Learn About Matrix Representations of Linear Transformations

Let X and Y be finite-dimensional vector spaces. Let $T:X\to Y$ be a linear transformation. Let $A=(e_1,\dots,e_n)$ and $B=(f_1,\dots,f_m)$ be ordered bases for X and Y respectively. (An ordered basis for an n-dimensional vector space is just an n-tuple whose components are the elements of a basis). For each $y\in Y$, there’s a unique m-tuple of scalars $y_1,\dots,y_m$ such that $y=\sum_{i=1}^m y_i f_i$. These scalars are called the components of y with respect to B.

Let $i\in\{1,\dots,m\}$ and $j\in\{1,\dots,n\}$ be arbitrary. $Te_j$ is by definition of $T$ an element of Y. Since we use the notation $y_i$ for the ith component of an arbitrary $y\in Y$ with respect to B, it’s natural to use the notation $(Te_j)_i$ for the ith component of $Te_j$ with respect to B. The mn (m times n) scalars $(Te_j)_i$ with $i\in\{1,\dots,m\}$ and $j\in\{1,\dots,n\}$ are called the components, or matrix elements, of T with respect to (A,B). The m×n matrix
$$\begin{pmatrix}(Te_1)_1 & \cdots & (Te_n)_1\\ \vdots & \ddots & \vdots\\ (Te_1)_m & \dots & (Te_n)_m\end{pmatrix}$$ is called the matrix representation of T with respect to (A,B). It is often denoted by the same symbol as the linear transformation, in this case T. In situations where you would prefer to use different notations for the linear transformation and its matrix representation, a notation like $[T]$ or $[T]_{B,A}$ can be used for the latter.

The standard notation for the scalar on row i, column j of a matrix $T$ is $T_{ij}$. In this notation, we have $T_{ij}=(Te_j)_i$. This is the formula you need to remember. An alternative notation for the scalar on row i, column j is $T^i_j$. In this notation, we have $T^i_j=(Te_j)^i$. You may find it easier to remember this version of the formula since the index that’s upstairs on the left is upstairs on the right.

Given an m×n matrix M, there’s a simple way to define a linear transformation $T:X\to Y$ such that the matrix representation of T with respect to (A,B) is M. We define T to be the unique linear $T:X\to Y$ such that $Te_j=\sum_{i=1}^n T_{ij}f_i$ for all $j\in\{1,\dots,n\}$. In the alternative notation, we would write $Te_j=\sum_{i=1}^n T^i_j f_i$. You may find it easier to remember this version of the formula since the summation is over an index that appears once upstairs and once downstairs.

The following observation provides some motivation for the definitions. Let $x\in X$ be arbitrary. Define $y\in Y$ by $y=Tx$. Let $x_1,\dots,x_n$ be the components of x with respect to A, and let $y_1,\dots,y_m$ be the components of y with respect to B. We have
\begin{align*}
y &=\sum_{i=1}^m y_i f_i\\
Tx &=T\left(\sum_{j=1}^n x_j e_j\right) =\sum_{j=1}^n x_jTe_j =\sum_{j=1}^n x_j \left(\sum_{i=1}^m(Te_j)_i f_i\right) =\sum_{j=1}^n \sum_{i=1}^m x_j (Te_j)_i f_i\\
&=\sum_{i=1}^m \left(\sum_{j=1}^n x_j (Te_j)_i\right) f_i.
\end{align*}
Since $\{f_1,\dots,f_m\}$ is linearly independent, these results and the equality y=Tx imply that
$$\sum_{j=1}^n x_j (Te_j)_i= y_i$$ for all $i\in\{1,\dots,m\}$. If you recall that the definition of matrix multiplication is $(AB)_{ij}=\sum_k A_{ik}B_{kj}$, you can recognize the above as the ith row of the matrix equation
$$\begin{pmatrix}(Te_1)_1 & \cdots & (Te_n)_1\\ \vdots & \ddots & \vdots\\ (Te_1)_m & \dots & (Te_n)_m\end{pmatrix} \begin{pmatrix}x_1\\ \vdots\\ x_n\end{pmatrix} =\begin{pmatrix}y_1\\ \vdots\\ y_m\end{pmatrix}.$$

The following is a simple example of how to find a matrix representation of a linear transformation. Define $S:\mathbb R^3\to\mathbb R^2$ by $S(x,y,z)=(3z-x,2y)$. This S is linear. Let $C=(g_1,g_2,g_3)$ and $D=(h_1,h_2)$ be the standard ordered bases for $\mathbb R^3$ and $\mathbb R^2$ respectively. We will denote the matrix of S with respect to (C,D) by $[ S]$.
\begin{align*}Sg_1 &=S(1,0,0)=(-1,0) =-1h_1+0h_2\\
Sg_2 &=S(0,1,0) =(0,2)=0h_1+2h_2\\
Sg_3 &=S(0,0,1) =(3,0)=3h_1+0h_2\\
[ S] &=\begin{pmatrix}(Sg_1)_1 & (Sg_2)_1 & (Sg_3)_1\\ (Sg_1)_2 & (Sg_2)_2 & (Sg_3)_2\end{pmatrix} =\begin{pmatrix}-1 & 0 & 3\\ 0 & 2 & 0 \end{pmatrix}.
\end{align*} Note that for all $x,y,z\in\mathbb R$,
$$[ S]\begin{pmatrix}x\\ y\\ z\end{pmatrix} = \begin{pmatrix}-1 & 0 & 3\\ 0 & 2 & 0 \end{pmatrix} \begin{pmatrix}x\\ y\\ z\end{pmatrix} =\begin{pmatrix}-x+3z\\ 2y\end{pmatrix}.$$

If Y is an inner product space, and B is an orthonormal ordered basis, the easiest way to find the matrix elements is often to use the inner product. We will use the physicist’s convention for inner products. This means that the term “inner product” is defined so that when we’re dealing with a vector space over ℂ (i.e. when the set of scalars is ℂ rather than ℝ), the map $v\mapsto\langle u,v\rangle$ is linear and the map $u\mapsto\langle u,v\rangle$ is antilinear (i.e. conjugate linear). Let $i\in\{1,\dots,m\}$ and $y\in Y$ be arbitrary. Let $y_1,\dots,y_m$ be the components of y with respect to B.
$$\left\langle f_i,y\right\rangle =\left\langle f_i,\sum_{j=1}^m y_j f_j \right\rangle =\sum_{j=1}^m y_j \left\langle f_i,f_j\right\rangle =\sum_{j=1}^m y_j \delta_{ij} =y_i.$$ Since i and y are arbitrary, this implies that for all $i\in\{1,\dots,m\}$ and all $j\in\{1,\dots n\}$,
$$\left\langle f_i,Te_j \right\rangle=(Te_j)_i .$$ If X=Y, it’s convenient to choose B=A, and to speak of the matrix representation of T with respect to A instead of with respect to (A,A), or (A,B). The formula for $T_{ij}$ can now be written as
$$T_{ij}=(Te_j)_i=\left\langle e_i,Te_j \right\rangle.$$ One final comment for those of you who have studied quantum mechanics. (If you haven’t, just ignore this). In bra-ket notation, we would usually write the ith basis vector as $\left|i\right\rangle$. This turns the last formula above into
$$T_{ij} =\left\langle i\right|T\left|j\right\rangle.$$

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