# Proving diag(A) is SPD

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.

• You are missing a justification as to why $a_{ii} \geq 0$ (not $a_{ii}$ > 0). – levap Jan 28 '16 at 10:59
• Because we know that A is symmetric positive definite, so by definiton $u^{T}Au>0$. – Roos Jansen Jan 28 '16 at 11:01
• 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 $a_{ii} > 0$. – levap Jan 28 '16 at 11:05
• Ah yes, I already did. But is the conclusion right? Diag(A) is positive semidefinite, so not positive definite? – Roos Jansen Jan 28 '16 at 11:06
• 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. – levap Jan 28 '16 at 11:14

$a_{ii} = e_i^TAe_i >0$ since $A$ is positive definite. Hence $\operatorname{diag}(A)$ is a diagonal matrix with positive diagonal entries. This is clearly positive definite.