Skip to main content
Mathematics LibreTexts

11.2: Matrix of a Linear Transformation (Redux)

  • Page ID
  • \( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \) \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)\(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\) \(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\)\(\newcommand{\AA}{\unicode[.8,0]{x212B}}\)

    Not only do bases allow us to describe arbitrary vectors as column vectors, they also permit linear transformations to be expressed as matrices. This is a very powerful tool for computations, which is covered in chapter 7 and reviewed again here.

    Suppose we have a linear transformation \(L \colon V\rightarrow W\) and ordered input and output bases \(E=(e_{1}, \ldots, e_{n})\) and \(F=(f_{1}, \ldots, f_{m})\) for \(V\) and \(W\) respectively (of course, these need not be the standard basis--in all likelihood \(V\) is \(\textit{not}\) \(\mathbb{R}^{n}\)). Since for each \(e_{j}\), \(L(e_{j})\) is a vector in \(W\), there exist unique numbers \(m^{i}_{j}\) such that
    L(e_j)=f_{1}m^{1}_{j} + \cdots + f_{m}m^{m}_{j} =(f_{1},\ldots, f_{m}) \begin{pmatrix}m^{1}_{j}\\\vdots\\m^{m}_{j}\end{pmatrix}\, .
    = \sum_{i=1}^{m} f_{i}M^{i}_{j}
    =\begin{pmatrix}f_{1} & f_{2} & \cdots & f_{m}\end{pmatrix}\begin{pmatrix}M^{1}_{j} \\ M^{2}_{j} \\ \vdots \\ M^{m}_{j}\end{pmatrix}\, .
    The number \(m^{i}_{j}\) is the \(i\)th component of \(L(e_{j})\) in the basis \(F\), while the \(f_{i}\) are vectors (note that if \(\alpha\) is a scalar, and \(v\) a vector, \(\alpha v=v\alpha\), we have used the latter---rather uncommon---notation in the above formula). The numbers \(m^{i}_{j}\) naturally form a matrix whose \(j\)th column is the column vector displayed above. Indeed, if $$v=e_{1}v^{1}+\cdots+e_{n} v^{n}\, ,$$

    \[L(v) = L(v^{1}e_{1} + v^{2}e_{2} + \cdots + v^{n}e_{n})\]

    \[~~~~= v^{1}L(e_{1}) + v^{2}L(e_{2}) + \cdots + v^{n}L(e_{n}) = \sum_{j=1}^{m} L(e_{j})v^{j}\]

    \[~~~~= \sum_{j=1}^{m} (f_{1}m_{j}^{1} + \cdots + f_{m}m_{j}^{m})v^{j} = \sum_{i=1}^{n}f_{i}[\sum_{j=1}^{m} M_{j}^{i}v^{j}]\]

    \[~~~~= (f_{1} f_{2} \cdots f_{m}) \begin{pmatrix} m_{1}^{1} & m_{2}^{1} & \cdots & m_{n}^{1} \\ m_{1}^{2} & m_{2}^{2} & & \\ \vdots & & \ddots & \vdots \\ m_{1}^{m} & & \cdots & m_{n}^{m}\end{pmatrix} \begin{pmatrix} v^{1} \\ v^{2} \\ \vdots \\ v^{n} \end{pmatrix}$$
    In the column vector-basis notation this equality looks familiar:
    L\begin{pmatrix}v^{1}\\ \vdots \\ v^{n}\end{pmatrix}_{E}
    m^{1}_{1} & \ldots & m^{1}_{n} \\
    \vdots & & \vdots \\
    m^{m}_{1} & \ldots & m^{m}_{n} \\
    \begin{pmatrix}v^{1}\\ \vdots \\ v^{n}\end{pmatrix}
    The array of numbers \(M=(m^{i}_{j})\) is called the matrix of \(L\) in the input and output bases \(E\) and \(F\) for \(V\) and \(W\), respectively. This matrix will change if we change either of the bases. Also observe that the columns of \(M\) are computed by examining \(L\) acting on each basis vector in \(V\) expanded in the basis vectors of \(W\).

    Example \(\PageIndex{1}\):

    Let \(L \colon P_{1}(t) \mapsto P_{1}(t)\), such that \(L(a+bt)=(a+b)t\). Since \(V=P_{1}(t)=W\), let's choose the same ordered basis \(B=(1-t, 1+t )\) for \(V\) and \(W\).
    L(1-t)&=&(1-1)t=\ 0\ =(1-t)\cdot 0 + (1+t)\cdot 0=
    \begin{pmatrix}1-t, 1+t\end{pmatrix}\begin{pmatrix}0\\0\end{pmatrix} \\
    L(1+t)&=&(1+1)t=2t\ =(1-t)\cdot -1 + (1+t)\cdot 1=
    \begin{pmatrix}1-t, 1+t\end{pmatrix}\begin{pmatrix}-1\\1\end{pmatrix}\\
    0 & -1 \\
    0 & 1 \\

    When the vector space is \(\mathbb{R^{n}}\) and the standard basis is used, the problem of finding the matrix of a linear transformation will seem almost trivial. It is worthwhile working through it once in the above language though:

    Example \(\PageIndex{2}\):

    Any vector in \(\Re^{n}\) can be written as a linear combination of the \(\textit{standard (ordered) basis}\) \((e_{1},\dots e_{n})\). The vector \(e_{i}\) has a one in the \(i\)th position, and zeros everywhere else. \(\textit{i.e.}\)
    e_{1}=\begin{pmatrix}1\\ 0\\ \vdots \\0\end{pmatrix}\, ,\quad e_{2}=\begin{pmatrix}0\\ 1\\ \vdots \\0\end{pmatrix}\, ,\ldots\quad e_{n}=\begin{pmatrix}0\\ 0\\ \vdots \\1 \end{pmatrix}\, .
    Then to find the matrix of any linear transformation \(L \colon \Re^{n} \rightarrow \Re^{n}\), it suffices to know what \(L(e_{i})\) is for every \(i\).

    For any matrix \(M\), observe that \(Me_{i}\) is equal to the $i$th column of \(M\). Then if the \(i\)th column of \(M\) equals \(L(e_{i})\) for every \(i\), then \(Mv=L(v)\) for every \(v\in \Re^{n}\). Then the matrix representing \(L\) in the standard basis is just the matrix whose \(i\)th column is \(L(e_{i})\).

    For example, if
    L\begin{pmatrix}1\\0\\0\end{pmatrix}=\begin{pmatrix}1\\4\\7\end{pmatrix}\, ,\quad
    L\begin{pmatrix}0\\1\\0\end{pmatrix}=\begin{pmatrix}2\\5\\8\end{pmatrix}\, ,\quad
    L\begin{pmatrix}0\\0\\1\end{pmatrix}=\begin{pmatrix}3\\6\\9\end{pmatrix}\, ,
    then the matrix of \(L\) in the standard basis is simply
    \begin{pmatrix}1&2&3\\4&5&6\\7&8&9\end{pmatrix}\, .
    Alternatively, this information would often be presented as
    L\begin{pmatrix}x\\y\\z\end{pmatrix}=\begin{pmatrix}x+2y+3z\\4x+5y+6z\\7x+8y+9z\end{pmatrix}\, .
    You could either rewrite this as
    L\begin{pmatrix}x\\y\\z\end{pmatrix}=\begin{pmatrix}1&2&3\\4&5&6\\7&8&9\end{pmatrix}\begin{pmatrix}x\\y\\z\end{pmatrix}\, ,
    to immediately learn the matrix of \(L\), or taking a more circuitous route:
    \begin{pmatrix}3\\6\\9\end{pmatrix}\:=\:\begin{pmatrix}1&2&3\\4&5&6\\7&8&9\end{pmatrix}\begin{pmatrix}x\\y\\z\end{pmatrix}\, .


    This page titled 11.2: Matrix of a Linear Transformation (Redux) is shared under a not declared license and was authored, remixed, and/or curated by David Cherney, Tom Denton, & Andrew Waldron.

    • Was this article helpful?