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Can someone help me with determining if the following statement is true: If A is symmetric positive definite then diag(A) is symmetric positive definite.

What I have done is:

$u^{T}(diag(A))u=\sum_{i=1}^{n} \sum_{j=1}^{n} a_{ij}u_{i}u_{j}=\sum _{i=1}^{n} a_{ii}u_{i}^2 + \sum_{i \neq j}^{n} a_{ij}u_{i}u_{j}$. The last term is equal to 0 because we have a diagonal matrix. The first term is bigger or equal 0 because $a_{ii}>0$ and $u_{i}^2 \geq 0$ So diag(A) is positive semidefinite.

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  • $\begingroup$ You are missing a justification as to why $a_{ii} \geq 0$ (not $a_{ii}$ > 0). $\endgroup$ – levap Jan 28 '16 at 10:59
  • $\begingroup$ Because we know that A is symmetric positive definite, so by definiton $u^{T}Au>0$. $\endgroup$ – Roos Jansen Jan 28 '16 at 11:01
  • $\begingroup$ But why does this imply that $a_{ii} > 0$? You need to choose an appropriate $u$ and use $u^T A u > 0$ to conclude that $a_{ii} > 0$. $\endgroup$ – levap Jan 28 '16 at 11:05
  • $\begingroup$ Ah yes, I already did. But is the conclusion right? Diag(A) is positive semidefinite, so not positive definite? $\endgroup$ – Roos Jansen Jan 28 '16 at 11:06
  • $\begingroup$ If $A$ is positive definite, then $\mathrm{diag}(A)$ will be positive definite. If $A$ is positive semidefinite, then $\mathrm{diag}(A)$ will be positive semidefinite. $\endgroup$ – levap Jan 28 '16 at 11:14
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$a_{ii} = e_i^TAe_i >0$ since $A$ is positive definite. Hence $\operatorname{diag}(A)$ is a diagonal matrix with positive diagonal entries. This is clearly positive definite.

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