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Given the eigendecompositions $AA^{\top}=Q \Lambda Q^{\top}$ and $A^{\top}A=P \Lambda P^{\top}$, where $\Lambda$ is a diagonal matrix (of eigenvalues) and $P$ and $Q$ are unitary eigenvectors matrices of $A^{\top}A$ and $AA^{\top}$, is there a nice way to show that $A^{\top}Q \Lambda^{-1} Q^{\top}=P \Lambda^{-1} P^{\top}A^{\top}$ ?

I originally obtained this by solving the following two optimization problems:


$A$ is an $m$ by $n$ matrix, $x$ a $n$ by 1 vector, and $y$,$\hat y$ are $m$ by $1$ vectors.

Assume the following two optimization problems:

a) Minimize $x^{\top}x$ given the constraint that $Ax=y$

b) Minimize $(y-\hat y)^{\top}(y-\hat y)$ where $\hat y=Ax$.

The solution to both of these problems can be shown to be $x_{opt}=Wy$ where $W$ is the Moore–Penrose pseudoinverse of matrix $A$.

what is a good reference (a textbook or published paper) that discusses the above problems ?

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Could you elaborate on what you mean by equivalent? For both the solution depends on the Moore-Penrose pseudo inverse and the solution is given by $x = A^{\perp} y$ – user17762 Dec 17 '11 at 22:59
Is there an easy way to explain why the solution to both problems is the same? – mghandi Dec 17 '11 at 23:03
Though $x=A^{\perp}y$ for both the problem, $A^{\perp}$ is different in the two cases. In the first case, you are finding the minimum norm solution where $A^{\perp} = A^T (AA^T)^{-1}$, while in the second case, you are finding the least square solution where $A^{\perp} = (A^TA)^{-1} A^T$. The derivation for the cases is fairly easy but I do not see if one is an equivalent of another (or) if there is a nice relationship between the two. – user17762 Dec 17 '11 at 23:12
Sivaram, to make it more clear, I added the alternative form using matrices $P,Q,\Lambda$. I guess this form works for both problems. Is there a neat way to visualize the two matrices $A^{\top}Q \Lambda^{-1} Q^{\top}$ and $P \Lambda^{-1} P^{\top}A^{\top}$ and show they are equal? – mghandi Dec 17 '11 at 23:41
See… – joriki Dec 17 '11 at 23:55

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