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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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  • $\begingroup$ Which definition of "generalized inverse" are you using? $\endgroup$
    – celtschk
    Commented Aug 20, 2012 at 10:27
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    $\begingroup$ this is more of a comment than an anser, I guess: You need to know that $b$ is in the range of $A$. The fact that b is orthogonal to the null space won't help very much, since, without additional knowledge about the structure of $A$ -- e.g. symmetry -- the image of $A$ need not be orthogonal to it's kernel. $\endgroup$
    – user20266
    Commented Aug 20, 2012 at 10:35
  • $\begingroup$ @Thomas See the edited question. $\endgroup$
    – user506901
    Commented Aug 20, 2012 at 10:52
  • $\begingroup$ @celtschk Generalized inverse corresponds to Moore-Penrose pseudoinverse. $\endgroup$
    – user506901
    Commented Aug 20, 2012 at 10:52
  • $\begingroup$ I don't understand what you mean by "give the exact solution to the system"; the distinguishing characteristic of the Moore-Penrose pseudoinverse is that it will give you the solution vector $\mathbf x^\ast$ that minimizes $\|\mathbf A\mathbf x-\mathbf b\|_2$ over all possible $\mathbf x$... $\endgroup$
    – J. M. ain't a mathematician
    Commented Aug 20, 2012 at 11:29

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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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    $\begingroup$ Or, simply, in case of a symmetric $A$, and $b$ being orthogonal to the null-space of $A$, $\bar{x}=A^+b$ is a solution to $Ax=b$. ? $\endgroup$
    – user506901
    Commented Aug 20, 2012 at 12:39
  • $\begingroup$ 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). $\endgroup$
    – celtschk
    Commented Aug 20, 2012 at 13:16
  • $\begingroup$ en.wikipedia.org/wiki/Conjugate_transpose $\endgroup$
    – HappyFace
    Commented May 29, 2023 at 4:17
  • $\begingroup$ The section "Complementarity of image and kernel" was helpful for me to understand this answer. en.wikipedia.org/wiki/Projection_(linear_algebra) $\endgroup$
    – HappyFace
    Commented May 29, 2023 at 4:26

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