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Let us consider a matrix $\mathbf{A} \in \mathbb{R}^{n \times n}$ and the block partitioning $$ \mathbf{A} = \begin{pmatrix} \mathbf{A}_{11} & \mathbf{A}_{12} \\ \mathbf{A}_{21} & \mathbf{A}_{22} \\ \end{pmatrix} $$ I have found some lecture notes that claim that the two conditions $$ \det(\mathbf{A}_{11}) \neq 0\quad \mathrm{and} \quad \det(\mathbf{A}_{22}) \neq 0 $$ are sufficient for the nonsingularity of $\mathbf{A}$. It seems to be false to me, as for example $$ \begin{pmatrix} 1 & 1 \\ 1 & 1 \\ \end{pmatrix} $$ verify the two conditions but is singular. Is there a missing condition or something?

Thanks!

gosbi
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$A$ is invertible if $\det(A_{11})\ne 0$, $\det(A_{22})\ne 0$, and $A_{12}=0$. Ie. $A$ is block upper-triagonal.

Invertibility of all four blockse does not imply invertibility of the matrix, as you showed.

Non-Invertibility of all four blocks also does not imply non-invertibility of $A$ $$ A=\left( \begin{array}{cc|cc} 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ \hline 0 & 0 & 1 & 0 \\ 0 & 1 & 0 & 0 \\ \end{array}\right) $$ all four blocks are singular, but $A$ is invertible.

daw
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