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

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The discussion centers on the mathematical properties of outer-product matrices, specifically whether the inverse of an outer-product matrix can also be expressed as an outer product. It is established that if matrix A is defined as A = v v^{\top}, where v is a nonzero vector, then A has rank 1. Consequently, the inverse A^{-1} does not exist for matrices of rank 1, as they are singular. The conversation emphasizes the need for understanding matrix rank and properties of outer products in linear algebra.

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omaradib
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Hi All,

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

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

Then can [tex]A^{-1}[/tex], inverse of [tex]A[/tex], 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 [tex]A[/tex] is an outer-product of a vector [tex]v[/tex] as : [tex]A = v v^{\top}[/tex]

Then can [tex]A^{-1}[/tex]
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.
 

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