
# 13.2: Change of Basis

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Suppose we have two ordered bases $$S=(v_{1}, \ldots, v_{n} )$$ and $$S'=(v'_{1}, \ldots, v'_{n} )$$ for a vector space $$V$$. (Here $$v_{i}$$ and $$v'_{i}$$ are $$\textit{vectors}$$, not components of vectors in a basis!) Then we may write each $$v'_{i}$$ uniquely as a linear combination of the $$v_{j}$$:

$v'_{j} = \sum_{i} v_{i}p^{i}_{j}\, ,$

or in matrix notation

$$\begin{pmatrix}v'_{1} , v'_{2} , \cdots , v'_{n}\end{pmatrix} = \begin{pmatrix}v_{1} , v_{2} , \cdots , v_{n}\end{pmatrix}\begin{pmatrix}p^{1}_{1} & p^{1}_{2} & \cdots & p^{1}_{n} \\ p^{2}_{1} & p^{2}_{2} && \\ \vdots &&& \vdots \\ p^{n}_{1} && \cdots & p^{n}_{n}\end{pmatrix}$$

Here, the $$p^{i}_{j}$$ are constants, which we can regard as entries of a square matrix $$P=(p^{i}_{j})$$. The matrix $$P$$ must have an inverse, since we can also write each $$v_{i}$$ uniquely as a linear combination of the $$v'_{j}$$:

$v_{j} = \sum_{k} v_{k}q^{k}_{j}.$

Then we can write:

$v_{j} = \sum_{k} \sum_{i} v'_{k}q^{k}_{i}p^{i}_{j}.$

But $$\sum_{i} q^{k}_{i}p^{i}_{j}$$ is the $$k,j$$ entry of the product matrix $$QP$$. Since the expression for $$v_{j}$$ in the basis $$S$$ is $$v_{j}$$ itself, then $$QP$$ maps each $$v_{j}$$ to itself. As a result, each $$v_{j}$$ is an eigenvector for $$QP$$ with eigenvalue $$1$$, so $$QP$$ is the identity, $$\textit{i.e.}$$

$$PQ=QP=I \leftrightarrow Q=P^{-1}\, .$$

The matrix $$P$$ is called a $$\textit{change of basis}$$ matrix. There is a quick and dirty trick to obtain it: Look at the formula above relating the new basis vectors $$v'_{1},v'_{2},\ldots v'_{n}$$ to the old ones $$v_{1},v_{2},\ldots,v_{n}$$. In particular focus on $$v'_{1}$$ for which

$$v'_{1}= \begin{pmatrix}v_{1} , v_{2} , \cdots , v_{n}\end{pmatrix} \begin{pmatrix}p^{1}_{1}\\p^{2}_{1}\\\vdots \\ p^{n}_{1} \end{pmatrix}\, .$$

This says that the first column of the change of basis matrix $$P$$ is really just the components of the vector $$v'_{1}$$ in the basis $$v_{1},v_{2},\ldots,v_{n}$$, so:

$$\textit{The columns of the change of basis matrix are the components of the new basis vectors in terms of the old basis vectors.}$$

Example 120

Suppose $$S'=(v'_{1},v'_{2})$$ is an ordered basis for a vector space $$V$$ and that with respect to some other ordered basis $$S=(v_{1}, v_{2})$$ for $$V$$
$$v'_{1}= \begin{pmatrix} \frac{1}{\sqrt{2}}\\\frac{1}{\sqrt{2}} \end{pmatrix} _{S} \quad \mbox{and} \quad v'_{2}= \begin{pmatrix} \frac{1}{\sqrt{3}}\\-\frac{1}{\sqrt{3}} \end{pmatrix}_{S} \, .$$
This means
$$v'_{1}=\begin{pmatrix}v_{1} , v_{2} \end{pmatrix}\begin{pmatrix} \frac{1}{\sqrt{2}}\\ \frac{1}{\sqrt{2}} \end{pmatrix} =\frac{v_{1}+v_{2}}{\sqrt{2}}\quad\mbox{and}\quad v'_{2}=\begin{pmatrix}v_{1} , v_{2} \end{pmatrix}\begin{pmatrix} \frac{1}{\sqrt{3}}\\-\frac{1}{\sqrt{3}} \end{pmatrix} =\frac{v_{1}-v_{2}}{\sqrt{3}}\, .$$
The change of basis matrix has as its columns just the components of $$v'_{1}$$ and $$v'_{2}$$;
$$P= \begin{pmatrix} \frac{1}{\sqrt{2}}&\frac{1}{\sqrt{3}}\\ \frac{1}{\sqrt{2}}&-\frac{1}{\sqrt{3}} \end{pmatrix}\, .$$

Changing basis changes the matrix of a linear transformation. However, as a map between vector spaces, $$\textit{the linear transformation is the same no matter which basis we use}$$. Linear transformations are the actual objects of study of this book, not matrices; matrices are merely a convenient way of doing computations.

Lets now calculate how the matrix of a linear transformation changes when changing basis. To wit, let $$L \colon V \longrightarrow W$$ with matrix $$M=(m^{i}_{j})$$ in the ordered input and output bases $$S=(v_{1}, \ldots, v_{n} )$$ and $$T=(w_{1},\ldots,w_{m})$$ so

$L(v_{i}) = \sum_{k} w_{k}m^{k}_{i}.$

Now, suppose $$S'=(v'_{1}, \ldots, v'_{n} )$$ and $$T'=(w'_{1},\ldots,w'_{m})$$ are new ordered input and out bases with matrix $$M'=({m'}_{i}^{k})$$. Then

$L(v'_{i})= \sum_{k} w_{k}m'^{k}_{i}\, .$

Let $$P=(p^{i}_{j})$$ be the change of basis matrix from input basis $$S$$ to the basis $$S'$$ and $$Q=(q^{j}_{k})$$ be the change of basis matrix from output basis $$T$$ to the basis $$T'$$. Then:

$L(v'_{j})=L\left(\sum_{i} v_{i} p^{i}_{j}\right) = \sum_{i} L(v_{i})p^{i}_{j} = \sum_{i} \sum_{k} w_{k} m^{k}_{i} p^{i}_{j}.$

Meanwhile, we have:

$L(v'_{i}) = \sum_{k}v_{k}m'^{k}_{i} = \sum_{k} \sum_{j} v_{j} q^{j}_{k}m^{k}_{i}.$

Since the expression for a vector in a basis is unique, then we see that the entries of $$MP$$ are the same as the entries of $$QM'$$. In other words, we see that $$MP = QM' \qquad \textit{or}\qquad M'=Q^{-1}MP.$$

Example 121

Let $$V$$ be the space of polynomials in $$t$$ and degree 2 or less and $$L:V\to \mathbb{R}^{2}$$ where

$$L(1)=\begin{pmatrix}1\\2\end{pmatrix}\, \quad L(t)=\begin{pmatrix}2\\1\end{pmatrix}\, ,\quad L(t^{2})=\begin{pmatrix}3\\3\end{pmatrix}\, .$$

From this information we can immediately read off the matrix $$M$$ of $$L$$ in the bases $$S=(1,t,t^{2})$$ and $$T=(e_{1},e_{2})$$, the standard basis for $$\mathbb{R}^{2}$$,
because

\begin{eqnarray*}\big(L(1),L(t),L(t^{2})\big)&=&(e_{1}+2 e_{2},2e_{1}+e_{2}, 3 e_{1}+3e_{2})\\&=&(e_{1},e_{2})\begin{pmatrix}1&2&3\\2&1&3\end{pmatrix}\, \Rightarrow \, M\ =\
\begin{pmatrix}1&2&3\\2&1&3\end{pmatrix}\, .\end{eqnarray*}

Now suppose we are more interested in the bases

$$S'=(1+t,t+t^{2},1+t^{2})\, , \quad T'=\left(\begin{pmatrix}1\\2\end{pmatrix},\begin{pmatrix}2\\1\end{pmatrix}\right)=:(w_{1}',w_{2}')\, .$$

To compute the new matrix $$M'$$ of $$L$$ we could simply calculate what $$L$$ does the the new input basis vectors in terms of the new output basis vectors:

\begin{eqnarray*}
\big(L(1+t)L(t+t^{2}),L(1+t^{2}))&=&\left(\begin{pmatrix}1\\2\end{pmatrix}+\begin{pmatrix}2\\1\end{pmatrix},
\begin{pmatrix}2\\1\end{pmatrix}+\begin{pmatrix}3\\3\end{pmatrix},\begin{pmatrix}1\\2\end{pmatrix}+\begin{pmatrix}3\\3\end{pmatrix}
\right)\\
&=&(w_{1}+w_{2},w_{1}+2w_{2},2w_{2}+w_{1})\\&=&(w_{1},w_{2})\begin{pmatrix}1&1&2\\1&2&1\end{pmatrix}\, \Rightarrow \, M'=\begin{pmatrix}1&1&2\\1&2&1\end{pmatrix}\, .
\end{eqnarray*}
Alternatively we could calculate the change of basis matrices $$P$$ and $$Q$$ by noting that

$$(1+t,t+t^{2},1+t^{2})=(1,t,t^{2})\begin{pmatrix}1&0&1\\1&1&0\\0&1&1\end{pmatrix}\, \Rightarrow\, P=\begin{pmatrix}1&0&1\\1&1&0\\0&1&1\end{pmatrix}$$

and

$$(w_{1},w_{2})=(e_{1}+2e_{2},2e_{1}+e_{2})=(e_{1},e_{1})\begin{pmatrix}1&2\\2&1\end{pmatrix}\, \Rightarrow\, Q=\begin{pmatrix}1&2\\2&1\end{pmatrix}\, .$$

Hence

$$M'=Q^{-1}MP = -\frac{1}{3}\begin{pmatrix}1&-2\\-2&1\end{pmatrix}\begin{pmatrix}1&2&3\\2&1&3\end{pmatrix} \begin{pmatrix}1&0&1\\1&1&0\\0&1&1\end{pmatrix}=\begin{pmatrix}1&1&2\\1&2&1\end{pmatrix}\, .$$

Notice that the change of basis matrices $$P$$ and $$Q$$ are both square and invertible. Also, since we really wanted $$Q^{-1}$$, it is more efficient to try and write $$(e_{1},e_{2})$$ in terms of $$(w_{1},w_{2})$$ which would yield directly $$Q^{-1}$$. Alternatively, one can check that $$MP=QM'$$.