# Computing eigenvalues of a specific block covariance matrix

In my studies of probability, I have recently come across the following problem on which i am stuck:

Let x be an $m$ dimensional random column vector, and let $a$ be a random scalar variable, both $x,a$ have zero mean, and have known distributions and a known joint distribution. We look at the concatenated vector $y = \begin{bmatrix} x \\ a \\ \end{bmatrix}$, we wish to look at the covariance matrix of this random vector, we thus obtain the covariance matrix: $\Sigma_{yy}= \left[ \begin{matrix} \Sigma_{xx} & E\{ax\} \\ E\{ax^T\} & \sigma_a ^ 2 \end{matrix} \right]$

Where $\Sigma_{xx}$ denotes the covariance matrix of $x$ with itself $x$, $E$ denotes expectation, $T$ denotes matrix transpose, and $\sigma_a ^ 2$ denotes variance of $a$. My intention was to somehow express the eigenvalues of this matrix given I know the exact joint distribution of $x,a$.

I have attempted to write the characteristic polynomial of this block matrix, but cannot proceed in finding this determinant, as the matrix has no special form and the Schur complement I tried doesn't seem to work for me here. I was hoping someone here could help me express the eigenvalues in a convenient or compact way , or at least in an explicit closed form using familiar quantities of this problem. All help is appreciated, even non trivial bounds on the eigenvalues or specific cases when this might be solvable. I of course know this can be handled when $a,x$ are statistically independent, but this case is uninteresting to me.

## 1 Answer

Suppose $A$ is a Hermitian matrix, $B$ is a column matrix, $B^{\dagger}$ is its conjugate, and $a$ is a real number. Set up the following eigenvalue equation, $$\begin{bmatrix} A & B \\ B^{\dagger} & a \end{bmatrix} \begin{bmatrix}u \\v \end{bmatrix}=\lambda\begin{bmatrix}u \\v \end{bmatrix} \tag1$$ $B\neq0$. Otherwise we have joining of two orthogonal linear function spaces and there is nothing to explore. We conclude that $v\neq0$.

Let $\{(\lambda_i, u_i)\}_{i=1}^m$ be the eigenvalue and orthonormal eigenvector basis pairs of $A$. Expand $$u=\sum_i a_iu_i$$ for some coefficients $\{a_i\}_{i=1}^m$. Left multiply the first eigen equation with $u_i^{\dagger}$ \begin{align} \sum_j a_j\lambda_ju_j +Bv = Au+Bv = \lambda u =\lambda\sum_j a_ju_j \end{align} There is an eigenvalue of the larger matrix distinct from all those of $A$ since $B\neq 0$. For that eivenvalue $\lambda$ we have $$\lambda_ia_i +u_i^{\dagger}Bv=\lambda a_i \Longleftrightarrow a_i=\frac{u_i^{\dagger}Bv}{\lambda-\lambda_i}$$ $v\neq0$. Otherwise, we have from Equation (1), $Au=\lambda u$ and thus $\lambda=\lambda_j$ for some $j$. Substitute the above expression into the second eigenvalue equation and divide by $v$, we obtain $$\lambda-a=\sum_i\frac{|u_i^{\dagger}B|^2}{\lambda-\lambda_i}. \tag2$$ The graph of the left hand side and right hand side of this equation as functions of $\lambda$ shows that the solution for $\lambda$ is the intersections of the two functions. It is easily seen and proved that eigenvalues $\beta_1\ge\beta_2\ge\cdots\ge\beta_{m+1}$ of the larger matrix interlace with those $\alpha_1\ge\alpha_2\ge\cdots\ge\alpha_m$ of the smaller matrix $A$ in the manner $$\beta_1\ge\alpha_1\ge\beta_2\ge\alpha_2\ge\cdots\ge\beta_m\ge\alpha_m\ge \beta_{m+1}.$$ We can easily and efficiently solve Equation (2) for all $m+1$ eigenvalues by Newton's method.