Computing S*S' and comparing to eye(5)

Question

In matlab, we asked to set A=rand(5,2)*rand(2,5)

then to set Q=orth(A), W=null(A'), S=[Q W]

Compute S*S' and compare your result with eye(5).

$$\text{In Theory $A^TW$ and $W^TA$ should consist entirely of zeros. Why?}$$

need help answering this question.

Answer 1

The columns $w_1,\dots,w_k$ of $W$ are in the nullspace of $A^T$. It follows that $$ A^TW = A^T[w_1\; \cdots \; w_k] = [A^T w_1 \; \cdots \; A^T w_k] = [0\; \cdots \; 0] $$ So, $A^TW = 0$. It follows that $W^TA = (A^TW)^T = 0^T = 0$.

Because $S = [Q\; W]$ is orthogonal (as I explain here), S*S' and eye(5) (the $5 \times 5$ identity matrix) should be the same (up to rounding errors in the computation).