91Ó°ÊÓ

: Let \(A = \begin{pmatrix} a_{11} & a_{12} & ... & a_{1n} \\ a_{21} & a_{22} & ... & a_{2n} \\ \vdots & \vdots & \ddots & \vdots \\ a_{m1} & a_{m2} & ... & a_{mn} \end{pmatrix}\) and \( B = \begin{pmatrix} b_{11} & b_{12} & ... & b_{1p} \\ b_{21} & b_{22} & ... & b_{2p} \\ \vdots & \vdots & \ddots & \vdots \\ b_{n1} & b_{n2} & ... & b_{np} \end{pmatrix} \). Now, represent column vectors of matrix A and row vectors of matrix B as \(A_{i} = \begin{pmatrix} a_{1i} \\ a_{2i} \\ \vdots \\ a_{mi} \end{pmatrix}\) and \(B_{j}^\top = \begin{pmatrix} b_{j1} & b_{j2} & ... & b_{jp} \end{pmatrix}\), where \(1 \le i \le m\) and \(1 \le j \le n\). Now, we can represent the product AB as a sum of n matrices of the form: \(A_i B_i^\top\): \[AB = A_1 B_1^\top + A_2 B_2^\top + \cdots + A_n B_n^\top\]

: Now, we want to show that each matrix \(A_i B_i^\top\) has a rank of at most one. Recall that the rank of a matrix is the maximum number of linearly independent columns (or rows) in the matrix. Observe that, for each \(A_i B_i^\top\), the columns of the resulting matrix are scalar multiples of the column vector of matrix A: \[A_i B_i^\top = A_i (b_{i1}, b_{i2}, ..., b_{ip}) = (b_{i1}A_i, b_{i2}A_i, ..., b_{ip}A_i)\] Since all the columns of the resulting matrix are scalar multiples of the same column vector, the maximum number of linearly independent columns is 1. Therefore, the rank of each \(A_i B_i^\top\) is at most one. Now, we have shown that the product matrix AB can be represented as a sum of n matrices of rank at most one: \[AB = A_1 B_1^\top + A_2 B_2^\top + \cdots + A_n B_n^\top\]