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Suppose $B$ is an $m \times n$ matrix. Prove that $BB^T$ is positive semidefinite.

Can someone give a fairly good proof?

Inputs are greatly appreciated. The question is listed above.

TMM
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sdfsdfsd
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    What does it mean for $BB^T$ to be positive semidefinite? What can you say about the product $x^T B B^T x$ for arbitrary vectors $x$? – TMM May 09 '13 at 23:46

2 Answers2

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Hint: For column vectors $a,b$, their inner (or scalar or dot) product is $$\langle a,b\rangle = a^Tb$$ using matrix product, and $a^Ta=\langle a,a\rangle=\|a\|^2$.

Berci
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$BB^T$ is positive semidefinite if for any vector $x$ $$ xBB^Tx^T\ge0 $$ Let $u=xB$. Then, $$ xBB^Tx^T=uu^T=|u|^2\ge0 $$

robjohn
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