Example2.2.1

Suppose we have the matrix \(A\) and vector \(\xvec\) as given below.

\begin{equation*} A = \left[\begin{array}{rr} -2 \amp 3 \\ 0 \amp 2 \\ 3 \amp 1 \\ \end{array}\right], \xvec = \left[\begin{array}{r} 2 \\ 3 \\ \end{array}\right] \text{.} \end{equation*}

Their product will be defined to be the linear combination of the columns of \(A\) using the componentts of \(\xvec\) as weights. This means that

\begin{equation*} \begin{aligned} A\xvec = \left[\begin{array}{rr} -2 \amp 3 \\ 0 \amp 2 \\ 3 \amp 1 \\ \end{array}\right] \left[\begin{array}{r} 2 \\ 3 \\ \end{array}\right] {}={} \amp 2 \left[\begin{array}{r} -2 \\ 0 \\ 3 \\ \end{array}\right] + 3 \left[\begin{array}{r} 3 \\ 2 \\ 1 \\ \end{array}\right] \\ \\ {}={} \amp \left[\begin{array}{r} -4 \\ 0 \\ 6 \\ \end{array}\right] + \left[\begin{array}{r} 9 \\ 6 \\ 3 \\ \end{array}\right] \\ \\ {}={} \amp \left[\begin{array}{r} 5 \\ 6 \\ 9 \\ \end{array}\right]. \\ \end{aligned} \end{equation*}

Let's take note of the dimensions of the matrix and vectors. The two components of the vector \(\xvec\) are weights used to form a linear combination of the columns of \(A\text{.}\) Since \(\xvec\) has two components, \(A\) must have two columns. In other words, the number of columns of \(A\) must equal the dimension of the vector \(\xvec\text{.}\)

In the same way, the columns of \(A\) are 3-dimensional so any linear combination of them is 3-dimensional as well. Therefore, \(A\xvec\) will be 3-dimensional.

We then see that if \(A\) is a \(3\times2\) matrix, \(\xvec\) must be a 2-dimensional vector and \(A\xvec\) will be 3-dimensional.

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