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A psuedo inverse of a matrix $A$ is defined as a matrix $A^+$ that satisfies $AA^+A=A, A^+AA^+=A^+$ such that $AA^+$ and $A^+A$ are symmetric. Show that the row space of $A^+$ and the row space of $A^T$ are equal.

I have no idea how to start this. I was thinking of showing that the row space of $A^+$ is a subset of the row space of $A^T$ and vice versa. I tried like this: Let $x$ be a member of the row space of $A^+$. Then $x$ can be expressed as a linear combination of the rows of $A^+$. I used the fact that $(A^+)^T=(A^T)^+$. But I cannot proceed further. Can someone please give me some hint?

Charlotte
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1 Answers1

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Actually for any of the four fundamental spaces of the two matrices are always the same.

Let the SVD decomposition of $A$ be $A = U\Sigma V^T$, then $A^{+} = V\Sigma^{+} U^T$. To make it more clear, we will turn to compare the spaces of $(A^{+})^T = U(\Sigma^{+})^TV^T$ and $A = U\Sigma V^T$, instead of comparing those of $A^{+}$ and $A^T$ directly.

Notice that, for any matrix $M$, if $M = U_0 \Sigma_0 V_0^T$ is an SVD decomposition(entries on the main diagonal of $\Sigma_0$ are in descending order), then:

  • the first $r$ columns of $U_0$ span the column space of $A$(because $MV_0^T = U_0\Sigma_0$), and the rest span the null space of $A^T$(because $U_0^TM = \Sigma_0 V_0$ and the corresponding entries in $\Sigma_0$ are zero).
  • the first $r$ rows of $V_0$ span the row space of $A$, and the rest span the null space of $A$.

Recall that

$$(A^{+})^T = U(\Sigma^{+})^TV^T$$ $$A = U\Sigma V^T$$

As $A, (A^{+})^T$ have the same rank and both their SVD decompositions involve the same $U$ and $V$, so the four fundamental spaces of $A, (A^{+})^T$(or $A^T, A^{+}$) are always the same.

There is a similar question years ago, and what I have just done is generalizing the idea briefly based on the original answers. For further information, see link.

ducati
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