Change-of-basis matrix

If we have a vector space V and two bases of it, we can write one vector of the space as a combination of the vectors from either basis. A change-of-basis matrix then helps us convert one representation into the other.

Motivation

Let V be an n-dimensional vector space with two different bases E = {e₁, …, e_n} and F = {f₁, …, f_n}. Now choose a vector x from V. Since x is in V and both E and F are bases of this space, there must be coefficients a₁, …, a_n such that:

$$\mathbf{x}=a_1\cdot {\mathbf{e}}_1+\dots +a_n\cdot {\mathbf{e}}_n$$

and at the same time there must be coefficients b₁, …, b_n such that:

$$\mathbf{x}=b_1\cdot {\mathbf{f}}_1+\dots +b_n\cdot {\mathbf{f}}_n$$

In the first case we express the vector x using the basis E, in the second case using the basis F. We use two different bases to express x, so we get two different sets of coefficients a_i and b_i. This raises a question: if we know the coefficients b_i, is there a way to get from them the coefficients a_i, that is, the representation with respect to the other basis?

Yes, and this is what a change-of-basis matrix is for.

How to find the change-of-basis matrix

We keep the space V and the two bases E = {e₁, …, e_n} and F = {f₁, …, f_n}. Since E is a basis of V, all the vectors f₁, …, f_n can certainly be written as linear combinations of vectors from the basis E. The vectors f₁, …, f_n are part of the basis F, but they are also ordinary elements of the space V, so we must have:

$$\begin{array}{rcl}{\mathbf{f}}_1& =& a_{11}\cdot {\mathbf{e}}_1+a_{21}\cdot {\mathbf{e}}_2+\dots +a_{n1}\cdot {\mathbf{e}}_n\\ {\mathbf{f}}_2& =& a_{12}\cdot {\mathbf{e}}_1+a_{22}\cdot {\mathbf{e}}_2+\dots +a_{n2}\cdot {\mathbf{e}}_n\\ & \dots & \\ {\mathbf{f}}_n& =& a_{1n}\cdot {\mathbf{e}}_1+a_{2n}\cdot {\mathbf{e}}_2+\dots +a_{nn}\cdot {\mathbf{e}}_n\end{array}$$

Now we rewrite these equations as a matrix by putting the coefficients of the vector $\mathbf{f}\mathrm{\_}1$, that is a_i1, in the first column, the coefficients of the vector $\mathbf{f}\mathrm{\_}2$, that is a_i2, in the second column, and so on. We get the matrix:

$$P=\left(\begin{array}{cccc}{a}_{11}& a_{12}& \dots & a_{1n}\\ a_{21}& a_{22}& \dots & a_{2n}\\ \dots \\ a_{n1}& a_{n2}& \dots & a_{nn}\end{array}\right)$$

Such a matrix is called the change-of-basis matrix from the basis E to the basis F. If we have a vector x = (x₁, …, x_n) from V whose coefficients with respect to the basis F are b₁, …, b_n, then we get the coefficients a₁, …, a_n with respect to the basis E from

$$\left(\begin{array}{c}{a}_1\\ ⋮\\ a_n\end{array}\right)=P\cdot \left(\begin{array}{c}{b}_1\\ ⋮\\ b_n\end{array}\right)$$

Example

We will work in the vector space ℝ³. Take the standard basis as one basis, E = {(1, 0, 0), (0, 1, 0), (0, 0, 1)}, and take F = {(2, 0, 0), (3, 2, 0), (1, 5, 4)} as the other. The change-of-basis matrix from the basis E to the basis F is

$$P=\left(\begin{array}{ccc}2& 3& 1\\ 0& 2& 5\\ 0& 0& 4\end{array}\right)$$

Now choose any vector x ∈ ℝ³, for example x = (12, 41, 28). With respect to the basis F it has the coefficients b₁ = −2, b₂ = 3, b₃ = 7. So we carry out the matrix multiplication:

$$\left(\begin{array}{ccc}2& 3& 1\\ 0& 2& 5\\ 0& 0& 4\end{array}\right)\cdot \left(\begin{array}{c}-2\\ 3\\ 7\end{array}\right)=\left(\begin{array}{c}12\\ 41\\ 28\end{array}\right)$$

With respect to the basis E, the vector x has the coefficients a₁ = 12, a₂ = 41, a₃ = 28. So:

$$(12,41,28)=12\cdot (1,0,0)+41\cdot (0,1,0)+28\cdot (0,0,1)$$