# A calculus of the absurd

##### 20.5.3 The matrix representation of a linear transformation

This is just an extension of the previous section.

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

$$\mathbf {v} = \sum _{1 \leqq i \leqq n} v_i b_i$$

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 20.123) is the co-ordinate vector (in the ordered base $$\mathcal {C}$$) of the result of $$T(\beta _j)$$.