# Change of Basis

Any vector space has multiple bases, so the question naturally arises: what are the relationships between bases of a vector space? In the first place, there must be the same number of elements in any basis of a vector space. Then, given two bases of a vector space, there is a way to translate vectors in terms of one basis into terms of the other; this is known as **change of basis**.

Change of basis is a technique applied to finite-dimensional vector spaces in order to rewrite vectors in terms of a different set of basis elements. It is useful for many types of matrix computations in linear algebra and can be viewed as a type of linear transformation.

## Definition

Suppose \(v_1, \, v_2, \, \dots, \, v_n\) is a basis of \(V\) and \(u_1, \, u_2, \, \dots, \, u_n\) is another basis of \(V\). Then, by the definition of basis, there is precisely one way to write each \(v_k\) in terms of \(u_i\)'s. Specifically, there are scalars \(a_{i,j}\) \((\)for integers \(1 \le i,j \le n)\) such that

\[\begin{align*} v_1 &= a_{1,1} u_1 + a_{1,2} u_2 + \dots + a_{1,n} u_n \\ &\ \ \vdots \hspace{1.3cm} \vdots \hspace{1.7cm} \vdots\\ v_n &= a_{n,1} u_1 + a_{n,2} u_2 + \dots + a_{n,n} u_n. \end{align*}\]

Then, there is an \(n \times n\) matrix \(M = (a_{i,j})_{i,j}\) that is known as the **change-of-basis matrix**. \(M\) must be invertible because it must be injective (and it is square), by the definition of basis. Any vector \(v = b_1 v_1 + b_2 v_2 + \dots + b_n v_n\) could then be represented in terms of the \(u_i\)'s by means of direct substitution:

\[ \begin{align*} v &= \begin{pmatrix} v_1 & v_2 & \dots & v_n \end{pmatrix} \begin{pmatrix} a_1 \\ a_2 \\ \vdots \\ a_n \end{pmatrix} \\ &= \begin{pmatrix} u_1 & u_2 & \dots & u_n \end{pmatrix} M \begin{pmatrix} a_1 \\ a_2 \\ \vdots \\ a_n \end{pmatrix}. \end{align*} \]

That is, the linear transformation \(T: V \to V\) defined by \(T(v) = Mv\) changes the coordinates from \(v_i\)'s to \(u_i\)'s.

Express the vector \(v = 15\hat{i} - 5\hat{j}\) as a linear combination of the vectors \(\begin{pmatrix} 1 \\ 1 \end{pmatrix}\) and \(\begin{pmatrix} 3 \\ -2 \end{pmatrix}\).

This comes down to finding the scalars \(a\) and \(b\) so that \(\begin{pmatrix} 15 \\ -5 \end{pmatrix} = a\begin{pmatrix} 1 \\ 1 \end{pmatrix} + b\begin{pmatrix} 3 \\ -2 \end{pmatrix},\) which can be written in matrix form as \(\begin{pmatrix} 1 & 3 \\ 1 & -2 \end{pmatrix}\begin{pmatrix} a \\ b \end{pmatrix} = \begin{pmatrix} 15 \\ -5 \end{pmatrix}\). From here, we can use Gauss-Jordan elimination to conclude that \(a = 3\) and \(b = 4\). \(_\square\)

## Rewriting a Linear Transformation

When using coordinates (and therefore matrices) for linear transformations, it is sometimes helpful to consider a linear transformation with a different basis.

Suppose \(T: V \to V\) is a linear transformation with matrix \(N\) in basis \(\mathcal{B}_1 = \{v_1, \, v_2, \, \dots, \, v_n\}\) and \(V\) has another basis \(\mathcal{B}_2 = \{u_1, \, u_2, \, \dots, \, u_n\}\). Then, if \(M\) is the change-of-basis matrix for \(\mathcal{B}_1\) to \(\mathcal{B}_2\) as above, the linear transformation has a different matrix in basis \(\mathcal{B}_2\), namely \[T(u) = MNM^{-1} u.\]

So, some vector \(u \in V\) has a representation with basis \(\mathcal{B}_2\). Then, \(M^{-1}u\) is the same vector whose representation is now with basis \(\mathcal{B}_1\); \(NM^{-1}u\) is the application of \(N\) to a vector expressed in basis \(\mathcal{B}_1\), yielding the output of \(T\) expressed in basis \(\mathcal{B}_1\); and \(MNM^{-1}\) is the output of \(T\) expressed in basis \(\mathcal{B}_2\).