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Math Help - Invertibility of the product matrix-transpose

  1. #1
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    Invertibility of the product matrix-transpose

    Why the matrix AA^T is invertible, if the generic real matrix A has full rank?
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  2. #2
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    If a matrix has full rank, then its transpose also has full rank.
    The product of two invertible matrices is invertible.
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  3. #3
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    I don't think that quite answers the question! The OP didn't say they are invertible, just of "full rank". A square matrix is invertible if and and only if it has full rank but I imagine this problem is for non-square matrices.

    If the matrix A is, say, m by n, then we can think of it as representing a linear transformation from a vector space of dimension n to a vector space of dimension m. If A has full rank, then m> n and A maps the vector space of dimension n one-to-one onto a subspace of dimension n.
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  4. #4
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Riesz View Post
    Why the matrix AA^T is invertible, if the generic real matrix A has full rank?
    Assuming that these are real matrices.


    In general \text{rk}\left(AA^{\top}\right)=\text{rk}\left(A\r  ight). To prove this we first note that in general \text{rk}(M)=\text{rk}\left(M^{\top}\right) so it suffices to prove that \text{rk}\left(A^{\top}A\right)=\text{rk}\left(A\r  ight). But, for this (by first isomorphism theorem [the rank-nullity theorem if that previous theorem is unfamiliar to you]) it is sufficent to prove that \ker A^{\top}A=\ker A. Evidently \ker A\subseteq\ker A^{\top}A and the reverse inclusion follows by noticing that if v\in\ker A^{\top}A then 0=\left\langle A^{\top}Av,v\right\rangle=\left\langle Av,Av\right\rangle and thus Av=\bold{0} and thus v\in\ker A.
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  5. #5
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    Sorry, I have to specify matrix dimensions:

    A \in \mathbb{R}^{nxm}, n \le m.
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