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I'm trying to solve an homework question but I got stuck.

Let A be a m x n matrix with the SVD $A = U \Sigma V^*$ and $A^+ = (A^* A)^{-1} A^*$ its pseudoinverse.

I'm trying to get $A^+ = V \Sigma^{-1} U^*$, but I'm missing something.

Can anyone help me with this please?

Thanks!

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If you are content with Sivaram Ambikasaran's or my answer you might want to accept one of them so that the question will be recognised as an answered one. –  Rasmus Feb 19 '11 at 3:44
    
Saying "SVD decomposition" is not quite unlike saying "enter your PIN number into the ATM machine"... –  J. M. Aug 3 '11 at 8:31
    
Fair enough! Thanks for the fix! =) –  paulochf Aug 24 '11 at 21:06
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2 Answers

up vote 0 down vote accepted

$$ \begin{align} A^+ &= (A^*A)^{-1}A^* \\ &=(V\Sigma U^*U\Sigma V^*)^{-1} V\Sigma U^* \\ &=(V\Sigma^2 V^*)^{-1} V\Sigma U^* \\ &=(V^*)^{-1} \Sigma^{-2} V^{-1} V\Sigma U^* \\ &= V \Sigma^{-2}\Sigma U^* \\ &= V\Sigma^{-1}U^* \end{align} $$ using the properties of the matrices $U,V,\Sigma$ in the Singular_value_decomposition.

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Hi Rasmus! I could get by myself until 3rd line. The 4th one was my point of doubt. I forgot to invert the $\left( \cdot \right)^{-1}$ sequence! Thanks in pointing that! =) –  paulochf Feb 2 '11 at 15:12
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Paulochf. It is SVD - Singular Value Decomposition and not SDV. I have edited the title. Also there is a homework tag which you need to apply when you are posting a homework question.

To get your answer, first you need to assume that the matrix $A^*A$ is invertible. For which you need $n \leq m$ and rank($A$) is $n$.

So when $n \leq m$ and when rank($A$) is $n$, then the reduced SVD of $A$ is $A = U \Sigma V^*$ where $U \in \mathbb{R}^{m \times n}$, $\Sigma \in \mathbb{R}^{n \times n}$ and $V \in \mathbb{R}^{n \times n}$ such that $U^* U = I_{n \times n}$, $V^* V = I_{n \times n}$, $V V^* = I_{n \times n}$ and $\Sigma$ is a square diagonal matrix and has only (positive) real entries.

Note that $V^{-1}=V^*$.

Also note that $A^* = V \Sigma^* U^* = V \Sigma U^*$ since $\Sigma^* = \Sigma$.

Further note that if $M_1,M_2 \text{and} M_3$ are invertible matrices then $(M_1 M_2 M_3)^{-1} = M_3^{-1} M_2^{-1} M_1^{-1}$.

Use these to get the final answer.

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Given that the question is homework - which I realise only now - it would have probably been better to restrict to hints like you did. Sorry. –  Rasmus Feb 1 '11 at 22:37
    
Hey Sivaram, I was in a hurry when I posted, so it was a (bad one) typo. Thanks for fixing the title! –  paulochf Feb 2 '11 at 15:05
    
I knew all these hints but I missed your "further note" and did not invert the sequence inside those parentheses. Thanks for your patience! –  paulochf Feb 2 '11 at 15:14
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