We have repeatedly seen the importance of forming linear combinations of the columns of a matrix. As one example of this, the oft-used Theorem SLSLC, said that every solution to a system of linear equations gives rise to a linear combination of the column vectors of the coefficient matrix that equals the vector of constants. This theorem, and others, motivate the following central definition.
Suppose $A$ is an $m\times n$ matrix with columns $\vectorlist{A}{n}$ and $\vect{u}$ is a vector of size $n$. Then the matrix-vector product of $A$ with $\vect{u}$ is the linear combination \begin{equation*} A\vect{u}= \vectorentry{\vect{u}}{1}\vect{A}_1+ \vectorentry{\vect{u}}{2}\vect{A}_2+ \vectorentry{\vect{u}}{3}\vect{A}_3+ \cdots+ \vectorentry{\vect{u}}{n}\vect{A}_n \end{equation*}
So, the matrix-vector product is yet another version of “multiplication,” at least in the sense that we have yet again overloaded juxtaposition of two symbols as our notation. Remember your objects, an $m\times n$ matrix times a vector of size $n$ will create a vector of size $m$. So if $A$ is rectangular, then the size of the vector changes. With all the linear combinations we have performed so far, this computation should now seem second nature.
We can now represent systems of linear equations compactly with a matrix-vector product (Definition MVP) and column vector equality (Definition CVE). This finally yields a very popular alternative to our unconventional $\linearsystem{A}{\vect{b}}$ notation.
The set of solutions to the linear system $\linearsystem{A}{\vect{b}}$ equals the set of solutions for $\vect{x}$ in the vector equation $A\vect{x}=\vect{b}$.
The matrix-vector product is a very natural computation. We have motivated it by its connections with systems of equations, but here is another example.
Later (much later) we will need the following theorem, which is really a technical lemma (see Proof Technique LC). Since we are in a position to prove it now, we will. But you can safely skip it for the moment, if you promise to come back later to study the proof when the theorem is employed. At that point you will also be able to understand the comments in the paragraph following the proof.
Suppose that $A$ and $B$ are $m\times n$ matrices such that $A\vect{x}=B\vect{x}$ for every $\vect{x}\in\complex{n}$. Then $A=B$.
You might notice from studying the proof that the hypotheses of this theorem could be “weakened” (i.e. made less restrictive). We need only suppose the equality of the matrix-vector products for just the standard unit vectors (Definition SUV) or any other spanning set (Definition SSVS) of $\complex{n}$ (Exercise LISS.T40). However, in practice, when we apply this theorem the stronger hypothesis will be in effect so this version of the theorem will suffice for our purposes. (If we changed the statement of the theorem to have the less restrictive hypothesis, then we would call the theorem “stronger.”)