# Eigenvalues of a $4 \times 4$ block matrix

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!

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

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