13 Matrices and Linear Maps from $\R^n$ to $\R^m$
$\newcommand{\mij}[1]{\begin{pmatrix} #1_{11} & #1_{12} & \cdots & #1_{1n} \\ #1_{21} & #1_{22} & \cdots & #1_{2n} \\ \vdots & \vdots & \ddots & \vdots\\ #1_{m1} & #1_{m2} & \cdots & #1_{mn} \end{pmatrix}} \newcommand{\mrc}[3]{\begin{pmatrix} #1_{11} & #1_{12} & \cdots & #1_{1#3} \\ #1_{21} & #1_{22} & \cdots & #1_{2#3} \\ \vdots & \vdots & \ddots & \vdots\\ #1_{#21} & #1_{#22} & \cdots & #1_{#2#3} \end{pmatrix}} $
Observe that the set, $M_{m\times n}$, of $m\times n$ matrices can be made into a vector space by defining vector addition and scalar multiplication entry-wise: $$\mij a + \mij b = \begin{pmatrix} a_{11} + b_{11} & \cdots & a_{1n} + b_{1n} \\ a_{21} + b_{21} & \cdots & a_{2n} + b_{2n} \\ \vdots & \dots & \vdots\\ a_{m1} + b_{m1} & \cdots & a_{mn} + b_{mn} \end{pmatrix} $$ and $$r\cdot\mij c = \mij{rc}$$ It is clear that $M_{m\times n}$ is isomorphic to $\R^{mn}$, since an $m\times n$ matrix has $mn$ entries.
We can also define another kind of operation on matrices called matrix multiplication. A matrix multiplication operation takes an $m\times k$ matrix and a $k\times n$ matrix, and it outputs an $m\times n$ matrix: $$\mrc amk \mrc bkn = \mrc cmn$$ where $$c_{ij} = a_{i1}b_{1j}+ a_{i2}b_{2j}+ a_{ik}b_{kj}$$ That is, the $i,j$ entry in the matrix product is the inner product of the $i^{\rm th}$ row vector of the first matrix with the $j^{th}$ column vector of the second matrix. Note that the number of rows in the first matrix must equal the number of columns in the second matrix.
The $n\times n$ identity matrix is defined to have 1's along the main (descending) diagonal, and 0's everywhere else: $$I_n\begin{pmatrix} 1 & 0 & 0 &\cdots &0 \\ 0 & 1 & 0 &\cdots &0 \\ 0 & 0 & 1 &\cdots &0 \\ \vdots&\vdots&\vdots&\ddots&\vdots\\ 0 & 0 & 0 &\cdots &1 \\ \end{pmatrix}$$ This matrix satisfies $AI_n=A$ and $I_nB=B$ for all matrices $A,B$ for which the products are defined.
Given an $n\times n$ matrix $A$, an inverse matrix for $A$ is a matrix $A^{-1}$ satisfying $AA^{-1}=I_n$ and $A^{-1}A=I_n$. Not every $n\times n$ matrix has an inverse. A matrix that has an inverse is said to be invertible. In fact, a matrix is invertible if and only if it is non-singular, which is true if and only if when it is put into reduced echelon form, the result is the identity matrix.
There is a straightforward technique for finding the inverse of a matrix, if it exists. For an $n\times n$ matrix $A$, form the augmented matrix $(A|I_n)$, where $I_n$ is the identity matrix. Put the matrix into reduced echelon form. If the resulting augmented matrix has the form $(I_n|B)$, then $A$ is is invertible and $B=A^{-1}$.
In particular, when the first matrix is $m\times n$ and the second is an $n\times 1$ matrix, we are looking at the multiplication of a matrix times a column vector of length $n$. The result is a column vector of length $m$ $$\mrc amn \begin{pmatrix}b_1\\b_2\\\vdots\\b_n\end{pmatrix} =\begin{pmatrix} a_{11}b_1 + a_{12}b_2 + \cdots + a_{1n}b_n\\ a_{21}b_1 + a_{22}b_2 + \cdots + a_{2n}b_n\\ \vdots\\ a_{m1}b_1 + a_{m2}b_2 + \cdots + a_{mn}b_n \end{pmatrix} $$ This defines a function, $h$, from $\R^n$ to $\R^m$, and that function is a homomorphism. If you compute $h(\vec e_i)$ for the standard basis vector $\vec e_i$, you can see that $h(\vec e_i)$ is just the $i^{\rm th}$ column of the matrix. We get a relationship between $m\times n$ matrices and homomorphisms from $\R^n$ to $\R^n$.
Theorem: Let $A$ be an $m\times n$ matrix. Define the function $h_A\colon \R^n \to \R^m$ by $h_A(\vec v)= A\vec v$ (where $A\vec v$ represents the matrix product of the $m\times n$ matrix $A$ with $\vec v$ considered as an $n\times 1$ matrix). Then $h_A$ is a homomorphism. For each standard basis vector $\vec e_i$ in $\R^n$, $h(\vec e_i)$ is the $i^{\rm th}$ column vector of $A$. The range space of $h_A$ is the column space of $A$. The null space of $A$ is the set of solutions of the homogeneous system corresponding to $A$. The rank of $h_A$ is the rank of $A$, which is also the number of leading variables in $A$. The nullity of $h_A$ is the number of free variables in $A$, which is also $n$ minus the rank of $A$.
Recall that $L(\R^n,\R^m)$ is the vector space of all homomorphisms from $\R^n$ to $\R^m$. We have associated each $m\times n$ matrix $A$ with a homomorphism $h_A\colon \R^n\to\R^m$. This association defines a function $\mathscr H\colon M_{m\times n} \to L(\R^n,\R^m)$, given by $\mathscr H(A) = h_A$. This function is in fact an isomorphism. Thus, the space of $m\times n$ matrices is vector space isomorphic to the space of linear maps $L(\R^n,\R^m)$.
But note that this isomorphism is defined using the standard bases of $\R^n$ and $\R^m$. Our goal is to use matrices to represent arbitrary homomorphisms between finite-dimensional vector spaces. To do that, we must choose a basis for each vector space, and the matrix representation that we get for a homomorphism will depend on those choices. Even for $\R^n$ and $\R^m$, if we use bases different from the standard bases, we get a different matrix representation than the one that we have considered here.
One final note: Suppose that $f\colon \R^n\to \R^k$ and $g\colon \R^k\to \R^m$ are homomorphisms. Then the composition $g\circ f\colon \R^n\to\R^m$ is also a homormorphism. Now, $f$ can be represented by a $k\times n$ matrix $B$, $g$ can be represented by an $m\times k$ matrix $A$, and $g\circ f$ can be represented by an $m\times n$ matrix $C$. If you follow the effects of these matrices on the standard bases, you will find that $C=AB$. That is the matrix for a composition of homomorphisms is the matrix product of the matrices for the individual homomorphisms.