The title says it all. Why does positive semi definiteness implies positivity on diaginal elements.
Why does positive definiteness implies positivity on diagnal
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matrices
1 Answers
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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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0Could you please elaborate more on $a_{ii} = e_i^T A e_i > 0$. I did not understand it. – 2012-11-15
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0I added the definition of $e_i$ to my answer. Use the definition of matrix multiplication to check that $a_{ii} = e_i^T A e_i$. – 2012-11-15
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0I 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. – 2012-11-16
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0Could you please elaborate more on your answer. – 2012-11-20
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0Explain what you don't understand. – 2012-11-20
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0How did you infer $a_{ii} = e_i^T A e_i > 0$ – 2012-11-21
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0$a_{ii} = e_i^T A e_i$ by the definition of matrix multiplication; $e_i^T A e_i > 0$ since $A$ is positive definite. – 2012-11-21