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

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