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Suppose $n \times n$ hermitian and positive semi-definite matrix $A$ is given. We can rewrite $A$ using its eigen decomposition, $$ A = U_A \Lambda_A U_A^H. $$ Now suppose matrix $B$ is also $n \times n$, hermitian, and positive semi-definite. However, the only information known about $B$ is that it has identical eigenvalues, i.e. if eigen decomposition of $B$ is $B = U_B \Lambda_B U_B^H$, then $\Lambda_B = c~I_n$ where $c$ is a scalar and $I_n$ is $n\times n$ unity matrix.

I need to know if the value of $$\mathrm{det}(I_n+BA)$$ is affected by the choice of eigenvector matrix, $U_B$. For example, I face a contradiction when I calculate it by two ways. First, relaying on the fact that $U_B$ is unitary, it is written as, $$\mathrm{det}(I_n +c~U_B~U_B^H~A) = \mathrm{det}(I_n+c~I_n~A)=\mathrm{det}(I_n+c~A).$$

A second interpretation is, $$\det(I_n + c~U_B~U_B^H~U_A~\Lambda_A~U_A^H) = \det\Big(I_n+c~\underbrace{(U_B^H~U_A)}_{V}~\Lambda_A~(U_B^H~U_A)^H\Big)\\ = \det(I_n +c~\underbrace{V~\Lambda_A~V^H}_{\bar{A}}).$$

I suspect that the choice of $U_B$ may alter eigenvalues of matrix $\bar{A}$. In the extreme, if $V = U_B^H~U_A = 0$ the expression reduces to zero which is not equal to $\det(I_n+c~A)$ necessarily.

How do you explain this contradiction?

Thank you, regards.

Mona Hajimomeni
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  • Maybe my problem is not clear enough, should I add more explanation? Any questions? – Mona Hajimomeni Oct 18 '13 at 04:44
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    @A good rule of thumb is that if it is not easy to quickly determine the answer to various practical questions, then you will lose many viewers. For example, it is difficult to understand the level of difficulty of this problem. Even locating the actual question requires significant effort. There are some acronyms, many confusing statements (e.g. you state that a couple of determinants "are true", which doesn't mean anything), and just lots of things that will slow readers down. You should aim to have people know 80% of what you are talking about within 10 seconds maximum. – Andrew Dudzik Oct 18 '13 at 10:09
  • @user33433 Thank you, I will reword it in a simpler way; maybe something close to the actual math question which is buried in unnecessary details of my application. – Mona Hajimomeni Oct 18 '13 at 10:41
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    How can you have $U_B^HU_A=0$ if they are both unitary? – Algebraic Pavel Oct 18 '13 at 22:52
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    Also, since $V=U_B^HU_A$ is unitary (and hence nonsingular), the matrix $\bar{A}=V\Lambda_A V^H$ has same eigenvalues as $A$ (both $A$'s are similar through a unitary transformation). Or did I miss something? – Algebraic Pavel Oct 18 '13 at 22:55
  • @AlgebraicPavel Yes, that's what I didn't notice (that $V$ is unitary). I thought I can find a unitary $U_B$ such that matrix $V$ turns out to have a lower rank than $A$. Is there any way that I can mask some eigenvalues of a matrix by some linear algebraic transformations? – Mona Hajimomeni Oct 19 '13 at 05:43
  • Supposing you mean $U_B$ to be unitary so that $U_B^H=U_B^{-1}$ (which is somewhat implied by calling $B = U_B \Lambda_B U_B^H$ and eigen decomposition) then $\Lambda_b=cI_n$ just means that $B=cI_n$. What am I missing? – Marc van Leeuwen Oct 19 '13 at 07:40
  • @MonaHajimomeni I guess you mean something like having a matrix $U$ with orthogonal columns such that the spectrum of $U^HAU$ is a subset of the spectrum of $A$? Then take $U$ to be equal to the eigenvectors of that part of the spectrum of $A$ which you want to preserve (still assuming that $A$ is Hermitian). – Algebraic Pavel Oct 19 '13 at 21:03

1 Answers1

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One has $B=cI_n$ and $\det(I_n+AB)=\det(I_n+cA)$ is correct. Also since $V$ is unitary $V^H=V^{-1}$, and $\det(I_n+cVAV^{-1})=\det(V(I_n+cA)V^{-1})=\det(I_n+cA)$ so there is no contradiction.

Marc van Leeuwen
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