# Spectrum of this block matrix

Consider the following block matrix

$$A = \left(\begin{matrix} B & T\\ T & 0 \end{matrix} \right)$$

where all submatrices are square and

• matrix $$B = \mbox{diag}\left(b_1 ,0,0,\dots,0,b_n \right)$$ with $$b_1, b_n > 0$$.

• matrix $$T$$ is self-adjoint and positive semidefinite.

What can one say about the lowest eigenvalue of this matrix $$A$$? In particular, how does it depend on the spectrum of $$T$$ and the entries of $$B$$? Are there any known results?

• $A$ may have negative eigenvalues, you want the smallest in absolute value or the most negative one? – Carlo Beenakker Feb 19 '19 at 22:57
• the lowest one (which may be negative)... – Sascha Feb 19 '19 at 23:47

If $$\lambda_\max$$ is the greatest eigenvalue of $$T$$, the least eigenvalue of $$A$$ is between $$-\lambda_\max$$ and $$\max(b_1, b_n) - \lambda_\max$$.
• The eigenvalues of $\pmatrix{0 & T\cr T & 0\cr}$ are the eigenvalues of $T$ and $-T$. You're adding a positive semidefinite matrix. Use the min-max theorem. – Robert Israel Feb 20 '19 at 0:34