On the vector space \(\mathbb{K}^n\) we have a linear coordinate system defined by the identity mapping, that is, we define \(\boldsymbol{\varphi}(\vec{v})=\vec{v}\) for all \(\vec{v} \in \mathbb{K}^n.\) We call this coordinate system the standard coordinate system of \(\mathbb{K}^n.\)

In Linear Algebra we only consider linear coordinate systems, but in other areas of mathematics non-linear coordinate systems are also used. An example are the so-called polar coordinates \[\boldsymbol{\rho}: \mathbb{R}^2\setminus\{0_{\mathbb{R}^2}\} \to (0,\infty)\times (-\pi,\pi]\subset \mathbb{R}^2, \qquad \vec{x}\mapsto \begin{pmatrix} r \\ \phi\end{pmatrix}=\begin{pmatrix} \sqrt{(x_1)^2+(x_2)^2} \\ \arg(\vec{x}) \end{pmatrix},\] where \(\arg(\vec{x})=\arccos(x_1/r)\) for \(x_2\geqslant 0\) and \(\arg(\vec{x})=-\arccos(x_1/r)\) for \(x_2<0.\) Notice that the polar coordinates are only defined on \(\mathbb{R}^2\setminus\{0_{\mathbb{R}^2}\}.\) For further details we refer to the Calculus module.

The vector \(\vec{v}=\begin{pmatrix} 2 \\ 1 \end{pmatrix}\) has coordinates \(\begin{pmatrix} 2 \\ 1 \end{pmatrix}\) with respect to the standard coordinate system of \(\mathbb{R}^2.\) The same vector has coordinates \(\boldsymbol{\varphi}(\vec{v})=\begin{pmatrix} 4 \\ -1 \end{pmatrix}\) with respect to the coordinate system \(\boldsymbol{\varphi}\left(\begin{pmatrix}v_1 \\ v_2\end{pmatrix}\right)=\begin{pmatrix}v_1+2v_2 \\ -v_1+v_2\end{pmatrix}.\)

Let \(V=\mathbb{K}^3\) and \(\mathbf{e}=(\vec{e}_1,\vec{e}_2,\vec{e}_3)\) denote the ordered standard basis. Then for all \(\vec{x}=(x_i)_{1\leqslant i\leqslant 3}\in \mathbb{R}^3\) we have \[\boldsymbol{\varepsilon}(\vec{x})=\vec{x}.\] where \(\boldsymbol{\varepsilon}\) denotes the linear coordinate system corresponding to \(\mathbf{e}.\) Notice that \(\boldsymbol{\varepsilon}\) is the standard coordinate system on \(\mathbb{K}^n.\) Considering instead the ordered basis \(\mathbf{b}=(\vec{v}_1,\vec{v}_2,\vec{v}_3)=(\vec{e}_1+\vec{e}_3,\vec{e}_3,\vec{e}_2-\vec{e}_1),\) we obtain \[\boldsymbol{\beta}(\vec{x})=\begin{pmatrix} x_1+x_2 \\ x_3-x_1-x_2 \\ x_2\end{pmatrix}\] since \[\vec{x}=\begin{pmatrix} x_1 \\ x_2 \\ x_3\end{pmatrix}=(x_1+x_2)\underbrace{\begin{pmatrix}1 \\ 0 \\ 1 \end{pmatrix}}_{=\vec{v}_1}+(x_3-x_1-x_2)\underbrace{\begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix}}_{=\vec{v}_2}+x_2\underbrace{\begin{pmatrix}-1 \\ 1 \\ 0 \end{pmatrix}}_{=\vec{v}_3}.\]

Let \(V=\mathsf{P}_{2}(\mathbb{R})\) and \(W=\mathsf{P}_{1}(\mathbb{R})\) and \(g=\frac{\mathrm{d}}{\mathrm{d}x}.\) We consider the ordered basis \(\mathbf{b}=(v_1,v_2,v_3)=((1/2)(3x^2-1),x,1)\) of \(V\) and \(\mathbf{c}=(w_1,w_2)=(x,1)\) of \(W.\)

1. Compute the image under \(g\) of the elements \(v_i\) of the ordered basis \(\mathbf{b}.\) \[\begin{aligned} g\left(\frac{1}{2}(3x^2-1)\right)&=\frac{\mathrm{d}}{\mathrm{d}x}\left(\frac{1}{2}(3x^2-1)\right)=3x\\ g\left(x\right)&=\frac{\mathrm{d}}{\mathrm{d}x}(x)=1\\ g\left(1\right)&=\frac{\mathrm{d}}{\mathrm{d}x}(1)=0. \end{aligned}\]

2. Write the image vectors as linear combinations of the elements of the ordered basis \(\mathbf{c}.\) \[\tag{8.4} \begin{aligned} 3x &=3\cdot w_1+ 0\cdot w_2 \\ 1 & = 0\cdot w_1+ 1 \cdot w_2 \\ 0 & = 0\cdot w_1+0 \cdot w_2 \end{aligned}\]

3. Taking the transpose of the matrix of coefficients appearing in (8.4) gives the matrix representation \[\mathbf{M}\left(\frac{\mathrm{d}}{\mathrm{d}x},\mathbf{b},\mathbf{c}\right)=\begin{pmatrix} 3 & 0 & 0 \\ 0 & 1 & 0 \end{pmatrix}.\] of the linear map \(g=\frac{\mathrm{d}}{\mathrm{d}x}\) with respect to the bases \(\mathbf{b},\mathbf{c}.\)

Let \(\mathbf{e}=(\vec{e}_1,\ldots,\vec{e}_n)\) and \(\mathbf{d}=(\vec{d}_1,\ldots,\vec{d}_m)\) denote the ordered standard basis of \(\mathbb{K}^n\) and \(\mathbb{K}^m,\) respectively. Then for \(\mathbf{A}\in M_{m,n}(\mathbb{K}),\) we have \[\mathbf{A}=\mathbf{M}(f_\mathbf{A},\mathbf{e},\mathbf{d}),\] that is, the matrix representation of the mapping \(f_\mathbf{A}: \mathbb{K}^n \to \mathbb{K}^m\) with respect to the standard bases is simply the matrix \(\mathbf{A}.\) Indeed, we have \[f_\mathbf{A}(\vec{e}_j)=\mathbf{A}\vec{e}_j=\begin{pmatrix} A_{1j} \\ \vdots \\ A_{mj} \end{pmatrix}=\sum_{i=1}^m A_{ij}\vec{d}_i.\]

Let \(\mathbf{e}=(\vec{e}_1,\vec{e}_2)\) denote the ordered standard basis of \(\mathbb{R}^2.\) Consider the matrix \[\mathbf{A}=\begin{pmatrix} 5 & 1 \\ 1 & 5 \end{pmatrix}=\mathbf{M}(f_\mathbf{A},\mathbf{e},\mathbf{e}).\] We want to compute \(\mathrm{Mat}(f_\mathbf{A},\mathbf{b},\mathbf{b}),\) where \(\mathbf{b}=(\vec{v}_1,\vec{v}_2)=(\vec{e}_1+\vec{e}_2,\vec{e}_2-\vec{e}_1)\) is not the standard basis of \(\mathbb{R}^2.\) We obtain \[\begin{aligned} f_\mathbf{A}(\vec{v}_1)&=A\vec{v}_1=\begin{pmatrix} 5 & 1 \\ 1 & 5 \end{pmatrix}\begin{pmatrix} 1 \\ 1 \end{pmatrix}=\begin{pmatrix} 6 \\ 6 \end{pmatrix}=6\cdot \vec{v}_1+ 0\cdot \vec{v}_2\\ f_\mathbf{A}(\vec{v}_2)&=A\vec{v}_2=\begin{pmatrix} 5 & 1 \\ 1 & 5 \end{pmatrix}\begin{pmatrix} -1 \\ 1 \end{pmatrix}=\begin{pmatrix} -4 \\ 4 \end{pmatrix}=0\cdot \vec{v}_1+ 4\cdot \vec{v}_2 \end{aligned}\] Therefore, we have \[\mathbf{M}(f_\mathbf{A},\mathbf{b},\mathbf{b})=\begin{pmatrix} 6 & 0 \\ 0 & 4\end{pmatrix}.\]

With respect to the ordered basis \(\mathbf{b}=\left(\frac{1}{2}(3x^2-1),x,1\right),\) the polynomial \(a_2x^2+a_1x+a_0 \in V=\mathsf{P}_2(\mathbb{R})\) is represented by the vector \[\boldsymbol{\beta}(a_2x^2+a_1x+a_0)=\begin{pmatrix} \frac{2}{3}a_2 \\ a_1 \\ \frac{a_2}{3}+a_0\end{pmatrix}\] Indeed \[a_2x^2+a_1x+a_0=\frac{2}{3}a_2\left(\frac{1}{2}(3x^2-1)\right)+a_1x+\left(\frac{a_2}{3}+a_0\right)1.\] Computing \(\mathbf{M}(\frac{\mathrm{d}}{\mathrm{d}x},\mathbf{b},\mathbf{c})\boldsymbol{\beta}(a_2x^2+a_1x+a_0)\) gives \[\begin{pmatrix} 3 & 0 & 0 \\ 0 & 1 & 0 \end{pmatrix}\begin{pmatrix} \frac{2}{3}a_2 \\ a_1 \\ \frac{a_2}{3}+a_0\end{pmatrix}=\begin{pmatrix} 2a_2 \\ a_1 \end{pmatrix}\] and this vector represents the polynomial \(2a_2\cdot x+a_1\cdot 1=\frac{\mathrm{d}}{\mathrm{d}x}(a_2x^2+a_1x+a_0)\) with respect to the basis \(\mathbf{c}=(x,1)\) of \(\mathsf{P}_1(\mathbb{R}).\)

Let \(V=\mathsf{P}_3(\mathbb{R})\) so that \(\dim(V)=4\) and \[U=\operatorname{span}\{x^3+2x^2-x,4x^3+8x^2-4x-3,x^2+3x+4,2x^3+5x+x+4\}.\] We want to compute a basis of \(U.\)

The obvious ordered basis of \(V\) is \(\mathbf{f} = \{x^3, x^2, x, 1\},\) and the corresponding coordinate system \(\boldsymbol{\phi}\) is the isomorphism \(V \to \mathbb{R}_4\) defined by \[\boldsymbol{\phi}(a_3x^3+a_2x^2+a_1x+a_0)=\begin{pmatrix} a_3 & a_2 & a_1 & a_0 \end{pmatrix}.\] The images of the given generators of \(U\) are the row vectors \(\vec{\nu}_1, \dots, \vec{\nu}_4\) given by \(\vec{\nu}_1=\begin{pmatrix} 1 & 2 & -1 & 0 \end{pmatrix},\) \(\vec{\nu}_2=\begin{pmatrix} 4 & 8 & -4 & -3 \end{pmatrix},\) \(\vec{\nu}_3=\begin{pmatrix} 0 & 1 & 3 & 4 \end{pmatrix}\) and \(\vec{\nu}_4=\begin{pmatrix} 2 & 5 & 1 & 4 \end{pmatrix}.\)

We form the matrix \(\mathbf{N}\) with the \(\vec{\nu}_i\) as rows: \[\mathbf{N}= \begin{pmatrix} 1 & 2 & -1 & 0 \\ 4 & 8 & -4 & -3 \\ 0 & 1 & 3 & 4 \\ 2 & 5 & 1 & 4\end{pmatrix}\] The RREF of \(\mathbf{N}\) is \[\begin{pmatrix} 1 & 0 & -7 & 0 \\ 0 & 1 & 3 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 \end{pmatrix}.\] (Here we applied Gauss-Jordan elimination, but Gaussian elimination would be enough.) The non-zero rows of the RREF matrix are the vectors \(\vec{\omega}_1=\begin{pmatrix} 1 & 0 & -7 & 0 \end{pmatrix},\) \(\vec{\omega}_2=\begin{pmatrix} 0 & 1 & 3 & 0 \end{pmatrix},\) and \(\vec{\omega}_3=\begin{pmatrix} 0 & 0 & 0 & 1 \end{pmatrix}.\)

So a basis of \(U\) is given by \[\{\boldsymbol{\phi}^{-1}(\vec{\omega}_1),\boldsymbol{\phi}^{-1}(\vec{\omega}_2),\boldsymbol{\phi}^{-1}(\vec{\omega}_3)\}=\left\{x^3-7x,x^2+3x,1\right\},\] where we use that \[\boldsymbol{\phi}^{-1}\left(\begin{pmatrix} a_3 & a_2 & a_1 & a_0\end{pmatrix}\right)=a_3x^3+a_2x^2+a_1x+a_0.\]

Let \(V=\mathbb{R}^2\) and \(\mathbf{e}=(\vec{e}_1,\vec{e}_2)\) be the ordered standard basis and \(\mathbf{b}=(\vec{v}_1,\vec{v}_2)=(\vec{e}_1+\vec{e}_2,\vec{e}_2-\vec{e}_1)\) another ordered basis. According to the recipe mentioned in Example 8.13, if we want to compute \(\mathbf{C}(\mathbf{e},\mathbf{b})\) we simply need to write each vector of \(\mathbf{e}\) as a linear combination of the elements of \(\mathbf{b}.\) The transpose of the resulting coefficient matrix is then \(\mathbf{C}(\mathbf{e},\mathbf{b}).\) We obtain \[\begin{aligned} \vec{e}_1&=\frac{1}{2}\vec{v}_1-\frac{1}{2}\vec{v}_2,\\ \vec{e}_2&=\frac{1}{2}\vec{v}_1+\frac{1}{2}\vec{v}_2, \end{aligned}\] so that \[\mathbf{C}(\mathbf{e},\mathbf{b})=\begin{pmatrix} \frac{1}{2} & \frac{1}{2} \\ -\frac{1}{2} & \frac{1}{2}\end{pmatrix}.\] Reversing the role of \(\mathbf{e}\) and \(\mathbf{b}\) gives \(\mathbf{C}(\mathbf{b},\mathbf{e})\) \[\begin{aligned} \vec{v}_1&=1\vec{e}_1+1\vec{e}_2,\\ \vec{v}_2&=-1 \vec{e}_1+1\vec{e}_2, \end{aligned}\] so that \[\mathbf{C}(\mathbf{b},\mathbf{e})=\begin{pmatrix} 1 & -1 \\ 1 & 1 \end{pmatrix}.\] Notice that indeed we have \[\mathbf{C}(\mathbf{e},\mathbf{b})\mathbf{C}(\mathbf{b},\mathbf{e})=\begin{pmatrix} \frac{1}{2} & \frac{1}{2} \\ -\frac{1}{2} & \frac{1}{2}\end{pmatrix}\begin{pmatrix} 1 & -1 \\ 1 & 1 \end{pmatrix}=\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}\] so that \(\mathbf{C}(\mathbf{e},\mathbf{b})^{-1}=\mathbf{C}(\mathbf{b},\mathbf{e}).\)

Let \(\mathbf{e}=(\vec{e}_1,\vec{e}_2)\) denote the ordered standard basis of \(\mathbb{R}^2\) and \[\mathbf{A}=\begin{pmatrix} 5 & 1 \\ 1 & 5 \end{pmatrix}=\mathbf{M}(f_\mathbf{A},\mathbf{e},\mathbf{e}).\] Let \(\mathbf{b}=(\vec{e}_1+\vec{e}_2,\vec{e}_2-\vec{e}_1).\) We computed that \[\mathbf{M}(f_\mathbf{A},\mathbf{b},\mathbf{b})=\begin{pmatrix} 6 & 0 \\ 0 & 4\end{pmatrix}\] as well as \[\mathbf{C}(\mathbf{e},\mathbf{b})=\begin{pmatrix} \frac{1}{2} & \frac{1}{2} \\ -\frac{1}{2} & \frac{1}{2}\end{pmatrix} \quad \text{and} \quad \mathbf{C}(\mathbf{b},\mathbf{e})=\begin{pmatrix} 1 & -1 \\ 1 & 1 \end{pmatrix}.\] According to Theorem 8.26 we must have \[\mathbf{M}(f_\mathbf{A},\mathbf{b},\mathbf{b})=\mathbf{C}(\mathbf{e},\mathbf{b})\mathbf{M}(f_\mathbf{A},\mathbf{e},\mathbf{e})\mathbf{C}(\mathbf{b},\mathbf{e})\] and indeed \[\begin{pmatrix} 6 & 0 \\ 0 & 4\end{pmatrix} =\begin{pmatrix} \frac{1}{2} & \frac{1}{2} \\ -\frac{1}{2} & \frac{1}{2}\end{pmatrix}\begin{pmatrix} 5 & 1 \\ 1 & 5 \end{pmatrix}\begin{pmatrix} 1 & -1 \\ 1 & 1 \end{pmatrix}.\]