# Shae's Ramblings

Stuff I find slightly meaningful and close to achievable

## A linear map and its matrix

Suppose we have a linear map $T\colon V \to W$ where $V,W$ are two $\mathbb{R}$-vector spaces with bases $\{v_i\}_{i=1}^n$ and $\{w_j\}_{j=1}^m$.
The matrix representation $\mathrm{Mat}(T) = \mathrm{A}\in\mathbb{R}^{m\times n}$ of the map $T$ is easy to find: first expand any $v\in V$ in terms of the basis vectors:

$$T( v ) = T \left(\sum_{i=1}^n \alpha_i\,v_i\right)$$

Then use the linearity of the map $T$ to obtain:

$$T( v ) = \sum_{i=1}^n \alpha_i\,T(v_i)$$

finally, expand each vector $T(v_i)$ in terms of the basis vectors of $W$:

$$T(v_i) = \sum_{j=1}^m a_{ji} w_j$$

This gives us the equality for any $v\in V$:
$$T(v) = \sum_{j=1}^m \sum_{i=1}^n \alpha_i a_{ji} w_j$$

The coefficients $\vec{\alpha} = [\alpha_1\,,\, \dots \,,\, \alpha_n]$ uniquely characterize the vector $v$, and similarly the coefficients $a_{ij}$ are uniquely characterize the map $T$ given the bases of $V,W$.
Because the expansion of $T(v)$ is unique in the basis $\{w_j\}_j$, we obtain that $T(v)$ has the coefficients $\beta_j = \sum_{k=1}^n a_{jk}\alpha_k$.
We can then only think of linear maps as multiplication and addition of the coefficients $\alpha_i$ with the entries $a_{ij}$.

We can then define the array $\mathrm{A} = \left[ a_{ij} \right]_{ij}$ and matrix multiplication $\mathrm{A} \vec{\alpha}$ as expected.

## Composition as matrix multiplication

Let $T,S,R$ be linear maps with matrices $\mathrm{A},\mathrm{B},\mathrm{C}$. The third map $R$ is the composition $S\circ T$.
Given bases for the spaces $V\xrightarrow{T} W \xrightarrow{S} Z$, let us denote

• $\vec{\alpha} = [\alpha_1\,,\, \dots \,,\, \alpha_n]$ the unique coefficients representing an arbitrary $v\in V$ using the basis for $V$,
• $\vec{\beta} = [\beta_1\,,\, \dots \,,\, \beta_m]$ the unique coefficients representing $w = T(v)\in W$
• $\vec{\gamma} = [\gamma_1\,,\, \dots \,,\, \gamma_p]$ the unique coefficients representing $z = S(w)\in W$

From the first section, we know that $T$ has matrix representation $\mathrm{A} = \left[a_{ij}\right]_{ij}$ so that the ouput coefficients are

$$\beta_j = \sum_{k=1}^n a_{jk} \alpha_k$$

and that $S$ has representation $\mathrm{B} = \left[b_{ij}\right]_{ij}$ so that

$$\gamma_q = \sum_{j=1}^m b_{qj} \beta_j$$

therefore, we have the equalities

\begin{align}
\gamma_q &= \sum_{j=1}^m b_{qj} \sum_{k=1}^n a_{jk} \alpha_k \\
&= \sum_{k=1}^n \left(\sum_{j=1}^m b_{qj}a_{jk}\right) \alpha_k \\
&= \sum_{k=1}^n c_{qk}\alpha_k
\end{align}

thus we obtain that the matrix $\mathrm{C}$ has coefficients $c_{qk} = \sum_{j=1}^m b_{qj}a_{jk} = \left[ \mathrm{BA} \right]_{qk}$.
We thusly proved that $\mathrm{C} = \mathrm{Mat}(S\circ T) = \mathrm{BA} = \mathrm{Mat}(S)\mathrm{Mat}(T)$.