Let $V$ be a vector space over the field $\mathbb{F}$ having dimension $n$. Recall that given an ordered basis $(v_1,\dots ,v_n)$ for $V$ and a vector $u\in V$, the tuple representation of $u$ with respect to $(v_1,\dots ,v_n)$, denoted by $[u{]}_{(v_1,\dots ,v_n)}$, is the tuple $\left[\begin{array}{c}{\lambda }_1\\ {\lambda }_2\\ ⋮\\ {\lambda }_n\end{array}\right]$ such that ${\displaystyle u=\sum _{i=1}^n{\lambda }_i{v}_i}$. (Note that ${\lambda }_1,\dots ,{\lambda }_n$ are uniquely determined since $v_1,\dots ,v_n$ form a basis for $V$.) As a result, one can work with tuples instead of the vectors in the original vector space.

We will now see that we can express linear transformations as matrices as well. Hence, one can simply focus on studying linear transformations of the form $T(x)=Ax$ where $A$ is a matrix.

As for tuple representations of vectors, matrix representations of a linear transformation will depend on the choice of the ordered basis for the domain and that for the codomain.

Matrix representations of linear transformations

Let $V$ and $W$ be vector spaces over some field $\mathbb{F}$. Let $\mathrm{\Gamma }=(v_1,\dots ,v_n)$ be an ordered basis for $V$ and let $\mathrm{\Omega }=(w_1,\dots ,w_m)$ be an ordered basis for $W$.

Let $T:V\to W$ be a linear transformation. We can give a matrix representation of $T$ as follows.

For each $j\in \{1,\dots ,n\}$, $T(v_j)$ is a vector in $W$. Hence, we can write $T(v_j)$ as a linear combination of $w_1,\dots ,w_m$. Therefore, there are scalars $a_{1,j},a_{2,j},\dots ,a_{m,j}$ such that $T(v_j)=a_{1,j}{w}_1+a_{2,j}{w}_2+\cdots +a_{m,j}{w}_m$.

Take an arbitrary $v\in V$. Then, there exist scalars ${\lambda }_1,\dots ,{\lambda }_n$ such that $v={\lambda }_1{v}_1+\cdots +{\lambda }_n{v}_n$. So $[v{]}_{\mathrm{\Gamma }}=\left[\begin{array}{c}{\lambda }_1\\ ⋮\\ {\lambda }_n\end{array}\right]$.

Hence, $$\begin{array}{rcl}T(v)& =& T({\lambda }_1{v}_1+\cdots +{\lambda }_n{v}_n)\\ & =& \sum _{j=1}^n{\lambda }_{j}T(v_j)\\ & =& \sum _{j=1}^n{\lambda }_j\left(\sum _{i=1}^m{a}_{i,j}{w}_i\right)\\ & =& \sum _{j=1}^n\sum _{i=1}^n(a_{i,j}{\lambda }_j)w_i\\ & =& \sum _{i=1}^m\left(\sum _{j=1}^n{a}_{i,j}{\lambda }_j\right)w_i\end{array}$$

Thus, $[T(v){]}_{\mathrm{\Omega }}={\displaystyle \left[\begin{array}{c}\sum _{j=1}^n{a}_{1,j}{\lambda }_j\\ \sum _{j=1}^n{a}_{2,j}{\lambda }_j\\ ⋮\\ \sum _{j=1}^n{a}_{m,j}{\lambda }_j\end{array}\right]=\left[\begin{array}{cccc}{a}_{1,1}& a_{1,2}& \cdots & a_{1,n}\\ a_{2,1}& a_{2,2}& \cdots & a_{2,n}\\ ⋮& ⋮& \ddots & ⋮\\ a_{m,1}& a_{m,2}& \cdots & a_{m,n}\end{array}\right]\left[\begin{array}{c}{\lambda }_1\\ {\lambda }_2\\ ⋮\\ {\lambda }_n\end{array}\right].}$ So, if we let $A$ be the $m\times n$ matrix such that $A_{i,j}=a_{i,j}$, then $[T(v){]}_{\mathrm{\Omega }}$ is precisely $A[v{]}_{\mathrm{\Gamma }}$.

Hence, given any $u\in V$, we can obtain the tuple representation of $T(u)$ with respect to $\mathrm{\Omega }$ by computing $A[u{]}_{\mathrm{\Gamma }}$. The matrix $A$ is called the matrix representation of $T$ and is denoted $[T{]}_{\mathrm{\Gamma }}^{\mathrm{\Omega }}$.

Note that column $i$ of $[T{]}_{\mathrm{\Gamma }}^{\mathrm{\Omega }}$ is given by $[T(v_i){]}_{\mathrm{\Omega }}$.

Example

Let $T:{\mathbb{R}}^3\to P_2$ be a linear transformation, where $P_2$ is the vector space of polynomials in $x$ with real coefficients having degree at most 2, given by $$T\left(\left[\begin{array}{c}a\\ b\\ c\end{array}\right]\right)=(a-b)x^2+cx+(a+b+c).$$ Let $\mathrm{\Gamma }=(\left[\begin{array}{c}1\\ 0\\ 0\end{array}\right],\left[\begin{array}{c}1\\ 1\\ 0\end{array}\right],\left[\begin{array}{c}0\\ -1\\ 1\end{array}\right])$ and $\mathrm{\Omega }=(x+1,x^2-x,x^2+x-1)$. We now find $[T{]}_{\mathrm{\Gamma }}^{\mathrm{\Omega }}$.

One can easily check that $\mathrm{\Gamma }$ and $\mathrm{\Omega }$ are ordered bases for ${\mathbb{R}}^3$ and $P_2$, respectively.

We first find scalars ${\alpha }_1,{\alpha }_2,{\alpha }_3,{\beta }_1,{\beta }_2,{\beta }_3,{\gamma }_1,{\gamma }_2,{\gamma }_3$ such that $$\begin{array}{rcl}T\left(\left[\begin{array}{c}1\\ 0\\ 0\end{array}\right]\right)& =& {\alpha }_1(x+1)+{\beta }_1(x^2-x)+{\gamma }_1(x^2+x-1)\\ T\left(\left[\begin{array}{c}1\\ 1\\ 0\end{array}\right]\right)& =& {\alpha }_2(x+1)+{\beta }_2(x^2-x)+{\gamma }_2(x^2+x-1)\\ T\left(\left[\begin{array}{c}0\\ -1\\ 1\end{array}\right]\right)& =& {\alpha }_3(x+1)+{\beta }_3(x^2-x)+{\gamma }_3(x^2+x-1).\end{array}$$ Then $[T{]}_{\mathrm{\Gamma }}^{\mathrm{\Omega }}=\left[\begin{array}{ccc}{\alpha }_1& {\alpha }_2& {\alpha }_3\\ {\beta }_1& {\beta }_2& {\beta }_3\\ {\gamma }_1& {\gamma }_2& {\gamma }_3\end{array}\right].$

Now, $T\left(\left[\begin{array}{c}1\\ 0\\ 0\end{array}\right]\right)=x^2+1$. Hence, we need $$x^2+1={\alpha }_1(x+1)+{\beta }_1(x^2-x)+{\gamma }_1(x^2+x-1),$$ or equivalently, $$x^2+1=({\beta }_1+{\gamma }_1)x^2+({\alpha }_1-{\beta }_1+{\gamma }_1)x+({\alpha }_1-{\gamma }_1).$$ Comparing coefficients gives $$\begin{array}{rcl}1& =& {\beta }_1+{\gamma }_1\\ 0& =& {\alpha }_1-{\beta }_1+{\gamma }_1\\ 1& =& {\alpha }_1-{\gamma }_1.\end{array}$$ Solving gives ${\alpha }_1=1$, ${\beta }_1=1$, ${\gamma }_1=0$.

Similarly, we obtain $$T\left(\left[\begin{array}{c}1\\ 1\\ 0\end{array}\right]\right)=2=\frac{4}{3}(x+1)+\frac{2}{3}(x^2-x)-\frac{2}{3}(x^2+x-1)$$ and $$T\left(\left[\begin{array}{c}0\\ -1\\ 1\end{array}\right]\right)=x^2+x=\frac{2}{3}(x+1)+\frac{1}{3}(x^2-x)+\frac{2}{3}(x^2+x-1).$$

Hence, $[T{]}_{\mathrm{\Gamma }}^{\mathrm{\Omega }}=\left[\begin{array}{ccc}1& \frac{4}{3}& \frac{2}{3}\\ 1& \frac{2}{3}& \frac{1}{3}\\ 0& -\frac{2}{3}& \frac{2}{3}\end{array}\right].$

We check this against a couple of vectors. Let $u=\left[\begin{array}{c}0\\ 1\\ 0\end{array}\right]$ and $v=\left[\begin{array}{c}0\\ 0\\ 1\end{array}\right]$. Then $T(u)=-x^2+1$ and $T(v)=x+1$ and $[u{]}_{\mathrm{\Gamma }}=\left[\begin{array}{c}-1\\ 1\\ 0\end{array}\right]$ and $[v{]}_{\mathrm{\Gamma }}=\left[\begin{array}{c}-1\\ 1\\ 1\end{array}\right]$.

Let $A=[T{]}_{\mathrm{\Gamma }}^{\mathrm{\Omega }}$.

And $A[v{]}_{\mathrm{\Gamma }}=\left[\begin{array}{c}1\\ 0\\ 0\end{array}\right]$, which is the tuple representation of $x+1=T(v)$.

Remark: It is perhaps a bit confusing that tuple representations of vectors in ${\mathbb{R}}^3$ are also vectors in ${\mathbb{R}}^3$. But why would one want to have $3$-tuple representations of vectors in ${\mathbb{R}}^3$? The answer to this question might not be obvious at first sight. However, when one works with ${\mathbb{R}}^n$ for large $n$, it might be possible to choose an ordered bases in such a way that the vectors that one is interested in have sparse tuple representations (i.e. tuple representations in which very few entries are nonzero.) Sparsity can drastically improve certain kinds of computations and is one of the main ingredients on which current digital audio and image compression algorithms are based.

Summary

With tuple representations for vectors and matrix representations for linear transformations, we have a unifying framework for computations over all finite-dimensional vector spaces. In other words, thinking about an $n$-dimensional vector space is essentially thinking about $n$-tuples and thinking about linear transformations is essentially thinking about matrices. The usefulness of this fact cannot be overstated, especially when it comes to developing computer software that works with vector spaces.

Quick Quiz

Exercises

Let $T:{\mathbb{R}}^2\to {\mathbb{R}}^2$ be a linear transformation. Let $\mathrm{\Gamma }=(\left[\begin{array}{c}1\\ 0\end{array}\right],\left[\begin{array}{c}1\\ 1\end{array}\right])$ and $\mathrm{\Omega }=(\left[\begin{array}{c}2\\ 1\end{array}\right],\left[\begin{array}{c}1\\ 0\end{array}\right])$ be ordered bases for ${\mathbb{R}}^2$

1. Suppose that $T$ is given by $T\left(\left[\begin{array}{c}{x}_1\\ x_2\end{array}\right]\right)=\left[\begin{array}{c}2x_1-x_2\\ x_1+x_2\end{array}\right]$. Find $[T{]}_{\mathrm{\Gamma }}^{\mathrm{\Omega }}$.  

   Note that $$T\left(\left[\begin{array}{c}1\\ 0\end{array}\right]\right)=\left[\begin{array}{c}2\\ 1\end{array}\right]=1\left[\begin{array}{c}2\\ 1\end{array}\right]+0\left[\begin{array}{c}1\\ 0\end{array}\right]$$ and $$T\left(\left[\begin{array}{c}1\\ 1\end{array}\right]\right)=\left[\begin{array}{c}1\\ 2\end{array}\right]=2\left[\begin{array}{c}2\\ 1\end{array}\right]+(-3)\left[\begin{array}{c}1\\ 0\end{array}\right]$$ Hence, $[T{]}_{\mathrm{\Gamma }}^{\mathrm{\Omega }}=\left[\begin{array}{cc}1& 2\\ 0& -3\end{array}\right].$

2. Suppose that $T\left(\left[\begin{array}{c}1\\ 0\end{array}\right]\right)=\left[\begin{array}{c}2\\ 1\end{array}\right]$ and $T\left(\left[\begin{array}{c}0\\ 1\end{array}\right]\right)=\left[\begin{array}{c}1\\ -1\end{array}\right]$. Find $[T{]}_{\mathrm{\Gamma }}^{\mathrm{\Omega }}$.  

   Based on the information given about $T$, we see that $$T\left(\left[\begin{array}{c}{x}_1\\ x_2\end{array}\right]\right)=T(x_1\left[\begin{array}{c}1\\ 0\end{array}\right]+x_2\left[\begin{array}{c}0\\ 1\end{array}\right])=x_{1}T\left(\left[\begin{array}{c}1\\ 0\end{array}\right]\right)+x_{2}T\left(\left[\begin{array}{c}0\\ 1\end{array}\right]\right)=\left[\begin{array}{c}2x_1+x_2\\ x_1-x_2\end{array}\right].$$

   Note that $$T\left(\left[\begin{array}{c}1\\ 0\end{array}\right]\right)=\left[\begin{array}{c}2\\ 1\end{array}\right]=1\left[\begin{array}{c}2\\ 1\end{array}\right]+0\left[\begin{array}{c}1\\ 0\end{array}\right],$$ and $$T\left(\left[\begin{array}{c}1\\ 1\end{array}\right]\right)=\left[\begin{array}{c}3\\ 0\end{array}\right]=0\left[\begin{array}{c}2\\ 1\end{array}\right]+3\left[\begin{array}{c}1\\ 0\end{array}\right].$$ Hence, $[T{]}_{\mathrm{\Gamma }}^{\mathrm{\Omega }}=\left[\begin{array}{cc}1& 0\\ 0& 3\end{array}\right].$