Given a trace-$0$ matrix $W \in \mathbb{R}^{n \times n}$ with an equal number of positive and negative eigenvalues, I want to find matrix $L$ such that $L^T L = W$. Can one do such a factorisation?Right now I am able to factorise it as $$D=A\otimes A^{T}-C\otimes C^{T}$$ The main goal is to prove that the matrix has got equal number of positive and negative Eigen values.
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Let $A^tA$ have eigenvector $v$ with eigenvalue $\lambda$. Then $(Av)^tAv=v^tA^tAv=v^t\lambda v=\lambda v^tv$, but $(Av)^tAv$ is nonnegative and $v^tv$ is positive so $\lambda\ge0$. – Gerry Myerson Aug 09 '21 at 02:50
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@GerryMyerson This means such a decomposition is not possible for a matrix with equal number of positive and negative Eigen values. Is that what you meant? – Jasmine Aug 09 '21 at 02:53
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It means it's not possible for such a matrix to have any negative eigenvalues at all. – Gerry Myerson Aug 09 '21 at 02:54
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If I got W=AA^T-BB^T, Is it possible to prove that W has got equal number of positive and negative eigen values – Jasmine Aug 09 '21 at 02:55
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I don't know. Have you tried any examples? – Gerry Myerson Aug 09 '21 at 02:55
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Have you tried letting $A$ and $B$ be different multiples of the identity? – Gerry Myerson Aug 09 '21 at 13:09
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@GerryMyerson I didn’t understand that. Can you explain? W is a matrix with equal number of positive and negative eigenvalues. That’s already proved numerically. W can be written as AA^T-BB^T. That is already proved. What needs to be proved is that from this structure, any analytical proof that it has got equal number of positive and negative eigen values – Jasmine Aug 09 '21 at 22:14
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@Jasmine This may help you. – Rodrigo de Azevedo Aug 10 '21 at 05:27
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@RodrigodeAzevedo what please – Jasmine Aug 10 '21 at 05:28
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@Jasmine Why don't you start with $D = \mbox{diag}(1,-1)$ and try to find a square root of $D$? – Rodrigo de Azevedo Aug 10 '21 at 05:30
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@RodrigodeAzevedo Did you mean just a 2 X 2 diagonal matrix with 1 and -1 in the diagonal? – Jasmine Aug 10 '21 at 05:32
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Related – Rodrigo de Azevedo Aug 10 '21 at 05:32
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Let us continue this discussion in chat. – Jasmine Aug 10 '21 at 05:32
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1If $A=2I$, and $B=I$, then $AA^t-BB^t$ has a positive eigenvalue, and no negative eigenvalues. If $A=I$, and $B=2I$, then $AA^t-BB^t$ has a negative eigenvalue, and no positive eigenvalues. So, being of the form $AA^t-BB^t$ tells you nothing about the number of positive or the number of negative eigenvalues. – Gerry Myerson Aug 10 '21 at 10:50
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1Any thoughts about my answer, Jasmine? – Gerry Myerson Aug 13 '21 at 10:23
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1Not polite to ask a question and then fail to engage with the user who provides an answer, jasmine. – Gerry Myerson Aug 14 '21 at 13:43
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@GerryMyerson Really sorry that I was busy. The reasoning you have said is awesome. – Jasmine Aug 16 '21 at 02:34
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1You have the option, Jasmine, of "accepting" the answer, by clicking in the check mark next to it. – Gerry Myerson Aug 16 '21 at 02:37
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Sure Gerry...... – Jasmine Aug 16 '21 at 02:38
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If a matrix $W$ has any negative eigenvalues at all, then it can't be written as $L^tL$ (for $L$ real). For let $v$ be an eigenvector of $W$, with eigenvalue $\lambda$. Then $v^tv>0$, and $$\lambda v^tv=v^tWv=v^tL^tLv=(Lv)^t(Lv)\ge0$$ so $\lambda\ge0$.
Gerry Myerson
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What if $L$ is complex-valued and one uses non-Hermitian transposition? – Rodrigo de Azevedo Aug 11 '21 at 14:06
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1@Rodrigo, then you can have negative eigenvalues, e.g., $L=\pmatrix{i&0\cr0&i}$, $W=L^tL=\pmatrix{-1&0\cr0&-1}$. But who would ever want to use non-Hermitian transposition on a matrix with nonreal entries? – Gerry Myerson Aug 11 '21 at 23:34
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No one, I assume. My somewhat cheeky comment's motivation was as follows. Given an indefinite real and symmetric matrix $A$, one can find matrix $R$ such that $R^\top R = A$, though not a real $R$. Hence, it would be wise to mention to which set $R$ must belong. – Rodrigo de Azevedo Aug 12 '21 at 06:38
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