# Moore-Penrose Inverse of $A^{T}A$

The Moore-Penrose inverse of a rectangular matrix is given by: $$A^{+}=V_{r}S_{r}^{-1}U_{r}^{T}$$ Here I am using the economical size SVD. Where $$U_{r}^{T}U_{r}=V_{r}^{T}V_{r}=I_{r}$$. My question is

It is known that $$(A^{T}A)^{+}=V_{r}S_{r}^{-2}V_{r}^{T}$$ so my question is how can we show it instead of proving $$(A^{T}A)^{+}(A^{T}A)=I_{r}$$?

Edit :

I would like to thank user @cangrejo for his suggesstion which I find useful as it confirms my assumption that this problem can be tackled by using properties of the pseudo-inverse : $$(A^{T}A)^{+}=A^{+}(A^{+})^{T}$$ However, the wikipedia article en.wikipedia.org/wiki/Moore%E2%80%93Penrose_inverse under the section properties lists one of the sufficient conditions for $$(AB)^{+}=B^{+}A^{+}$$ is that $$B=A^{T}$$ which is similar to my case but how can we prove it to be a sufficient condition to the main result $$(AB)^{+}=B^{+}A^{+}$$ for which this would help me complete the main objective of finding pseudoinverse of $$A^{T}A$$? In other words, how do we show that if the expression $$B=A^{T}$$ is true, then this will allow me to write $$(AB)^{+}=B^{+}A^{+}$$

I will award the bounty for a complete answer.

• Do the shapes match? Jun 9 at 16:48
• Yes $A$ is $m\times n$ rectangular matrix. Jun 9 at 16:51
• Are the shapes of $(A^TA)^+$ and $(A^T)^+A^+$ equal? Jun 9 at 16:54
• In any case, I think the Wikipedia page might have an answer for you: en.wikipedia.org/wiki/Moore%E2%80%93Penrose_inverse Jun 9 at 16:58
• Note that in general $(A^T A)^+(A^T A) \neq I$, for example if $A=0$. Instead, one needs to verify the 4 Moore-Penrose conditions $(A^T A)(A^T A)^+(A^T A) = (A^T A)$, $(A^T A)^+(A^T A)(A^T A)^+ = (A^T A)^+$, etc. Jun 11 at 23:13

A general strategy for proving statements about the Moore-Penrose inverse is to (1) use a SVD to reduce statements about general matrices to statements about diagonal matrices, and (2) use the 4 Moore-Penrose conditions:

1. $$X Y X = X$$
2. $$Y X Y = Y$$
3. $$(XY)^T = XY$$
4. $$(YX)^T = Y X$$

If $$Y$$ satisfies these 4 equations, then $$Y=X^+$$. Note that the pseudo-inverse is always unique. So, all we need to do to prove $$B=A^T\implies (AB)^+ = B^+ A^+$$ is to verify that $$Y=B^+A^+$$ satisfies these 4 equations for $$X=AB$$.

Given the economical SVD $$A=USV^T$$, and $$B=A^T = V^T S^T U = V^T S U$$ we have

• $$A^+ = VS^{-1}U^T$$
• $$B^+ = US^{-1}V^T$$
• $$AB = (USV^T)(VSU^T) = US^2 U^T$$
• $$B^+A^+ = (US^{-1}V^T)(VS^{-1}U^T) = US^{-2}U^T$$

Plugging these into the 4 MP conditions yields straightforward computations

1. $$(AB)(B^+A^+)(AB) = \big(US^2U^T\big)\big(U(S^{-2}U^T\big)\big(US^2U^T\big) = US^2 U^T = AB$$

2. $$(B^+A^+)(AB)(B^+A^+) = \big(U(S^{-2}U^T\big)\big(US^2U^T\big)\big(U(S^{-2}U^T\big) = US^{-2} U^T = B^+A^+$$

3. $$\big((AB)(B^+A^+)\big)^T = \Big(\big(US^2U^T\big)\big(U(S^{-2}U^T\big)\Big)^T = (UU^T)^T = UU^T= (AB)(B^+A^+)$$

4. $$\big((B^+A^+)(AB)\big)^T = \Big(\big(U(S^{-2}U^T\big)\big(US^2U^T\big)\Big)^T = (UU^T)^T = UU^T= (B^+A^+)(AB)$$

Notice how, due to the cancellation of the (semi-)orthogonal matrices, all statements are effectively reduced to statements about diagonal matrices.