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Section 4.7 Ordered basis and linear maps

In this section we give a matrix representation of a linear transformation. We consider only finite-dimensional vector spaces in this section.

Subsection 4.7.1 Ordered basis

In this subsection, we define notion of an ordered basis of a finite-dimensional vector space. Once an ordered basis for a vector space is fixed one may then define “coordinates” for the vector space.
Definition 4.7.1. (Ordered basis).
Let \(V\) be a finite-dimensional vector space over a field \(F\text{.}\) An ordered basis for \(V\) is a finite sequence of vectors which is linearly independent and spans \(V\text{.}\)
Remark 4.7.2.

If a sequence \(v_1,v_2,\ldots,v_n\) is an ordered basis for \(V\) then we write it as \((v_1,v_2,\ldots,v_n)\text{.}\) Thus, we consider ordered basis \((v_1,v_2,\ldots,v_n)\) and ordered basis \(\big(v_{\sigma(1)},v_{\sigma(2)},\ldots,v_{\sigma(n)}\big)\) to be distinct for every non-identity permutation \(\sigma\text{.}\)

To put it in different words, in an ordered basis the order in which basis vectors occur is important. For instance, the ordered basis \((v,w)\) is different from \((w,v)\text{.}\) However, as sets \(\{v,w\}\) and \(\{w,v\}\) are the same.

Definition 4.7.3.
Let \((v_1,v_2,\ldots,v_n)\) be an ordered basis of \(V\text{.}\) Given \(v\in V\) there is a unique \(n\)-tuple \((\alpha_1,\alpha_2,\ldots,\alpha_n)\) of scalars such that \(v=\sum_i\alpha_iv_i\) (see Exercise 3.5.11). We call \(\alpha_i\) the \(i\)-th coordinate of \(v\) relative to the ordered basis\((v_1,v_2,\ldots,v_n)\text{.}\)

Subsection 4.7.2 Matrix representation of a linear transformation

Now we are in a position to give matrix representation for linear transformation with respect to ordered bases.

Let \(V\) and \(W\) be finite-dimensional vector spaces over a field \(F\text{.}\) Suppose that \(\mathfrak{B}_V=(v_1,v_2,\ldots,v_n)\) and \(\mathfrak{B}_W=(w_1,w_2,\ldots,w_m)\) are ordered bases of \(V\) and \(W\text{,}\) respectively. Let \(T\colon V\to W\) be an \(F\)-linear transformation.

Since \(\mathfrak{B}_V\) is a basis, for a given \(v\in V\) there are uniquely determined \(\alpha_i\in F\) such that \(v=\sum\alpha_iv_i\) (see Exercise 3.5.11). Then,

\begin{equation*} T(v)=T\big(\sum\alpha_iv_i\big)=\sum\alpha_iT(v_i). \end{equation*}

Hence, \(T\) is completely determined by \(T(v_i)\) for \(1\leq i\leq n\text{.}\)

Because the ordered set \(\mathfrak{B}_W\) is a basis of \(W\) we get, for every \(i\in\{1,2,\ldots,n\}\text{,}\)

\begin{equation*} T(v_i)=\sum_{k=1}^{m}\beta_{ki}w_k. \end{equation*}

Therefore,

\begin{equation*} T(v)=\sum_{i=1}^{n}\alpha_i\sum_{k=1}^{m}\beta_{ki}w_k=\sum_{k=1}^{m}\big(\sum_{i=1}^{n}\alpha_i\beta_{ki}\big)w_k. \end{equation*}

Therefore, the \(k\)-th coordinate of \(T(v)\) with respect to \(\mathfrak{B}_W\) is

\begin{equation} \sum_{i=1}^{n}\alpha_i\beta_{ki}\text{.}\label{coordinate-of-Tv}\tag{4.1} \end{equation}

We obtain the matrix of \(T\) relative to ordered bases \(\mathfrak{B}_V\) and \(\mathfrak{B}_W\text{,}\)

\begin{equation*} [T]_{\mathfrak{B}_V}^{\mathfrak{B}_W}=\begin{pmatrix}\beta_{11}\amp\beta_{12}\amp\cdots\amp\beta_{1n}\\\beta_{21}\amp\beta_{22}\amp\cdots\amp\beta_{2n}\\\vdots\amp\vdots\amp\ddots\amp\vdots\\\beta_{m1}\amp\beta_{m2}\amp\cdots\amp\beta_{mn}\end{pmatrix}. \end{equation*}

We observe the following.

\begin{equation} \begin{pmatrix}\beta_{11}\amp\beta_{12}\amp\cdots\amp\beta_{1n}\\\beta_{21}\amp\beta_{22}\amp\cdots\amp\beta_{2n}\\\vdots\amp\vdots\amp\ddots\amp\vdots\\\beta_{m1}\amp\beta_{m2}\amp\cdots\amp\beta_{mn}\end{pmatrix}\begin{pmatrix}\alpha_1\\\alpha_2\\\vdots\\\alpha_n\end{pmatrix}=\begin{pmatrix}\sum_{i=1}^{n}\alpha_i\beta_{1i}\\\sum_{i=1}^{n}\alpha_i\beta_{2i}\\\vdots\\\sum_{i=1}^{n}\alpha_i\beta_{mi}\end{pmatrix}\label{matrix-rep-coordinates}\tag{4.2} \end{equation}

In particular, if \(e_k=(\delta_{k1},\delta_{k2},\ldots,\delta_{kn})\) then

\begin{equation} \begin{pmatrix}\beta_{11}\amp\beta_{12}\amp\cdots\amp\beta_{1n}\\\beta_{21}\amp\beta_{22}\amp\cdots\amp\beta_{2n}\\\vdots\amp\vdots\amp\ddots\amp\vdots\\\beta_{m1}\amp\beta_{m2}\amp\cdots\amp\beta_{mn}\end{pmatrix}\cdot e_k^t=\begin{pmatrix}\beta_{1k}\\\beta_{2k}\\\vdots\\\beta_{mk}\end{pmatrix}\label{e_i-gives-i-th-column}\tag{4.3} \end{equation}

Thus, the action of \([T]_{\mathfrak{B}_V}^{\mathfrak{B}_W}\) on \(e_k^t\) gives the \(k\)-th column of the matrix.

Remark 4.7.4.
We emphasize that matrix of a linear transformation depends on the choice of bases.