0
$\begingroup$

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

1 Answers 1

3

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.

  • 0
    Could you please elaborate more on $a_{ii} = e_i^T A e_i > 0$. I did not understand it.2012-11-15
  • 0
    I added the definition of $e_i$ to my answer. Use the definition of matrix multiplication to check that $a_{ii} = e_i^T A e_i$.2012-11-15
  • 0
    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.2012-11-16
  • 0
    Could you please elaborate more on your answer.2012-11-20
  • 0
    Explain what you don't understand.2012-11-20
  • 0
    How did you infer $a_{ii} = e_i^T A e_i > 0$2012-11-21
  • 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