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.
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.