Summary

4.4 Coordinate Systems

Theorem 1 (Theorem 8) Let \(\mathcal{B}=\left\{\mathbf{b}_1, \ldots, \mathbf{b}_n\right\}\) be a basis for a vector space \(V\). Then the coordinate mapping \(\mathbf{x} \mapsto[\mathbf{x}]_{\mathcal{B}}\) is a one-to-one linear transformation from \(V\) onto \(\mathbb{R}^n\).

Remark

Remark. What this theorem means is that for a transformation \(T: V \to \mathbb{R}^n\) which is defined as \(\mathbf{x} \mapsto [\mathbf{x}]_{\mathcal{B}}\), \(T\) is a linear transformation and is both one-to-one and onto.

Exercise 1 (Practice Problem 2) The set \(\mathcal{B}=\left\{1+t, 1+t^2, t+t^2\right\}\) is a basis for \(\mathbb{P}_2\). Find the coordinate vector of \(\mathbf{p}(t)=6+3 t-t^2\) relative to \(\mathcal{B}\).

Remark

Remark. One way to solve this is to equate the coefficients of the equation \[c_1(1+t) + c_2(1+t^2) + c_3(t+t^2)=6+3 t-t^2\] and solve for \(c_1,c_2,c_3\).

However, by observation, we can also first write \(1+t,1+t^2,t+t^2,6+3t-t^2\) as \(\begin{bmatrix}1\\ 1\\ 0\end{bmatrix}, \begin{bmatrix}1\\ 0\\ 1\end{bmatrix}, \begin{bmatrix}0\\ 1\\ 1\end{bmatrix}, \begin{bmatrix}6\\ 3\\ -1\end{bmatrix}\). Then, as for subspaces in \(\mathbb{R}\), we solve the coefficients \(\mathbf{c}\) using row reduction.

Why does this method work?

Let \(T: \mathbf{x} \mapsto [\mathbf{x}]_\mathcal{B}\), so \(T\) is a bijective linear transformation.

\[\begin{align} T(c_1\mathbf{b}_1 + \cdots + c_n\mathbf{b}_n) &= c_1 T(\mathbf{b}_1) + \cdots + c_n T(\mathbf{b}_n) \\ &= T(\mathbf{p}) \end{align}\]

where \(T(\mathbf{b}_1),\dots,T(\mathbf{b}_n)\) are the coordinate vectors of the basis. Since \(T\) is bijective, we can apply \(T^{-1}\) to both side of the equation. Reversing the steps gives the same coefficients.