I have a matrix of the form: $X=\Delta \Delta^T (\Phi+\Phi^T) P + P (\Phi+\Phi^T) \Delta \Delta^T $, where $\Delta$ is $N\times 1$ real, $P=P^T$. I know that such matrix is rank two, but after doing some simulations, i found that the largest eigenvalue is always positive. Is this a coincidence or are there any conditions that can ensure $\lambda_{max}(X)>0$? Thanks.

$\begingroup$ What is $\Phi?$ $\endgroup$ – J. E. Pascoe Aug 3 '14 at 4:39

$\begingroup$ $\Phi$ is a real $N\times N$ matrix, no special properties. $\endgroup$ – Michael Fan Zhang Aug 3 '14 at 4:51

2$\begingroup$ Then it would seem that taking $\Phi = I$ and $P = I,$ where I, would imply that the phenomenon is indeed a coincidence. Although, with some extra special structure, the phenomenon may hold. $\endgroup$ – J. E. Pascoe Aug 3 '14 at 4:56

$\begingroup$ I mean $I$ to be the N by N identity matrix. $\endgroup$ – J. E. Pascoe Aug 3 '14 at 5:06
Consider the following matrix
$a b^T + b a^T$ where $a=\Delta$, $b = P(\Phi+\Phi^T)\Delta$. Both $a$ and $b$ are N by 1 matrix.
Consider a vector $v = a + x b$ where $x$ is a real number. we have
$ (a b^T + b a^T )v = [(a b^T + b a^T )a] + x[(a b^T + b a^T )b] $
$ =[a (b^T a) + b (a^T a)] + x[a (b^T b) + b (a^T b)] $
$ = [ (b^T a) + x (b^T b) ] a + [ (a^T a)+ x(a^T b)]b $
$ =(b^T a +x b^T b) a + (a^T a+ x b^T a) b$
Note that $(b^T a) = (a^T b) $ , $a^T a$, and $(b^T b)$ are all real numbers. The idea to construct $v$ this way is because the column space of $X$ is spanned from $a,b$, so the eigenvector must be the linear combination of $a$ and $b$.
Let
$ (b^T a +x b^T b) = \lambda $
and
$ (a^T a+ x b^T a) = \lambda x $
We have $(a^T a+ x b^T a) = x (b^T a +x b^T b) $
$ x^2(b^T b)  a^T a = 0 $
So we have the two roots of $x= \pm \sqrt{\frac{a^T a}{b^T b}}$
So $\lambda = b^T a \pm \sqrt{a^T a \times b^T b}$ and by cauchy schwarz
$\lambda_{max} = b^T a + \sqrt{a^T a \times b^T b} > 0$
$\lambda_{min} = b^T a  \sqrt{a^T a \times b^T b} < 0$
So the conclusion is that the two nonzero eigenvalues has the opposite sign.


$\begingroup$ hi dehua, thanks for your answer. Seems your answer matches my simulations result. Can i ask how u calculate the third line, any wrong? $\endgroup$ – Michael Fan Zhang Aug 3 '14 at 5:10

$\begingroup$ @MichaelFanZhang I updated the answer. I think it is correct. If you spot any mistake, please tell me, thanks. $\endgroup$ – cdh Aug 3 '14 at 5:21

1$\begingroup$ In other words, when $a$ and $b$ are linearly independent, one can complete them to a basis, and when written in that basis the matrix has $\begin{bmatrix}0 & 1\\ 1 & 0\end{bmatrix}$ in the top left corner and then all zeros. Computing eigenvalues is easy then. $\endgroup$ – Federico Poloni Aug 3 '14 at 12:06