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Exercise 3.2.22

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We know that $B^{t}$ is an $m \times n$ matrix with rank $n$ . By the previous exercise we have some $n \times m$ matrix $C$ such that $B^{t} C = I_{m}$ . We may pick $A = C^{t}$ . Now we have the fact that $AB = C^{t} B = {(B^{t} C)}^{t} = {(I_{m})}^{t} - I_{m}$ .

jephianlin

2011-06-27 00:00

Exercise 3.2.22

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