largest eigenvalue of a symmetric matrix

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.

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What is $\Phi?$ – J. E. Pascoe Aug 3 '14 at 4:39
$\Phi$ is a real $N\times N$ matrix, no special properties. – Michael Fan Zhang Aug 3 '14 at 4:51
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. – J. E. Pascoe Aug 3 '14 at 4:56
I mean $I$ to be the N by N identity matrix. – 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.

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Obvious counter example is when $X = 0$. – cdh Aug 3 '14 at 5:02
hi dehua, thanks for your answer. Seems your answer matches my simulations result. Can i ask how u calculate the third line, any wrong? – Michael Fan Zhang Aug 3 '14 at 5:10
@MichaelFanZhang I updated the answer. I think it is correct. If you spot any mistake, please tell me, thanks. – cdh Aug 3 '14 at 5:21
thank you so much for your clarifications! – Michael Fan Zhang Aug 3 '14 at 5:35
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. – Federico Poloni Aug 3 '14 at 12:06