Is the Inverse of an Outer-Product Matrix Also an Outer-Product?

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Hi All,

This is not a homework or coursework question. Yet I have a curiosity.

If a matrix A is an outer-product of a vector v as : A = v v^{\top}

Then can A^{-1}, inverse of A, be also expressed as an outer-product of some other vector?

Please point me how to approach the question, how to find the answer or how to say it is possible or not.

Thanks.
 
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omaradib said:
If a matrix A is an outer-product of a vector v as : A = v v^{\top}

Then can A^{-1}
If A is 0x0 or 1x1, inverting it is easy. Otherwise...

Exercise: If v is nonzero, prove that A has rank 1.


You can do better:

Exercise: A matrix has rank 1 if and only if it is the outer product of two nonzero vectors.
 
oops! Thanks.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...

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