largest eigenvalue of a symmetric matrix

Question

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.

Answer 1

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.