Is the condition number of a 2x2 block symmetric matrix greater than the condition number of its upper left hand block?

Question

Is there any known relation between cond(M) and cond(Q) when

$$M=\begin{bmatrix}Q&A^T\\A&0\end{bmatrix}$$ and Q is symmetric positive definite and A is rectangular full row rank?

From the above conditions M is indefinite and invertible. From numerical experiments it seems that cond(M) > cond(Q), but I have yet to find a proof. I know that cond(M) > cond(A) from: $$M^2=\begin{bmatrix}Q^2+A^TA&QA^T\\AQ&AA^T\end{bmatrix}$$ and from the eigenvalue interlacing theorem $$\lambda_{min}(M^2)\le\lambda_{min}(AA^T)$$ $$\lambda_{max}(AA^T)\le\lambda_{max}(M^2)$$ therefore $$cond(M^2)\ge cond(AA^T)=\frac{\sigma^2_{max}(A)}{\sigma^2_{min}(A)}=cond(A)^2$$ and $$cond(M)\ge cond(A)$$ But this method doesn't seem to show a clear relation to cond(Q)

Answer 1

We give an example in which $\operatorname{cond}(M)>\operatorname{cond}(Q)$ and an example in which $\operatorname{cond}(M)<\operatorname{cond}(Q)$, showing there is no trivial relation between $\operatorname{cond}(M)$ and $\operatorname{cond}(Q)$.

For $a\in(0,1]$, we define matrices $Q$ and $A$ with $$Q = \begin{bmatrix}2a&0\\0&2\end{bmatrix}\,,\quad A=\begin{bmatrix}1&0\\0&1\end{bmatrix}\,.$$ For these $Q$ and $A$, matrix $M$ has characteristic polynomial $$p_M(\lambda)=(\lambda^2-2a\lambda-1)(\lambda^2-2\lambda-1)\,.$$ It follows that eigenvalues of $M$ are $1\pm\sqrt{2}$ and $a\pm\sqrt{1+a^2}$. Since $0<a\leq1$, the eigenvalue with the largest modulus is $1+\sqrt{2}$, and the one with the smallest modulus is $1-\sqrt{2}$. Therefore, $$\operatorname{cond}(M)=\frac{1+\sqrt{2}}{\sqrt{2}-1}=(1+\sqrt{2})^2\,.$$

Now, for $a=1$ we get $\operatorname{cond}(Q)=1<\operatorname{cond}(M)$, and for $a=0.1$ we get $\operatorname{cond}(Q)=10>\operatorname{cond}(M)$.