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Suppose I'm given a linear system $Ax=b,$ with unknown $x\in\mathbb{R}^n$, and some symmetric $A\in\mathbb{R}^{n\times n}$ and $b=\in\mathbb{R}^n$. Furthermore, it is known that $A$ is not full-rank matrix, and that its rank is $n-1$; therefore, $A$ is not invertible. However, to compute the "solution" $x$, one may use $x=A^+b$, where $A^+$ is a generalized inverse of $A$, i.e., Moore-Penrose inverse.

What is the characteristic of such solution? More precisely, under which conditions will $x=A^+b$ give the exact solution to the system (supposing the exact solution exists)? Could one state that in the above case, with additional note that $b$ is orthogonal to null-space of $A$, the generalized inverse will yield the exact solution to the system?

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    @J.M. Can a solution to $Ax=b$ exist in case symmetric $A$ is not full rank? I'm interested under which conditions (if a solution $x$ to $Ax=b$ exists) does the Moore-Penrose inverse yield $x$ such that $\|Ax-b\|=0$ (i.e., the exact solution).2012-08-20

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Let $\tilde x = A^+b$. Then obviously $A\tilde x = AA^+b$. But since $AA^+$ is an orthogonal projector, and specifically $I-AA^+$ is the projector to the null space of the Hermitian transpose of $A$, $\tilde x$ is a solution iff $b$ is orthogonal to the null space of $AA^+$, that is, orthogonal to the null space of the Hermitian transpose of $A$.

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    Yes, if $A$ is a real symmetric matrix, then it's Hermitian, and thus its null space is the same as the null space of the Hermitian transposed (which for real matrices is simply the transposed). (Note: I overlooked the $\mathbb R^{n\times n}$ in your question, therefore I wrote it for general complex matrices; of course my answer is also true for real matrices, where the Hermitian transpose equals the normal transpose).2012-08-20