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

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The discussion centers on whether the inverse of an outer-product matrix, defined as A = vv^T, can also be expressed as an outer product of another vector. It is noted that if matrix A is 0x0 or 1x1, its inversion is straightforward, but the situation is more complex for larger matrices. Participants emphasize that if vector v is nonzero, matrix A has rank 1, which is a crucial property in understanding the inversion. Additionally, it is highlighted that a matrix has rank 1 if and only if it is the outer product of two nonzero vectors. The conversation invites further exploration into the conditions under which the inverse retains the outer-product form.
omaradib
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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.
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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