Eigenvalues of matrix based on submatrices

Question

If I have the eigenvalues of matrix A, is there anything I can say about the eigenvalues of the following matrix:

\begin{bmatrix}A_{11}+w&A_{12}&A_{13}&x\\A_{21}&A_{22}&A_{23}&0\\A_{31}&A_{32}&A_{33}&0\\y&0&0&z\end{bmatrix}

Answer 1

The study can be split into two tractable cases

• if $w=0$: you are in the case of "matrix bordering". In this case, see the Cauchy interlacing theorem ("Sandwich theorem" for eigenvalues of symmetric matrices) which assumes the symmetry of the matrix.

• if $w > 0$, you can write the new matrix as the following sum, the second matrix being a rank-one matrix:

$$\begin{bmatrix}A_{11}&A_{12}&A_{13}&0\\A_{21}&A_{22}&A_{23}&0\\A_{31}&A_{32}&A_{33}&0\\0&0&0&(z-\dfrac{xy}{w})\end{bmatrix}+\begin{bmatrix}w&0&0&x\\0&0&0&0\\0&0&0&0\\y&0&0&\dfrac{xy}{w}\end{bmatrix}$$

Explanation : the 2nd matrix is a rank one matrix because it has the form:

$$\tag{*} \begin{bmatrix}\sqrt{w}\\0\\0\\ \dfrac{y}{\sqrt{w}}\end{bmatrix}*\begin{bmatrix}\sqrt{w}&0&0&\dfrac{x}{\sqrt{w}}\end{bmatrix}$$

Now, you can apply a theorem on the determinant of a matrix updated by the addition of a rank one matrix (Determinant of rank-one perturbations of (invertible) matrices), not by taking the numerical determinant but the determinant giving the characteristic polynomial.

Remark: We have assumed $w>0$. If $w<0$, replace every $\sqrt{w}$ by $\sqrt{-w}$, and take the opposite of the column vector in (*).