Discussion
If \(T : \mathbf{R}^{n} \to \mathbf{R}^{m}\) is a linear map, then it is represented by a matrix of order :
Answer: A
The matrix is of order \(m \times n\).
The matrix of \(T\) has one column for each basis vector of the domain \(\mathbf{R}^{n}\), namely \(T(e_{j})\), and each column is a vector in \(\mathbf{R}^{m}\), so it has \(m\) entries. That gives \(m\) rows and \(n\) columns, and \(T(x) = Ax\) is defined because an \(m \times n\) matrix times an \(n \times 1\) vector is \(m \times 1\).
\(n \times m\) is the transpose, which maps \(\mathbf{R}^{m}\) to \(\mathbf{R}^{n}\). Square orders \(m \times m\) or \(n \times n\) would fit only when \(m = n\).
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