Matrix as a Linear Transformation

As a quick review, a linear transformation is any transformation that is additive and first-degree homogeneous.

It turns out that most linear transformations may be expressed by a matrix. This holds as long a basis is provided for each of the two vector spaces, which is almost always the case.

Suppose that we have the linear transformation $f:V\mapsto W$, where $V$ is an $m$-dimensional vector space and $W$ is an $n$-dimensional vector space.

Let us say that we have a vector $X$, and we can express it in terms of the basis vectors of $V$ as follows: $$X=a_1V_1+a_2V_2+...+a_mV_m$$ Then, using the properties of a linear transformation: $$f(X)=a_1f(V_1)+a_2f(V_2)_...+a_mf(V_m)$$ This tells us that the value of $f(X)$, for any vector $X$ in $m$-dimensional space, is determined completely by the value of the function for all of the $m$ basis vectors.

Notice that $f(X)$ is expressed as a linear combination of $f(V_1)$, $f(V_2)$, $...$, $f(V_m)$. This may be expressed perfectly as this matrix with somewhat confusing notation: $$\left[ \begin{array}{ccc} f(V_1) & f(V_2) & ... & f(V_m) \end{array} \right] \left[ \begin{array}{ccc} a_1 \\ a_2 \\ ... \\ a_m \end{array} \right] =f(X)$$ Note that each one of $f(V_1)$, $f(V_2)$, $...$, $f(V_m)$ is meant to be a column vector of length $n$, and that $f(X)$ is a column vector of length $n$. This is only possible if we also have basis vectors for $W$, so this is where the other condition comes into play.

Thus, we have that the product of an $n\times m$ matrix and an $m\times 1$ matrix is an $n\times 1$ matrix, so the above formula represents a valid matrix multiplication.

A linear transformation is any mapping with the properties of additivity and first-degree homogeneity. In other words, a function $f$ is a linear transformation if and only if the following holds for all vectors $X$, $Y$ and scalars $\alpha$: $$f(X+Y)=f(X)+f(Y)$$ $$f(\alpha X)=\alpha f(X)$$

In the end, what this means for a function is as follows: if we have a function that maps $m$-dimensional space to $n$-dimensional space, then: $$f(\left<a_1,a_2,...,a_m\right>)=\left<\sum_{i=1}^{m}k_{i1}a_i,\sum_{i=1}^{m}k_{i2}a_i,...,\sum_{i=1}^{m}k_{in}a_i\right>$$ for some real numbers $k_{ij}, 1\le i\le m, 1\le j\le n$. Basically, every component of the value of the function is some linear combination of the components of the input.

This actually isn't so hard to prove, although it certainly looks a bit scary. But don't worry, you will soon see a much more concise and appealing way of dealing with all of those summations that clog up our expressions.