There are times when expressing vectors in a different basis is desirable. For example, in a high-dimensional vector space, if we have an ordered basis such that the tuple representations of the vectors that we have at hand have very few nonzero components, we can potentially save memory using a sparse representation of these tuples. We now want look at a systematic way to convert the tuple representation of a vector in a given ordered basis to the tuple representation of the vector in another given ordered basis. We first consider an example.

Let \(P_1\) denote the vector space of linear polynomials in \(x\) with real coefficients. Let \(\Gamma = (x-1,x+1)\) and \(\Omega = (1,2x+1)\) be ordered bases for \(P_1\).

Recall that if the tuple representation of \(u \in P_1\) with respect to \(\Gamma\) is \(\begin{bmatrix} a\\ b\end{bmatrix}\), then \(u = a(x-1) + b(x+1) = (a+b)x + (-a +b )\). Is there an easy way to determine \([u]_{\Omega}\)?

The answer is “yes”. But before we look at a systematic way of handling questions like this, let's work through the algebra to find out what \([u]_{\Omega}\) is.

What we need to do is to find \(\delta\) and \(\phi\) such that \(\delta(1) + \phi(2x+1) = u = (a+b)x + (-a+b) \). Then we will have \([u]_{\Omega} = \begin{bmatrix} \delta\\\phi \end{bmatrix}\).

Note that \[\delta(1) + \phi(2x+1) = 2\phi x + (\delta + \phi).\] Hence, comparing the coefficients with \((a+b)x + (-a+b)\), we obtain \begin{eqnarray*} 2\phi & = & (a+b) \\ \delta + \phi & = & -a + b \end{eqnarray*} The first equation gives, \(\phi = \frac{a+b}{2}\). Substituting for \(\phi\) in the second equation gives \(\delta = -\frac{3}{2} a + \frac{b}{2}\). Hence, \[[u]_{\Omega} = \begin{bmatrix} -\frac{3}{2} a + \frac{b}{2} \\ \frac{a}{2} + \frac{b}{2} \end{bmatrix}.\] We can also write this as \[[u]_{\Omega} = \begin{bmatrix} -\frac{3}{2} & \frac{1}{2} \\ \frac{1}{2} & \frac{1}{2} \end{bmatrix} \begin{bmatrix} a \\ b\end{bmatrix}.\]

Hence, the matrix \(A = \begin{bmatrix} -\frac{3}{2} & \frac{1}{2} \\ \frac{1}{2} & \frac{1}{2} \end{bmatrix}\) satisfies the property that \[ [u]_{\Omega} = A [u]_{\Gamma}.\] In casual terms, it transforms the tuple representation with respect to \(\Gamma\) to the tuple representation with respect to \(\Omega\).

Matrix representation of the identity linear transformation

In fact, for any \(n\)-dimensional vector space \(V\) and ordered bases \(\Gamma\) and \(\Omega\) of \(V\), there is a unique matrix \(A\) such that for every \(u \in V\), \([u]_\Omega = A [u]_\Gamma\).

To see this, let \(\operatorname{id}\) denote the identity linear transformation. That is, \(\operatorname{id}(u) = u\) for all \(u \in V\). Then \([\operatorname{id}]_{\Gamma}^{\Omega}\) is precisely the matrix we desire because \(u\) is the output of \(\operatorname{id}(u)\) and the product \([\operatorname{id}]_{\Gamma}^{\Omega} [u]_{\Gamma}\), as we saw in the previous segment, gives the tuple representation of the output with respect to the ordered basis \(\Omega\).

We call \([\operatorname{id}]_{\Gamma}^{\Omega}\) the change-of-basis matrix from \(\Gamma\) to \(\Omega\). Note that this matrix is necessarily invertible since \(\operatorname{id}\) is a bijection.

Example

Let \(\Gamma = \left ( \begin{bmatrix} 2\\0\end{bmatrix}, \begin{bmatrix} 1\\3\end{bmatrix} \right)\) and \(\Omega = \left ( \begin{bmatrix} 1\\1\end{bmatrix}, \begin{bmatrix} -1\\1\end{bmatrix} \right )\ \) be ordered bases of \(\mathbb{R}^2\). Determine the change-of-basis matrix from \(\Gamma\) to \(\Omega\).

Solution. We need to first write each vector in \(\Gamma\) as a linear combination of vectors in \(\Omega\). Note that \[\begin{bmatrix} 2\\0\end{bmatrix} = 1\begin{bmatrix} 1 \\ 1\end{bmatrix} + (-1) \begin{bmatrix} -1\\1\end{bmatrix} \] and \[\begin{bmatrix} 1\\3\end{bmatrix} = 2\begin{bmatrix} 1 \\ 1\end{bmatrix} + 1 \begin{bmatrix} -1\\1\end{bmatrix} \]

Hence, the change-of-basis matrix from \(\Gamma\) to \(\Omega\) is \(\begin{bmatrix} 1 & 2 \\ -1 & 1\end{bmatrix}\).

Quick Quiz

Exercises

1. Let \(\Gamma = \left ( \begin{bmatrix} 1 \\ 0 \end{bmatrix}, \begin{bmatrix} 1\\ 1 \end{bmatrix}\right) \) and \(\Omega = \left ( \begin{bmatrix} 2 \\ 1 \end{bmatrix}, \begin{bmatrix} 1\\ 0 \ \end{bmatrix}\right) \) be ordered bases for \(\mathbb{R}^2\) Give the change-of-basis matrix from \(\Gamma\) and \(\Omega\).  

   We first write each vector in \(\Gamma\) as a linear combination of vectors in \(\Omega\) by finding \(\alpha_1,\beta_1,\alpha_2,\beta_2\) such that \[\begin{bmatrix} 1 \\ 0 \end{bmatrix} = \alpha_1 \begin{bmatrix}2 \\ 1\end{bmatrix} + \beta_1 \begin{bmatrix}1 \\ 0\end{bmatrix}\] and \[\begin{bmatrix} 1 \\ 1 \end{bmatrix} = \alpha_2 \begin{bmatrix}2 \\ 1\end{bmatrix} + \beta_2 \begin{bmatrix}1 \\ 0\end{bmatrix}.\] Then, the change-of-basis matrix is given by \(\begin{bmatrix} \alpha_1 & \alpha_2\\ \beta_1 & \beta_2\end{bmatrix}.\) Note that \[\begin{bmatrix} 1 \\ 0 \end{bmatrix} = 0\begin{bmatrix}2 \\ 1\end{bmatrix} + 1\begin{bmatrix}1 \\ 0\end{bmatrix}\] and \[\begin{bmatrix} 1 \\ 1 \end{bmatrix} = 1\begin{bmatrix}2 \\ 1\end{bmatrix} + (-1)\begin{bmatrix}1 \\ 0\end{bmatrix}.\] Hence, the change-of-basis matrix is \(\begin{bmatrix} 0 & 1\\ 1 & -1\end{bmatrix}.\)

2. Let \(\Gamma = \left ( \begin{bmatrix} 1 \\ -1 \end{bmatrix}, \begin{bmatrix} 1 \\ 0 \end{bmatrix}\right )\) be an ordered basis for \(\mathbb{R}^2\). If \([\operatorname{id}]_\Gamma^\Omega = \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}\), what must \(\Omega\) be?  

   Let \(\Omega = (u, v).\) Thus, we need \[\begin{bmatrix} 1 \\ -1 \end{bmatrix} = 0u + 1 v\] and \[\begin{bmatrix} 1 \\ 0 \end{bmatrix} = 1 u + 0 v.\] Thus, \(u = \begin{bmatrix} 1\\ 0\end{bmatrix}\) and \(v = \begin{bmatrix} 1 \\ -1 \end{bmatrix}.\)