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Let $A \in \mathbb{C}^{n \times n}$ be a Hermitian, positive-definite matrix, i.e. $A=A^*$ and $\langle Ax,x \rangle >0$ for all $x \neq 0$. Moreover, one can show that this implies that all the diagonal entries are real and positive.

My question is: if I change all of the diagonal entries of my matrix $A$ by multiplying them by some positive number (let's say $2$ or $\frac{1}{2}$), will my matrix remain positive-definite? In other words, I want to multiply all of the diagonal entries by the same positive number while keeping the off diagonal entries unchanged.

Any help would be greatly appreciated.

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    $\begingroup$ If you multiply the diagonal with a number $c>1$, then yes. If you multiply it with a number $0<c<1$, then not necessarily. Counterexample: $A=\begin{bmatrix}1&0.9\\0.9&1\end{bmatrix}, c=0.1$. $\endgroup$ – Rahul Nov 16 '17 at 1:45
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As mentioned in the comments this holds for multiplication with a number $c\geq 1$, because if you take $P_i$ to be the diagonal matrix where $(P_i)_{ii}=1$ and all other entries are zero then $P_i$ is a positive matrix. Then for any positive definite matrix $A$ you can define $A'$ as the matrix you described, i.e. multiplying all diagonal entries of $A$ with some $c \geq 1$. Then we have $$ A' = A + \sum_{i=1}^n(c-1)A_{ii}P_i. $$ Now note that $$ \langle A'x, x\rangle = \langle Ax, x\rangle + (c-1) \sum_{i=1}^n A_{ii} |x_i|^2>0 $$ because $A_{ii} \geq 0$. It doesn't however hold for $0<c <1$, note for example $$ A = \left( \begin{matrix} 1 & 1/2 \\ 1/2 & 1\end{matrix} \right) $$ with eigenvalues $3/2,1/2$. If $c=1/2$, then $$ A' = \left( \begin{matrix} 1/2 & 1/2 \\ 1/2 & 1/2\end{matrix} \right) $$ which has eigenvalues $1, 0$ and thus isn't positive definite.

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  • $\begingroup$ Thank you for your answer. However, how is your matrix $A$ positive-definite if one of the eigenvalues is 0? $\endgroup$ – Pawel Nov 16 '17 at 2:13
  • $\begingroup$ @Pawel I overlooked the definitive part, I will edit it. $\endgroup$ – Demophilus Nov 16 '17 at 2:16
  • $\begingroup$ @Pawel I edited the answer, this should do. $\endgroup$ – Demophilus Nov 16 '17 at 2:41
  • $\begingroup$ thank you for your edit. I will look through it this weekend and then I will upvote it and accept it. $\endgroup$ – Pawel Nov 16 '17 at 14:27

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