A calculus of the absurd

22.4.2 The matrix representation of a linear transformation

The best way to understand this is to do a lot of examples, with specific linear transformations and vector spaces. It’s easy to get lost, sinking, “not waving but drowning” in the steaming soup of generality. As they don’t say, a little reification ¹³⁵¹³⁵ Meaning turning something abstract into something concrete. every day keeps the doctor away.

Let’s assume that we have a linear transformation $T: V \to W$, and we would like to find its matrix representation. It’s really easy to get confused here, but don’t lose sight of the goal. We need some information about $V$ and $W$, specifically

• A basis for $V$, denoted as $\mathcal {B} = \{\beta _i\}_{1 \leqq i \leqq n}$.
• A basis for $W$, denoted as $\mathcal {C} = \{\gamma _i\}_{1 \leqq i \leqq m}$.

We then pick an arbitrary vector, $\mathbf {v} \in V$, and finds its representation as a linear combination of $\beta $, that is we find

\begin{equation} \mathbf {v} = \sum _{1 \leqq i \leqq n} v_i b_i \end{equation}

But we’re not after $\mathbf {v}$, we’re after $T(\mathbf {v})$! Therefore, we apply $T$ to both sides, giving

\begin{align} T(\mathbf {v}) &= T\left (\sum _{1 \leqq j \leqq n} v_j \beta _j\right ) \\ \end{align}

We now use liberally the fact that $T$ is linear.

\begin{align} T(\mathbf {v}) &= \sum _{1 \leqq j \leqq n} v_j T(\beta _j) \end{align}

We’re not dealing with a concrete linear transformation, so “all” we can say is that for each $i$, $T(\beta _j)$ will give us a vector in $W$ and that we can certainly write this as a linear combination of $\mathcal {C}$, as it is a basis for $W$. Every $T(\beta _j)$ is a linear combination of the $m$ vectors in $\mathcal {C}$, i.e. $T(\beta _j) = \sum _{1 \leqq i \leqq m} \left (a_{i,j}\right ) \gamma _i$. Substituting this in, we get

\begin{align} T(\mathbf {v}) &= \sum _{1 \leqq j \leqq n} v_j \left (\sum _{1 \leqq i \leqq m} a_{i,j} \gamma _i\right ) \\ &= \sum _{1 \leqq i \leqq m} \gamma _i \left (\sum _{1 \leqq j \leqq n} a_{i, j} v_j\right ) \\ &= \sum _{1 \leqq i \leqq m} \left (\sum _{1 \leqq j \leqq n} a_{i, j} v_j\right ) \gamma _i \end{align}

Now, from the definition of a co-ordinate vector, as $\gamma _1, ..., \gamma _m$ are the basis vectors for $\mathcal {C}$, the representation of $T(v)$ as a co-ordinate vector in this basis is just

\begin{align} [T(v)]_{\mathcal {C}} &= \begin{pmatrix} \sum _{1 \leqq j \leqq n} a_{1, j} v_j \\ \sum _{1 \leqq j \leqq n} a_{2, j} v_j \\ ... \\ \sum _{1 \leqq j \leqq n} a_{m, j} v_j \\ \end {pmatrix} \\ &= \underbrace { \begin{pmatrix} a_{1, 1} & a_{1, 2} & ... & a_{1, n} \\ a_{2, 1} & a_{2, 2} & ... & a_{2, n} \\ ... & ... & ... & ... \\ a_{m, 1} & a_{1, 2} & ... & a_{1, n} \end {pmatrix} }_{\text {Let this be $\mathbf {A}$.}} \begin{pmatrix} v_1 \\ v_2 \\ ... \\ v_3 \end {pmatrix} \label {where we defined A} \end{align}

Which is exactly what we wanted to find. Specifically, the $j$th column of the matrix $\mathbf {A}$ (as defined in Equation 22.123) is the co-ordinate vector (in the ordered base $\mathcal {C}$) of the result of $T(\beta _j)$.