Let \(A=(a_{ij})\in F^{m\times n}\), let \(\mathbf x=(x_1,\dots,x_n)^t\in F^n\), and let \(\mathbf b=(b_1,\dots,b_m)^t\in F^m\). Prove that the matrix equation \[ A\mathbf x=\mathbf b \] is equivalent to the system of linear equations \[ a_{i1}x_1+\cdots+a_{in}x_n=b_i, \qquad i=1,\dots,m. \]
Solution
Let \(\mathbf a_i=(a_{i1},\dots,a_{in})\) be the \(i\)th row of \(A\).
1. The \(i\)th coordinate of \(A\mathbf x\) is\[ \sum_{j=1}^n a_{ij}x_j. \]
This is the definition of matrix-vector multiplication: the \(i\)th coordinate is the dot product of the \(i\)th row of \(A\) with the column vector \(\mathbf x\).
2. If \(A\mathbf x=\mathbf b\), then \(x_1,\dots,x_n\) satisfy the displayed system.
Equality of the two vectors implies equality of their \(i\)th coordinates for every \(i\). By 1, \[ \sum_{j=1}^n a_{ij}x_j=b_i \] for each \(i=1,\dots,m\).
3. Conversely, if \(x_1,\dots,x_n\) satisfy the displayed system, then \(A\mathbf x=\mathbf b\).
By assumption, for every \(i\) the \(i\)th coordinate of \(A\mathbf x\) equals \(b_i\). Hence the vectors \(A\mathbf x\) and \(\mathbf b\) have equal coordinates, so they are equal.