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Let $A$ be a tall matrix, $n\times m$ with $n>m$. Suppose it has full rank. It is a fact that $A$ will have infinitely many left inverses. I would like to know if there are interesting conditions that we can impose on $B$ to make it unique.

For instance the condition that $AB$ be symmetric seems fairly strong, and is at least satisfied by the pseudo-inverse. Does this in fact completely determine $B$? I haven't been able to find a counterexample.

I'm also interested in other interesting ideas along these lines, though this is more open ended and less well-defined a question.

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My answer below attempts to show how symmetry does imply uniqueness, let me know if you would like me to re-word and use SVD (as user1551 suggests) to attempt further clarification. Also here (step by step pinv) I have written a proof regarding the fact that if $B^\top$ shares the same column span as $A$, then it is unique (the pseudo-inverse). –  adam W Mar 7 '13 at 18:46

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Using the SVD decomposition of the matrices you can see that $N_s=0$ must be true for symmetry to be true: $$A=U\begin{pmatrix}\Sigma\\0\end{pmatrix}V^T$$

$$B=V\begin{pmatrix}\Sigma^{-1} & N_s\end{pmatrix}U^T$$ $$BA = I$$ $$AB=U\begin{pmatrix}I& \Sigma N_s \\ 0 & 0 \end{pmatrix}U^T$$ $$(AB)^\top=U\begin{pmatrix}I& 0 \\ N_s^\top \Sigma & 0 \end{pmatrix}U^T$$ The following is another path to see it. Let $N\ne 0$ and $NA = 0$ so that $N$ is in the null space of $A$. Then $(B + N)A = BA + 0= I$ and $$A(B+N) = AB + AN$$ Since $AB$ is symmetric, only the symmetry of $AN$ needs to be inspected. Compare it to its transpose $N^\top A^\top$. They must be equal for symmetry to be true.

$$N^\top A^\top \overset{?}{=} AN$$ If they both are left multiplied with $N$:

$$NN^\top A^\top \overset{?}{=} 0$$

Since $NN^\top \ne 0$ (unless we are dealing with complex numbers) then $NN^\top$ would have to be the null column space of $A$ for this to be true (the null row space of $A^\top$). This contradicts that $A$ has full span.

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If $A$ has full column rank, then there is no $N \ne 0$ s.t. $NN^\top A^\top = 0$, and thus the only left inverse $B$ s.t. $BA= I$ that gives $BA$ symmetric is the psuedo inverse. –  adam W Mar 7 '13 at 3:27
    
@adamW $B=A^+$ if $AB$ is symmetric (or Hermitian in the complex case). Think SVD. –  user1551 Mar 7 '13 at 7:37
    
The direct point is that $NN^TA^T=0$ implies $A(NN^T)=0$, contradicting that $A$ is full rank. I would be curious how SVD shows this. I've accepted your answer based on what you've written in the comments, but I would appreciate it if you incorporated them into your answer. Future passersby will also probably appreciate it. –  Lepidopterist Mar 10 '13 at 5:11
    
Yes, I forgot to even use that $A$ has full rank, so it made the argument a bit difficult to complete. SVD can be used to show it also. If it (user1551 gave it) does not seem as direct an argument to you, remember it is a useful tool that is always applicable, thus worth the effort to understand. –  adam W Mar 10 '13 at 12:21

Let $C=B-A^+$ (the symbol $A^+$ means the Moore-Penrose pseudoinverse of $A$; hence $CA=0$). To make the left inverse $B$ unique means to fix the choice of $C$ for each $A$.

One condition that fixes this choice is to simply pick $C=0$. This is equivalent to the condition that $AB$ is real symmetric (or Hermitian in the complex case). Indeed, if $A=U\begin{pmatrix}\Sigma\\0\end{pmatrix}V^T$ is a singular value decomposition of $A$, then all left inverses of $A$ can be written in the form of $B=V(\Sigma^{-1},R)U^T$ (here $C=V(0,R)U^T$). If $AB$ is symmetric, $R$ must be zero and therefore $B=A^+$.

Since every function $f:M_{n\times m}(\mathbb{R})\to M_{m\times n}(\mathbb{R})$ that satisfies $f(A)A=0$ gives rise to a left inverse function $B=A^++f(A)$, the previous condition is not the sole condition that one can impose. However, apart from the zero function, I don't know any simple way to explicitly specify $f$ (especially when $A$ has repeated singular values).

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