hfill {\displaystyle -\frac{14}{13}}\hfill & \hfill {\displaystyle 1}\hfill \\ \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 1}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle \frac{1}{13}}\hfill & \hfill {\displaystyle -\frac{11}{13}}\hfill & \hfill {\displaystyle 1}\hfill \\ \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 1}\hfill & \hfill {\displaystyle \frac{1}{13}}\hfill & \hfill {\displaystyle \frac{2}{13}}\hfill & \hfill {\displaystyle 0}\hfill \end{array}\right]}$$

The three columns on the right give the combinations of the vectors in $B\prime$ that produce ${\overrightarrow{e}}_1$, ${\overrightarrow{e}}_2$, and ${\overrightarrow{e}}_3$, in that order.

The resulting matrix on the right maps a vector relative to $B$ to a vector relative to $B\prime$.

Consider $\overrightarrow{v}=\left[\begin{array}{c}-1\\ 7\\ 2\end{array}\right]$. This is $[\overrightarrow{v}{]}_B$. Now:

$${\displaystyle \left[\begin{array}{ccc}\frac{6}{13}& -\frac{14}{13}& 1\\ \frac{1}{13}& -\frac{11}{13}& 1\\ \frac{1}{13}& \frac{2}{13}& 0\end{array}\right]\left[\begin{array}{c}-1\\ 7\\ 2\end{array}\right]=\left[\begin{array}{c}-6\\ -4\\ 1\end{array}\right]=[\overrightarrow{v}{]}_{B\prime }}$$

These are the coordinates of $\overrightarrow{v}$ relative to $B\prime$. That is,

$$-6\left[\begin{array}{c}2\\ -1\\ -1\end{array}\right]-4\left[\begin{array}{c}-2\\ 1\\ 2\end{array}\right]+1\left[\begin{array}{c}3\\ 5\\ 4\end{array}\right]=\left[\begin{array}{c}-1\\ 7\\ 2\end{array}\right].$$

Thus,

$${\displaystyle \left[\begin{array}{c}-6\\ -4\\ 1\end{array}\right]=[\overrightarrow{v}{]}_{B\prime }=\left[\begin{array}{ccc}\frac{6}{13}& -\frac{14}{13}& 1\\ \frac{1}{13}& -\frac{11}{13}& 1\\ \frac{1}{13}& \frac{2}{13}& 0\end{array}\right]\left[\begin{array}{c}-1\\ 7\\ 2\end{array}\right]=P_{B\to B\prime }[\overrightarrow{v}{]}_B}$$

Definition

If we change the basis for a vector space $V$ from an old basis $B=\{{𝐮}_1,{𝐮}_2,\dots ,{𝐮}_n\}$ to a new basis $B\prime =\{{𝐮\prime }_1,{𝐮\prime }_2,\dots ,{𝐮\prime }_n\}$, then $[{𝐮}_i{]}_{B\prime }$ are the coordinate vectors for the old basis vectors relative to the new basis. The transition matrix from $𝐁$ to $𝐁\prime$ is the matrix having these vectors as its columns and is written [by this author] as $P_{B\to B\prime }=[[{𝐮}_1{]}_{B\prime }|[{𝐮}_2{]}_{B\prime }|\dots |[{𝐮}_n{]}_{B\prime }]$. Note that this is a partitioned matrix.

We use this as follows: For every vector $𝐯$ in $V$,

$$[𝐯{]}_{B\prime }=P_{B\to B\prime }[𝐯{]}_B$$

#3

Consider the bases $B=\{{𝐮}_1,{𝐮}_2,{𝐮}_3\}$ and $B\prime =\{{𝐮\prime }_1,{𝐮\prime }_2,{𝐮\prime }_3\}$ for $R^3$, where

a. Find the transition matrix $B$ to $B\prime$.
b. Compute the coordinate vector $[𝐰{]}_B$, where $𝐰=\left[\begin{array}{c}-5\\ 8\\ -5\end{array}\right]$ and use the transition matrix in part (a) to compute $[𝐰{]}_{B\prime }$.
c. Check your work by computing $[𝐰{]}_{B\prime }$ directly.

Solution

a. Put the new basis vectors on the left and the old basis vectors on the right, and row reduce until the identity $I$ is on the left; what remains on the right is $P_{B\to B\prime }$:

$${\displaystyle \left[\begin{array}{cccccc}\hfill {\displaystyle 3}\hfill & \hfill {\displaystyle 1}\hfill & \hfill {\displaystyle -1}\hfill & \hfill {\displaystyle 2}\hfill & \hfill {\displaystyle 2}\hfill & \hfill {\displaystyle 1}\hfill \\ \hfill {\displaystyle 1}\hfill & \hfill {\displaystyle 1}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 1}\hfill & \hfill {\displaystyle -1}\hfill & \hfill {\displaystyle 2}\hfill \\ \hfill {\displaystyle -5}\hfill & \hfill {\displaystyle -3}\hfill & \hfill {\displaystyle 2}\hfill & \hfill {\displaystyle 1}\hfill & \hfill {\displaystyle 1}\hfill & \hfill {\displaystyle 1}\hfill \end{array}\right]⟶\left[\begin{array}{cccccc}\hfill {\displaystyle 1}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 3}\hfill & \hfill {\displaystyle 2}\hfill & \hfill {\displaystyle \frac{5}{2}}\hfill \\ \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 1}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle -2}\hfill & \hfill {\displaystyle -3}\hfill & \hfill {\displaystyle -\frac{1}{2}}\hfill \\ \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 1}\hfill & \hfill {\displaystyle 5}\hfill & \hfill {\displaystyle 1}\hfill & \hfill {\displaystyle 6}\hfill \end{array}\right]}$$

b.

$$\left[\begin{array}{cccc}\hfill {\displaystyle 2}\hfill & \hfill {\displaystyle 2}\hfill & \hfill {\displaystyle 1}\hfill & \hfill {\displaystyle -5}\hfill \\ \hfill {\displaystyle 1}\hfill & \hfill {\displaystyle -1}\hfill & \hfill {\displaystyle 2}\hfill & \hfill {\displaystyle 8}\hfill \\ \hfill {\displaystyle 1}\hfill & \hfill {\displaystyle 1}\hfill & \hfill {\displaystyle 1}\hfill & \hfill {\displaystyle -5}\hfill \end{array}\right]⟶\left[\begin{array}{cccc}\hfill {\displaystyle 1}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 9}\hfill \\ \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 1}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle -9}\hfill \\ \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 1}\hfill & \hfill {\displaystyle -5}\hfill \end{array}\right]$$

$${\displaystyle [\overrightarrow{w}{]}_B=\left[\begin{array}{c}9\\ -9\\ -5\end{array}\right],\text{so}[\overrightarrow{w}{]}_{B\prime }=\left[\begin{array}{ccc}3& 2& \frac{5}{2}\\ -2& -3& -\frac{1}{2}\\ 5& 1& 6\end{array}\right]\left[\begin{array}{c}9\\ -9\\ -5\end{array}\right]}$$

c.

$${\displaystyle \left[\begin{array}{cccc}\hfill {\displaystyle 3}\hfill & \hfill {\displaystyle 1}\hfill & \hfill {\displaystyle -1}\hfill & \hfill {\displaystyle -5}\hfill \\ \hfill {\displaystyle 1}\hfill & \hfill {\displaystyle 1}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 8}\hfill \\ \hfill {\displaystyle -5}\hfill & \hfill {\displaystyle -3}\hfill & \hfill {\displaystyle 2}\hfill & \hfill {\displaystyle -5}\hfill \end{array}\right]\text{yields}[\overrightarrow{w}{]}_{B\prime }=\left[\begin{array}{c}-\frac{7}{2}\\ \frac{23}{2}\\ 6\end{array}\right]}$$

Why the Transition Matrix Works

This process is generalized in the $2\times 2$ case thus: Let $B=\{{\overrightarrow{u}}_1,{\overrightarrow{u}}_2\}$ be called the old basis and $B\prime =\{{\overrightarrow{u}\prime }_1,{\overrightarrow{u}\prime }_2\}$ be a new basis.

First, find the coordinates that map $B\prime$ to $B$. That is, when applied to vectors in $B\prime$, they return vectors in $B$. In the $2\times 2$ case:

$$\begin{array}{cc}\hfill {\displaystyle {\overrightarrow{u}}_1}& {\displaystyle =a{\overrightarrow{u}\prime }_1+b{\overrightarrow{u}\prime }_2}\hfill \\ \hfill {\displaystyle {\overrightarrow{u}}_2}& {\displaystyle =c{\overrightarrow{u}\prime }_1+d{\overrightarrow{u}\prime }_2}\hfill \end{array}$$

Now let $\overrightarrow{v}$ be a fixed vector in $V$ (for which $B$ and $B\prime$ are bases). In terms of $B$,

$$\begin{array}{cc}\hfill {\displaystyle \overrightarrow{v}}& {\displaystyle =k_1{\overrightarrow{u}}_1+k_2{\overrightarrow{u}}_2}\hfill \\ \hfill {\displaystyle \overrightarrow{v}}& {\displaystyle =k_1(a{\overrightarrow{u}\prime }_1+b{\overrightarrow{u}\prime }_2)+k_2(c{\overrightarrow{u}\prime }_1+d{\overrightarrow{u}\prime }_2)}\hfill \\ \hfill {\displaystyle \overrightarrow{v}}& {\displaystyle =(k_{1}a+k_{2}c){\overrightarrow{u}\prime }_1+(k_{1}b+k_{2}d){\overrightarrow{u}\prime }_2}\hfill \end{array}$$

So the new coordinate vector for $\overrightarrow{v}$ relative to $B\prime$ is

$$\begin{array}{cc}\hfill {\displaystyle [\overrightarrow{v}{]}_{B\prime }}& {\displaystyle =\left[\begin{array}{c}{k}_{1}a+k_{2}c\\ k_{1}b+k_{2}d\end{array}\right]=\left[\begin{array}{cc}a& c\\ b& d\end{array}\right]\left[\begin{array}{c}{k}_1\\ k_2\end{array}\right]}\hfill \\ \hfill {\displaystyle }& {\displaystyle =\left[\begin{array}{cc}a& c\\ b& d\end{array}\right][\overrightarrow{v}{]}_B}\hfill \end{array}$$

The first column $\left[\begin{array}{c}a\\ b\end{array}\right]$ is $[{\overrightarrow{u}}_1{]}_{B\prime }$, and the second column $\left[\begin{array}{c}c\\ d\end{array}\right]$ is $[{\overrightarrow{u}}_2{]}_{B\prime }$. This matrix transitions from $B$ to $B\prime$!

Theorem 4.7.1

If $P$ is the transition matrix from a basis $B$ to a basis $B\prime$ for a finite-dimensional vector space $V$, then $P$ is invertible, and $P^{-1}$ is the transition matrix from $B\prime$ to $B$.

#13

If $P$ is the transition matrix from a basis $B\prime$ to a basis $B$, and $Q$ is the transition matrix from $B$ to a basis $C$, what is the transition matrix from $B\prime$ to $C$? What is the transition matrix from $C$ to $B\prime$?

Solution

Three bases $B\prime$, $B$, $C$ of $V$. $P$ goes from $B\prime$ to $B$ and $Q$ goes from $B$ to $C$. The combined arrow from $B\prime$ to $C$ is $QP=P_{B\prime \to C}$. The arrow back from $C$ to $B\prime$ is $(QP{)}^{-1}$.

Going from $B\prime$ to $B$ uses $P$, and then going from $B$ to $C$ uses $Q$, so $QP=P_{B\prime \to C}$. Going back the other way, $(QP{)}^{-1}=P^{-1}{Q}^{-1}$.