Question

Let \(A\) be an \(m \times n\) matrix of rank \(r > 0\) and let \(U\) be an echelon form of \(A\). Explain why there exists an invertible matrix \(E\) such that \(A = EU\), and use this factorization to write \(A\) as the sum of \(r\) rank 1 matrices. [Hint: See Theorem 10 in Section 2.4.]

Short answer

There exists an invertible matrix such that \(A = EU\).

Step-by-step solution

State the column-row expansion of \(AB\)

Theorem 10 states that if \(A\) is \(m \times n\) and \(B\) is \(n \times p\), then

\(\begin{array}{c}AB = \left[ {\begin{array}{*{20}{c}}{{{{\mathop{\rm col}\nolimits} }_1}\left( A \right)}&{{{{\mathop{\rm col}\nolimits} }_2}\left( A \right)}& \cdots &{{{{\mathop{\rm col}\nolimits} }_n}\left( A \right)}\end{array}} \right]\left[ {\begin{array}{*{20}{c}}{{{{\mathop{\rm row}\nolimits} }_1}\left( B \right)}\\{{{{\mathop{\rm row}\nolimits} }_2}\left( B \right)}\\ \vdots \\{{{{\mathop{\rm row}\nolimits} }_n}\left( B \right)}\end{array}} \right]\\ = {{\mathop{\rm col}\nolimits} _1}\left( A \right){{\mathop{\rm row}\nolimits} _1}\left( B \right) + \cdots + {{\mathop{\rm col}\nolimits} _n}\left( A \right){{\mathop{\rm row}\nolimits} _n}\left( B \right).\end{array}\)

Explain the existence of an invertible matrix \(E\)

Consider \(A\) is an \(m \times n\) matrix with rank \(r > 0\), and \(U\) is an echelon form of \(A\). Since \(A\) can be reduced to \(U\) by row operations, there exist invertible elementary matrices \({E_1},...,{E_p}\) with \(A = U\) because the product of invertible matrices is invertible. Therefore, \(A = {\left( {{E_p},...,{E_1}} \right)^{ - 1}}U\).

Consider \(E = {\left( {{E_p},...,{E_1}} \right)^{ - 1}}\), then \(A = EU\).

Let \({c_1},...,{c_n}\) represent the column of \(E\). \(U\) has \({\mathop{\rm r}\nolimits} \) non-zero rows that can be represented as \({\mathop{\rm d}\nolimits} _1^T,...,{\mathop{\rm d}\nolimits} _r^T\) because the rank of \(A\) is \({\mathop{\rm r}\nolimits} \) (according to theorem 10).

\(\begin{aligned} A &= EU\\ &= \left[ {\begin{array}{*{20}{c}}{{c_1}}& \cdots &{{c_m}}\end{array}} \right]\left[ {\begin{array}{*{20}{c}}{{\mathop{\rm d}\nolimits} _1^T}\\ \vdots \\{{\mathop{\rm d}\nolimits} _r^T}\\0\\ \vdots \\0\end{array}} \right]\\ &= {c_1}{\mathop{\rm d}\nolimits} _1^T + \cdots + {c_r}{\mathop{\rm d}\nolimits} _r^T\end{aligned}\)

The sum of the \(r\) rank-1 matrices is \(A = {c_1}{\mathop{\rm d}\nolimits} _1^T + \cdots + {c_r}{\mathop{\rm d}\nolimits} _r^T\).

Thus, there exists an invertible matrix such that \(A = EU\).