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As the question says, D is a diagonally positive matrix(i.e all the diagonal elements are > 0 and remaining are 0) and A is a skew symmetric matrix. Is there a way to prove this. I understand that I have to prove that the eigen values of (D+A) are non-zero

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  • $\begingroup$ I realized that. But I am trying to find the eigen values of (D+A). how can i get a non zero eigen value of the matrix(D+A) to show that it is non singular $\endgroup$ – Morpheus Oct 9 '15 at 14:44
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Yes, it is invertible, because it is positive definite: Let $B=D+L-L^T$ where $D$ is the diagonal matrix with positive entries and $L$ is the lower part of the skew symmetric matrix $A$ , and $-L^T$ is the upper part. Then you have $$(Bx,x)=((D+L-L^T)x,x)=(Dx,x)+(Lx,x)-(L^Tx,x)=(Dx,x)+(L^Tx,x)-(L^Tx,x)=(Dx,x)>0$$ for each nonzero vector $x$. We used $(Lx,x)=(Lx)^Tx=x^TL^Tx=(x,L^Tx)=(L^Tx,x)$

Addition: Why from $(Bx,x)>0,\,\,\forall x\neq 0$ follows that $B$ is invertible ? Because, if you assume that it is not, then there is $x\neq 0: Bx=0$. Then, you get $(Bx,x)=(0,x)=0$ which is a contradiction.

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