Ordered Basis #
Definition: #
Let \(\bf V \) be finite dimensional vector space over a field \(\bf F \). An \(\fcolorbox{red}{white}{\bf ordered basis}\) for \(\bf V\) is a basis of \(\bf V \) provided with a specific order.
Example #
Let us consider the vector space \(\bf \real^{3} \) and let \(\bf \alpha_{1}=(1,0,0) \), \(\bf \alpha_{2}=(0,1,0) \) and \(\bf \alpha_{3}=(0,0,1) \). Let \(\bf S=\{\alpha_{1},\alpha_{2},\alpha_{3} \} \) and \(\bf T=\{\alpha_{3},\alpha_{1},\alpha_{2} \} \) then \(\bf \\ \) \(\fcolorbox{red}{white}{\bf \(\bf S\) and \(\bf T\) are the same basis but they are different ordered bases.}\)
Coordinate Matrix of a Vector #
Definition: #
Let \(\bf V \) be \(\bf n \)-dimensional vector space over a field \(\bf F \) and \(\bf \alpha=\{\alpha_{1},\alpha_{2},…,\alpha_{n} \} \) be a ordered basis of \(\bf V \) and \(\bf x\in V\) such that $$\bf x=c_{1}\alpha_{1}+c_{2}\alpha_{2}+…+c_{n}\alpha_{n} $$ \(\bf \\ \) Then the \(\fcolorbox{red}{white}{\bf coordinate matrix}\) of \(\bf x \) relative to the ordered basis \(\bf \alpha \) is denoted by \(\bf [x]_{\alpha}\) and is defined by the \(\bf n\times 1 \) matrix such that $$\bf [x]_{\alpha}=\begin{pmatrix} \bf c_{1} \\ \bf c_{2} \\ \bf …\\ \bf c_{n} \\ \end{pmatrix}$$
Example #
Let us consider the vector space \(\bf \real^{3} \) and let \(\bf \alpha_{1}=(1,0,0) \), \(\bf \alpha_{2}=(0,1,0) \) and \(\bf \alpha_{3}=(0,0,1) \). Let \(\bf \gamma=\{\alpha_{1},\alpha_{2},\alpha_{3} \} \) and \(\bf \delta=\{\alpha_{3},\alpha_{1},\alpha_{2} \}.\\ \) Let \(\bf x=(2,3,-6)\in \real^{3} \) then \(\bf \\ \) $$\bf x=2\alpha_{1}+3\alpha_{2}-6\alpha_{3}$$
Explanation:
\begin{align} \bf x & \bf =(2,3,-6) \nonumber\\ & \bf =2(1,0,0)+3(0,1,0)-6(0,0,1) \nonumber\\ & \bf =2\alpha_{1}+3\alpha_{2}-6\alpha_{3}\nonumber\\ \end{align}
And also $$\bf x=-6\alpha_{3}+2\alpha_{1}+3\alpha_{2}$$ Then $$\bf [x]_{\gamma}=\begin{pmatrix} \bf ~~~2\\ \bf ~~~3\\ \bf -6\\ \end{pmatrix} $$ and $$\bf [x]_{\delta}=\begin{pmatrix} \bf -6\\ \bf ~~~2\\ \bf ~~~3\\ \end{pmatrix} $$
Matrix Representation of a Linear Transformation #
Definition: #
Let \(\bf V \) and \(\bf W \) be two finite dimensional vector space over a same field \(\bf F \) with dimensions \(\bf m\) and \(\bf n\) respectively and let \(\bf \alpha=\{\alpha_{1},\alpha_{2},…,\alpha_{m} \} \) and \(\bf \beta=\{\beta_{1},\beta_{2},…,\beta_{n} \} \) be two ordered bases of \(\bf V \) and \(\bf W \) respectively. Let \(\bf T:V\to W\) be a linear transformation such that \(\bf T(\alpha_{j})=\displaystyle\sum_{i=1}^{n}a_{ij}~\beta_{i} \) where \(\bf j=1,2,…,m \). Then the \(\fcolorbox{red}{white}{\bf matrix representation}\) of \(\bf T\), with respect to the ordered bases \(\bf \alpha\) and \(\bf \beta\), is denoted by \(\bf [T]_{\alpha}^{\beta} \) and is defined by the \(\bf n\times m \) matrix defined the $$\bf [T]_{\alpha}^{\beta} = \begin{pmatrix} \bf a_{11} & \bf a_{12} & \bf … & \bf a_{1m} \\ \bf a_{21} & \bf a_{22} & \bf … & \bf a_{2m} \\ \bf …& \bf …& \bf … & \bf …\\ \bf a_{n1} & \bf a_{n2} &\bf … & \bf a_{nm} \\ \end{pmatrix}$$
Example #
Let us consider the vector spaces \(\bf \real^{3} \) and \(\bf \real^{2} \). Let \(\bf \alpha=\{\alpha_{1},\alpha_{2},\alpha_{3} \} \) and \(\bf \beta=\{\beta_{1},\beta_{2}\} \) such that \(\bf \alpha_{1}=(1,0,0) \), \(\bf \alpha_{2}=(0,1,0) \), \(\bf \alpha_{3}=(0,0,1) \), \(\bf \beta_{1}=(1,0) \) and \(\bf \beta_{2}=(0,1). \\ \) Also consider the linear transformation \(\bf T:\real^{3}\to \real^{2}\) such that \(\bf T(x,y,z)=(x-y+2z,2x+3y-4z) \) \(\bf \forall (x,y,z)\in\real^{3}.\\ \) Then \begin{align*} & \bf T(\alpha_{1})=\beta_{1}+2\beta_{2}\\ & \bf T(\alpha_{2})=-\beta_{1}+3\beta_{2}\\ & \bf T(\alpha_{3})=2\beta_{1}-4\beta_{2} \end{align*} Therefore the matrix of \(\bf T\) is $$\bf [T]_{\alpha}^{\beta} = \begin{pmatrix} \bf 1 & \bf -1 &\bf ~~~2 \\ \bf 2 &\bf ~~~3 & \bf-4\\ \end{pmatrix}$$
Explanation:
\begin{align*} & \bf T(\alpha_{1})=T(1,0,0) \\ \implies & \bf T(\alpha_{1})=(1,2)\\ \implies & \bf T(\alpha_{1})=(1,0)+2(0,1)\\ \implies & \bf T(\alpha_{1})=\beta_{1}+2\beta_{2}\\ \end{align*}
\begin{align*} & \bf T(\alpha_{2})=T(0,1,0) \\ \implies & \bf T(\alpha_{2})=(-1,3)\\ \implies & \bf T(\alpha_{2})=-1(1,0)+3(0,1)\\ \implies & \bf T(\alpha_{2})=-\beta_{1}+3\beta_{2}\\ \end{align*}
\begin{align*} & \bf T(\alpha_{3})=T(0,1,0) \\ \implies & \bf T(\alpha_{3})=(2,-4)\\ \implies & \bf T(\alpha_{3})=2(1,0)-4(0,1)\\ \implies & \bf T(\alpha_{3})=2\beta_{1}-4\beta_{2} \end{align*}
