# Definiteness of symmetrix block matrix

Let $$A$$ and $$D$$ be symmetric positive definite matrices and consider the symmetric block matrix

$$M := \begin{pmatrix} A & \alpha B \\ \alpha B^\top & D \end{pmatrix}$$

where $$\alpha \in \mathbb{R}$$ is a scalar parameter. Is it possible to say something about the positive definiteness of $$M$$ as a function of $$\alpha$$?

Because of the facts that (i) the eigenvalues are continuous w.r.t. the matrix parameters, and (ii) for $$\alpha = 0$$, $$M$$ is positive definite and its eigenvalues are those of $$A$$ and $$D$$ (since block diagonal), it seems that if $$\alpha$$ is "small enough", the matrix $$M$$ will be positive definite. Hence, it seems possible to relate the positive definiteness of $$M$$ to conditions on the smallest eigenvalue $$\lambda_\min$$ of $$A$$ and $$D$$ and some metric of $$\alpha B$$ (maybe some norm). Does someone have an idea?

I was thinking on the Schur complement and looking at the matrix

$$A - \alpha^2 B D^{-1} B^\top$$

but I have difficulties showing the positive definiteness of that as a function of $$\alpha$$. However, again it is clear that if $$\alpha=0$$, everything works out. Does someone have some idea?

• Well, certainly if $\|\alpha B\|$ is smaller than the smallest eigenvalue of $A \oplus D$, then $M$ will still be positive definite. Commented Jul 18, 2023 at 9:46
• Why is that like this, which theorem or properties did you apply?
– Trb2
Commented Jul 18, 2023 at 10:57
• As it is a bit too long for a comment, I posted it as an answer below. Commented Jul 18, 2023 at 11:27

Let me write $$M_{\alpha}$$ for $$M$$ to stress the $$\alpha$$-dependence. Denote the smallest eigenvalue of $$M_{\alpha}$$ by $$\lambda_{\alpha}$$ and note that $$\lambda_{\alpha} = \inf\limits_{\|x\| = 1} \|M_{\alpha}x\|$$. From this it is easy to see that $$|\lambda_{\alpha} - \lambda_0| \leq \|M_{\alpha}-M_0\| = \|\alpha B\|.$$ Hence, if $$\|\alpha B\| < \lambda_0$$, then $$\lambda_{\alpha} > 0$$.

• Thank you a lot, I understand! I suppose that $\Vert \cdot \Vert$ is the 2-norm (for $\lambda_{\alpha}=\inf_{\Vert x \Vert=1}\Vert M_{\alpha} x\Vert$ to hold) and for $\Vert \alpha B \Vert$ it is the matrix norm induced by the 2 norm, right? If that would hold any vector norm, that would be great.
– Trb2
Commented Jul 18, 2023 at 14:40
• @Trb2 Yes, this only works for the $2$-norm. You sort of need the Hilbert space structure to talk about positive definiteness. Commented Jul 19, 2023 at 9:06

Firstly, there exists an invertible matrix $$P$$ such that $$M=P^T\left( \begin{matrix} A& 0\\ 0& D-\alpha ^2B^TA^{-1}B\\ \end{matrix} \right) P.$$ Secondly,because $$D$$ is positive-definite and $$\alpha ^2B^TA^{-1}B$$ is self-adjoint, there exists an invertible matrix $$C$$ such that $$C^{T}DC=I_s$$ and $$C^{T}(B^TA^{-1}B)C=\mathrm{diag}\left\{ \lambda _1,\cdots ,\lambda _s \right\}$$

where $$\lambda _1,\cdots ,\lambda _s$$ are all the eigenvalues of $$D^{-1}(B^TA^{-1}B)$$.

Then you can just calculate whether $$1-\alpha ^2\lambda_1,\cdots,1-\alpha ^2\lambda_s$$ are greater than $$0$$ or less than $$0$$.

Some explanations:

If $$T$$ is $$n\times n$$ positive-definite and $$U$$ is $$n\times n$$ self-adjoint, then there exists an invertible matrix $$C$$ such that $$C^{T}TC=I_n$$ and $$C^{T}UC=\mathrm{diag}\left\{ \lambda _1,\cdots ,\lambda _n \right\},$$

where $$\lambda _1,\cdots ,\lambda _n$$ are all the eigenvalues of $$T^{-1}U$$.

proof:

Because $$T$$ is positive-definite, then there exist an invertible matrix $$P$$ such that $$P^{T}TP=I_n$$. Because $$P^{T}UP$$ is self-adjoint then there exist an orthogonal matrix $$Q$$ such that $$Q^{T}P^{T}UPQ=\mathrm{diag}\left\{ \lambda _1,\cdots ,\lambda _n \right\}$$.

We can choose $$C=PQ$$.

$$C^{T}(\lambda T-U)C=\mathrm{diag}\left\{ \lambda-\lambda _1,\cdots ,\lambda-\lambda _n \right\}$$. So $$\lambda_i$$ is the root of $$|\lambda T-U|$$. Because $$T$$ is invertible, $$\lambda_i$$ is also the root of $$|\lambda I-T^{-1}U|$$.

Back to the question, choose $$T=D$$, $$U=B^TA^{-1}B$$.

• Thank you for the answer! I just can't follow why there exists a matrix $C$ which those two properties, and why $\lambda_1,\dots,\lambda_s$ are the eigenvalues of $D^{-1}(B^\top A^{-1}B)$ and not those of $(B^\top A^{-1}B)$.
– Trb2
Commented Jul 18, 2023 at 10:56
• I have added some explanations and you can see it. Commented Jul 18, 2023 at 11:17

The following Schur complement is positive semidefinite.

$${\bf A} - \alpha^2 {\bf B} \, {\bf D}^{-1} {\bf B}^\top \succeq {\bf O}$$

Since the matrix $$\bf A$$ is symmetric and positive definite, it has a (symmetric and invertible) square root $${\bf A}^\frac12$$. The inequality above can be rewritten as follows

$${\bf I} - \alpha^2 {\bf A}^{-\frac12} {\bf B} \, {\bf D}^{-1} {\bf B}^\top {\bf A}^{-\frac12} \succeq {\bf O}$$

and, thus,

$$\alpha^2 \leq \color{blue}{\frac{1}{\lambda_{\max} \left( {\bf A}^{-\frac12} {\bf B} \, {\bf D}^{-1} {\bf B}^\top {\bf A}^{-\frac12} \right)}} =: \alpha_{\max}^2$$

and $$\alpha \in [-\alpha_{\max}, \alpha_{\max}]$$. Using the other Schur complement, $${\bf D} - \alpha^2 {\bf B}^\top {\bf A}^{-1} {\bf B}$$, yields the same bound, as

\begin{aligned} \alpha^2 \leq \color{blue}{\frac{1}{\lambda_{\max} \left( {\bf D}^{-\frac12} {\bf B}^\top {\bf A}^{-1} {\bf B} \, {\bf D}^{-\frac12} \right)}} &= \frac{1}{\lambda_{\max} \left( \left( {\bf A}^{-\frac12} {\bf B} \, {\bf D}^{-\frac12} \right)^\top \left( {\bf A}^{-\frac12} {\bf B} \, {\bf D}^{-\frac12} \right) \right)} \\ &= \frac{1}{\lambda_{\max} \left( \left( {\bf A}^{-\frac12} {\bf B} \, {\bf D}^{-\frac12} \right) \left( {\bf A}^{-\frac12} {\bf B} \, {\bf D}^{-\frac12} \right)^\top \right)} \\ &= \frac{1}{\lambda_{\max} \left( {\bf A}^{-\frac12} {\bf B} \, {\bf D}^{-1} {\bf B}^\top {\bf A}^{-\frac12} \right)} = \alpha_{\max}^2 \end{aligned}

Related: Matrix relations involving trace and eigenvalues