If A is any real mxn-matrix consider the block matrix $\begin{pmatrix} E&A^T \\ A&0\end{pmatrix}$. This matrix seems to have close connections with pseudo inverse, svd etc. which are probably well known. Does anybody know a name for this block matrix and/or have hints where to find more details?
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$\begingroup$ Is $E$ the identity? $\endgroup$– Federico PoloniCommented Aug 2, 2014 at 8:32
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$\begingroup$ Yes, E means identity. Your reference is very helpful! Thx! Actually I came across this type of matrix by one of the applications mentioned there. $\endgroup$– Willi WernerCommented Aug 3, 2014 at 8:13
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2 Answers
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If I had to pick a name, I'd go with saddle-point matrix. It's not a completely fitting name, but it is a matrix that I associate immediately with saddle-point problems in optimization. You can start from the review Numerical solution of saddle-point problems, Benzi, Golub, and Liesen, which focuses on solving large-scale linear systems with similar matrices.
answered Aug 2, 2014 at 8:32
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I did not know the reference given by Federico. It is a great paper. Yet, it is not necessary to read it to see that this question is linked to the SVD decomposition.
Let $U=\begin{pmatrix}I_n&A^T\\A&0\end{pmatrix}$ which is a symmetric matrix. Clearly $\det(U-\lambda I_{m+n})=\lambda^{m-n}\det((\lambda^2-\lambda)I_n-A^TA)$ (even if $m-n<0$). Then an eigenvalue $\lambda$ of $U$ is $0$ or satisfies $\lambda^2-\lambda=\sigma$ where $\sigma$ is a singular value of $A$. Moreover if $[x,y]^T$ is an eigenvector associated to $\lambda$, then $AA^Ty=\sigma y,A^TAx=\sigma x$ and $x,y$ are eigenvectors of $A^TA,AA^T$. Then the SVD decomposition of $A$ gives explicitly the diagonalization of $U$.
