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The title says it all. Why does positive semi definiteness implies positivity on diaginal elements.

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If $A = (a_{ij})$ is a positive definite matrix then $v^T A v > 0$ for every vector $v\neq 0$. In particular, $a_{ii} = e_i^T A e_i > 0$, where $$e_i= \begin{pmatrix} 0\\ \vdots\\ 0\\ 1\\ 0\\ \vdots\\ 0 \end{pmatrix} $$ is the vector whose $i$-th coordinate is 1, and all other coordinates are 0.

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Could you please elaborate more on $a_{ii} = e_i^T A e_i > 0$. I did not understand it. –  Aman Deep Gautam Nov 15 '12 at 23:21
    
I added the definition of $e_i$ to my answer. Use the definition of matrix multiplication to check that $a_{ii} = e_i^T A e_i$. –  Yury Nov 15 '12 at 23:27
    
I do not get it, this what my question says. Actually it is just mathematical way of expressing my question. What I would like to know is the proof for this. –  Aman Deep Gautam Nov 16 '12 at 14:51
    
Could you please elaborate more on your answer. –  Aman Deep Gautam Nov 20 '12 at 16:35
    
Explain what you don't understand. –  Yury Nov 20 '12 at 17:56
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