# The Unapologetic Mathematician

## Matrices I

Yesterday we talked about the high-level views of linear algebra. That is, we’re discussing the category $\mathbf{Vec}(\mathbb{F})$ of vector spaces over a field $\mathbb{F}$ and $\mathbb{F}$-linear transformations between them.

More concretely, now: we know that every vector space over $\mathbb{F}$ is free as a module over $\mathbb{F}$. That is, every vector space has a basis — a set of vectors so that every other vector can be uniquely written as an $\mathbb{F}$-linear combination of them — though a basis is far from unique. Just how nonunique it is will be one of our subjects going forward.

Now if we’ve got a linear transformation $T:V\rightarrow W$ from one finite-dimensional vector space to another, and if we have a basis $\{f_j\}_{j=1}^{\mathrm{dim}(V)}$ of $V$ and a basis $\{g_k\}_{k=1}^{\mathrm{dim}(W)}$ of $W$, we can use these to write the transformation $T$ in a particular form: as a matrix. Take the transformation and apply it to each basis element of $V$ to get vectors $T(f_j)\in W$. These can be written uniquely as linear combinations $\displaystyle T(f_j)=\sum\limits_{k=1}^{\mathrm{dim}(W)}t_j^kg_k$

for certain $t_j^k\in\mathbb{F}$. These coefficients, collected together, we call a matrix. They’re enough to calculate the value of the transformation on any vector $v\in V$, because we can write $\displaystyle v=\sum\limits_{j=1}^{\mathrm{dim}(V)}v^jf_j$

We’re writing the indices of the components as superscripts here, just go with it. Then we can evaluate $T(v)$ using linearity $\displaystyle T(v)=T\left(\sum\limits_{j=1}^{\mathrm{dim}(V)}v^jf_j\right)=\sum\limits_{j=1}^{\mathrm{dim}(V)}v^jT(f_j)=$ $\displaystyle=\sum\limits_{j=1}^{\mathrm{dim}(V)}v^j\sum\limits_{k=1}^{\mathrm{dim}(W)}t_j^kg_k=\sum\limits_{k=1}^{\mathrm{dim}(W)}\left(\sum\limits_{j=1}^{\mathrm{dim}(V)}t_j^kv^j\right)g_k$

So the coefficients $v^j$ defining the vector $v\in V$ and the matrix coefficients $t_j^k$ together give us the coefficients defining the vector $T(v)\in W$.

If we have another finite-dimensional vector space $U$ with basis $\{e_i\}_{i=1}^{\mathrm{dim}(U)}$ and another transformation $S:U\rightarrow V$ then we have another matrix $\displaystyle S(e_i)=\sum_{j=1}^{\mathrm{dim}(V)}s_i^jf_j$

Now we can compose these two transformations and calculate the result on a basis element $\displaystyle \left[T\circ S\right](e_i)=T\left(S(e_i)\right)=T\left(\sum_{j=1}^{\mathrm{dim}(V)}s_i^jf_j\right)=\sum_{j=1}^{\mathrm{dim}(V)}s_i^jT(f_j)=$ $\displaystyle=\sum_{j=1}^{\mathrm{dim}(V)}s_i^j\sum\limits_{k=1}^{\mathrm{dim}(W)}t_j^kg_k=\sum\limits_{k=1}^{\mathrm{dim}(W)}\left(\sum_{j=1}^{\mathrm{dim}(V)}t_j^ks_i^j\right)g_k$

This last quantity in parens is then the matrix of the composite transformation $T\circ S$. Thus we can represent the operation of composition by this formula for matrix multiplication.

May 20, 2008 Posted by | Algebra, Linear Algebra | 7 Comments