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4.G Change of basis matrices

Up until now, our method for finding the coordinates of a vector in some new basis has been to set up a system of equations and solve for the coefficient variables. In this section, we will describe a different way, using matrices.

In fact, the first Theorem in this section is essentially a special case of Section 4.A, when applied to the identity transformation. One of the uses of the following matrix is to use Corollary 4.9 to convert the basis in the domain or codomain to something else.

Definition 4.48:

For any vector space $V$ , the identity linear transformation is the function $\operatorname{Id}\nolimits:V\to V$ defined by $\operatorname{Id}\nolimits(\vec{x})=\vec{x}$ . If $\mathcal{B}$ and $\mathcal{C}$ are bases for $V$ , then the change of basis matrix from $\mathcal{B}$ to $\mathcal{C}$ is:

{}_{\mathcal{C}}[\operatorname{Id}\nolimits]_{\mathcal{B}}.

The following theorem justifies this name.

Theorem 4.49.

Let $\mathcal{B}=(\vec{b_{1}},\cdots,\vec{b_{n}})$ and $\mathcal{C}=(\vec{c_{1}},\cdots,\vec{c_{n}})$ be two bases of the vector space $V$ over a field $F$ . Let $P:={{}_{\mathcal{C}}[}\operatorname{Id}\nolimits]_{\mathcal{B}}\in% \operatorname{M}_{{n}}({F})$ . Then

i.

The columns of $P$ are the vectors $[\vec{b_{i}}]_{\mathcal{C}}$ ,
ii.

For any $\vec{v}\in V$ we have $[\vec{v}]_{\mathcal{C}}=P[\vec{v}]_{\mathcal{B}}$ ,
iii.

$P^{-1}={{}_{\mathcal{B}}[}\operatorname{Id}\nolimits]_{\mathcal{C}}$ .

So when the target basis $\mathcal{C}$ is the standard basis of $F^{n}$ , then columns of $P$ are simply the vectors in $\mathcal{B}$ written in the standard coordinates.

Proof.

(i) follows directly from Definition 4.4. (ii) is a direct application of Theorem 4.7. By Corollary 4.9, we have $({{}_{\mathcal{C}}[}\operatorname{Id}\nolimits]_{\mathcal{B}})({{}_{\mathcal{B% }}[}\operatorname{Id}\nolimits_{\mathcal{C}})={{}_{\mathcal{C}}[}\operatorname% {Id}\nolimits]_{\mathcal{C}}$ , and the right hand side is the identity matrix. So (iii) follows. ∎

Example 4.50.

Let $\mathcal{C}$ be the standard basis of ${\mathbb{R}}^{3}$ , and $\mathcal{B}=((1,0,0),(2,2,0),(3,3,3))$ . Find the change of basis matrix $P:={{}_{\mathcal{C}}}[\operatorname{Id}\nolimits]_{\mathcal{B}}$ , and hence find $[(1,0,-1)]_{\mathcal{B}}$ .

Solution: By Theorem 4.49, the change of basis matrix from $\mathcal{B}$ to $\mathcal{C}$ has columns equal to the basis vectors in $\mathcal{B}$ written in the standard basis. So

P={{}_{\mathcal{C}}[}\operatorname{Id}\nolimits]_{\mathcal{B}}=\begin{bmatrix}% 1&2&3\\ 0&2&3\\ 0&0&3\end{bmatrix}.

If we set $\vec{v}:=(1,0,-1)$ , then our goal is to find $[\vec{v}]_{\mathcal{B}}$ . According to Theorem 4.49, we have $[\vec{v}]_{\mathcal{B}}=P^{-1}[\vec{v}]_{\mathcal{C}}$ . We compute $P^{-1}$ using methods of MATH105, and then we have:

[\vec{v}]_{\mathcal{B}}=P^{-1}[\vec{v}]_{\mathcal{C}}=\begin{bmatrix}1&-1&0\\ 0&1/2&-1/2\\ 0&0&1/3\end{bmatrix}\begin{bmatrix}1\\ 0\\ -1\end{bmatrix}=\begin{bmatrix}1\\ 1/2\\ -1/3\end{bmatrix}.

So the coordinate matrix is $\begin{bmatrix}1&1/2&-1/3\end{bmatrix}^{T}$ in the basis $\mathcal{B}$ .

Exercise 4.51:

Let $\mathcal{B}=((3,-1),(-2,1))$ be a basis of ${\mathbb{R}}^{2}$ , and $\mathcal{C}$ the standard basis of ${\mathbb{R}}^{2}$ .

i.

Find the change of basis matrix $P:={{}_{\mathcal{C}}[}\operatorname{Id}\nolimits]_{\mathcal{B}}$ from $\mathcal{B}$ to $\mathcal{C}$ .
ii.

Find the inverse of $P$ , using the formula for the inverse of a $2\times 2$ matrix.
iii.

Compute ${{}_{\mathcal{B}}[}\operatorname{Id}\nolimits]_{\mathcal{C}}$ by finding $[(1,0)]_{\mathcal{B}}$ and $[(0,1)]_{\mathcal{B}}$ .
iv.

Verify Theorem 4.49(iii) by comparing your answers to (ii) and (iii) above.
v.

For $\vec{v}=(2,5)$ , compute $[\vec{v}]_{\mathcal{B}}$ by using Theorem 4.49(iii) and $P^{-1}$ .

[End of Exercise]

Theorem 4.52.

Let $\mathcal{B}$ and $\mathcal{C}$ be two bases of a vector space $V$ and assume $T:V\to V$ is a linear transformation. Then the matrices associated to $T$ are related as follows:

{}_{\mathcal{B}}[T]_{\mathcal{B}}=P^{-1}({{}_{\mathcal{C}}[}T]_{\mathcal{C}})P

, where

P:={{}_{\mathcal{C}}[}\operatorname{Id}\nolimits]_{\mathcal{B}}.

Proof.

This follows directly from Corollary 4.9. ∎