# Block symmetric matrix

Given is symmetric block matrices $$M$$ of the form $$M=\left(\begin{array}{cc} A & B \\ B^{\top} & C \end{array}\right)$$ where $$A$$ is a constant, $$B$$ is $$1 \times n$$ vector and $$C$$ is a $$n \times n$$ symmetric matrix. Under which conditions on eigenvalues of $$C$$ is this matrix negative definite? Can I say if matrix C is negative definite, M will be negative definite too since A is a constant?

• Notice, just to start, that in order for $M$ to be negative definite, the constant $A$ must be negative. $e_1^\top M e_1 = A$, after all. Commented Jan 3, 2023 at 18:51
• math.stackexchange.com/questions/3428318/… What about this one? Here we can only check D? @TedShifrin Commented Jan 3, 2023 at 18:53
• Be careful with your logic. I'm saying that $M$ negative definite implies $A<0$, not vice-versa (the converse is false). In the case of the one you linked, if the big matrix $M$ is positive definite, then $D$ must be positive definite, not vice-versa. Commented Jan 3, 2023 at 18:55
• Note: Consider the case $n=1$. Even if $A$ and $C$ are negative scalars, $M$ will often be indefinite. A simple example is this: $$\begin{bmatrix} -1&2\\2&-2\end{bmatrix}.$$ Then with $v=e_1+e_2$, we have $v^\top Mv = 1 > 0$. Commented Jan 3, 2023 at 22:40

Schur complement implies that $$M \prec 0$$ iff $$A \prec 0$$ and $$C-B^TA^{-1}B\prec 0$$. Equivalently, $$M \prec 0 \Leftrightarrow \begin{cases}A < 0\\ C\prec \frac{1}{A}B^TB\end{cases}.$$ Since $$\frac{1}{A}B^TB$$ is a rank-one matrix, It has one eigenvalue equal to $$\frac{1}{A}BB^T$$ and other eigenvalues equal to zero. Therefore, if $$A < 0$$ and eigenvalues of $$C$$ are less than $$\frac{1}{A}BB^T$$, then $$M \prec 0$$.

$$\textbf{Edit:}$$

This is only a sufficient condition. A necessary condition is $$A <0$$ and $$C \prec 0$$. That is, eigenvalues of $$C$$ should be negative. However, I do not know the necessary and sufficient condition in terms of the eigenvalues of Matrix $$C$$.

• I think it would be worth mentioning that your last implication is only sufficient.
– KBS
Commented Jan 3, 2023 at 17:00
• What would be necessary condition? @KBS Commented Jan 3, 2023 at 18:18
• This is now completely muddled. Please edit it to make it correct. You start out with stating iff two times. Commented Jan 3, 2023 at 19:00
• @Maica A necessary and sufficient condition is that $M$ is negative definite if and only if $A$ and $C-B^TA^{-1}B$ are negative definite. This is also equivalent to say that $C$ and $A-BC^{-1}B^T$ are negative definite.
– KBS
Commented Jan 3, 2023 at 19:12

A necessary and sufficient condition is that $$M$$ is negative definite if and only if $$A$$ and $$C-B^TA^{-1}B$$ are negative definite. This is also equivalent to say that $$C$$ and $$A-BC^{-1}B^T$$ are negative definite. This follows from the Schur complement formula.

There is, in general, no condition on $$C$$ alone unless in some very particular cases.

Let $$A$$ be Hermitian with eigenvalues $$\lambda_1\geq\cdots\geq \lambda_n$$ and let $$B=\begin{pmatrix}A&c\\c^*&t\end{pmatrix},$$ with eigenvalues $$\beta_1\geq\cdots\geq\beta_n$$. Then we have the interlacing inequalities $$\beta_1\geq\lambda_1\geq\cdots\geq\beta_n\geq\lambda_n\geq\beta_{n+1}$$.

Thus is if $$B\succ 0$$, then so is $$A$$. Conversely if $$A\succ 0$$, then it suffices that $$\det B > 0$$ (Sylvester’s criterion). But $$\det B = \det(A)(t -c^*A^{-1}c),$$ so it suffices that $$t -c^*A^{-1}c> 0$$. The negative definite case is treated by considering the negatives of $$A$$ and $$B$$. Also, you can move $$A$$ at the bottom right by a suitable permutation similarity.

Edit: we can expand a bit more on this. If we scale $$t$$ and $$c$$ by $$\|c\|_2^{-2}$$, then the last inequality can be lower bounded as so: $$t>c^*A^{-1}c\geq \min_{\|x\|_2=1}x^*A^{-1}x=\lambda_1^{-1}>0.$$

• I guess there is a typo here. I should be $\det B = \det(A)(t -c^*A^{-1}c)$, which in the end gives the Schur complement condition.
– KBS
Commented Jan 3, 2023 at 19:14
• @KBS Yes, you’re correct! Commented Jan 3, 2023 at 19:17