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I am working on the following problem:

Let $B$ be a $4 \times 4$ matrix over $\mathbb{R}$ of the form \begin{bmatrix} 0 & 0 & a & b \\ 0 & 0 & c & d \\ a & c & 0 & 0 \\ b & d & 0 & 0 \\ \end{bmatrix}

Prove that $B$ has $4$ real eigenvalues (counting multiplicity), $B$ does not have 4 positive eigenvalues, and $B$ does not have $3$ positive and $1$ negative eigenvalue.

I attempted to compute the eigenvalues the usual way, by obtaining the corresponding characteristic equation for $B$. My computations led to the characteristic equation $\lambda^4 - (a^2 + b^2 + c^2 + d^2)\lambda^2 + a^2d^2 + b^2c^2 - 2abcd = 0$. How can I show the eigenvalues $\lambda$ occuring as roots of this characteristic equation are all real, not all positive, and not $3$ positive and $1$ negative ?

I noticed after my computation of the characteristic equation that this is a block matrix \begin{bmatrix} A_1 & A_2 \\ A_3 & A_4 \end{bmatrix}, composed of $2 \times 2$ blocks $A_1,A_2,A_3,A_4$ , where $A_2 = A_3^T$. Does this buy us anything as far as conditions on the eigenvalues that may help me prove the result ?

Thanks!

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    $\begingroup$ Your characteristic polynomial is quadratic in $\lambda^2$. $\endgroup$ Nov 15, 2019 at 23:59

2 Answers 2

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Since $B$ is symmetric, by spectral theorem, it has four real eigenvalues and since the trace is equal to zero we can exclude that it has $4$ positive eigenvalues and since the determinant is

$$ad(ad-bc)-bc(ad-bc)=(ad-bc)^2 \ge 0$$

we can't have only one negative eigenvalue.

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Consider your characteristic equation as a quadratic in $\lambda^2$. It's discriminant is $$(a^2+b^2+c^2+d^2)^2-4(ad-bc)^2$$$$=(a^2+b^2+c^2+d^2+2ad-2bc)(a^2+b^2+c^2+d^2-2ad+2bc)\ge 0.$$ The quadratic equation therefore has real roots, in fact two non-negative roots.

There are then four real solutions for $\lambda$, occurring in pairs which sum to zero.

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