- #1

MathematicalPhysicist

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well rank AA^t<=rank A, but how do i show that rankAA^t>=rankA, i mean i need to show if x is a solution of AA^tx=0 then x is a solution of Ax=0 or of A^tx=0, but how?

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- Thread starter MathematicalPhysicist
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- #1

MathematicalPhysicist

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well rank AA^t<=rank A, but how do i show that rankAA^t>=rankA, i mean i need to show if x is a solution of AA^tx=0 then x is a solution of Ax=0 or of A^tx=0, but how?

- #2

Mindscrape

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- #3

radou

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well rank AA^t<=rank A, but how do i show that rankAA^t>=rankA, i mean i need to show if x is a solution of AA^tx=0 then x is a solution of Ax=0 or of A^tx=0, but how?

Let U = {x : AA^t x = 0}, and V = {x : A^t x = 0}. Assume you have already shown that dim V <= dim U (which implies r(A) >= r(AA^t) ), and now assume dim U > dim V (which would imply r(A) > r(AA^t) ) and find a counterexample (it's simple). When you have shown that dim U > dim V can't hold, then dim U = dim V must hold, and hence r(A^t) = r(A) = r(AA^t). Hope this works.

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MathematicalPhysicist

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- #5

radou

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Actually, that's what's bothering me, too. If we assume it holds for

- #6

AlephZero

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Let the SVD of A be U.D.V^T where D is the diagonal matrix of singular values. Write the SVD of A.A^T in terms of U D and V, and show that A and A.A^T have the same number of zero singular values, hence the same rank.

There is a similar proof involving eigenvalues and eigenvectors.

There is probably a similar but more "elementary" proof involving choosing a set of basis vectors where a subset of the basis vectors spans the null space of A - but I haven't thought idea that right through.

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