Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
Which definition of "generalized inverse" are you using? – celtschk Aug 20 '12 at 10:27
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. – user20266 Aug 20 '12 at 10:35
@Thomas See the edited question. – user506901 Aug 20 '12 at 10:52
@celtschk Generalized inverse corresponds to Moore-Penrose pseudoinverse. – user506901 Aug 20 '12 at 10:52
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 $\mathbf x$... – J. M. Aug 20 '12 at 11:29
up vote 2 down vote accepted

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 $A$.

share|cite|improve this answer
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 $Ax=b$. ? – user506901 Aug 20 '12 at 12:39
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). – celtschk Aug 20 '12 at 13:16

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.