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From Dmitriy Lyubimov <dlie...@gmail.com>
Subject Re: SSVD compute U * Sigma
Date Mon, 10 Sep 2012 21:05:14 GMT
Yep that's my understanding too. V cannot have inverse since it is not
square (in common case).

But since it is orthonormal, transpose produces a pseudoinverse which
we can use just the same.

On Mon, Sep 10, 2012 at 2:03 PM, Sean Owen <srowen@gmail.com> wrote:
> Yes, doesn't the V V' != I bit mean it's not an inverse? I thought
> only square matrices had a real inverse. This is a one-sided inverse,
> which is (IIRC?) slightly stronger than being a pseudo-inverse.
>
> On Mon, Sep 10, 2012 at 9:59 PM, Ted Dunning <ted.dunning@gmail.com> wrote:
>> No.  It is the inverse.  V' V = I
>>
>> On the other hand, V V' != I.  We do know that norm(A V V' - A) is small.

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