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For any matrix $A$, $AA^T$ is symmetric, and thereby diagonalizable (Spectral Theorem). This implies that $AA^T$ has n linearly independent eigenvectors and therefore full rank.

Does this imply that for any matrix $A$, $AA^T$ is always invertible ? I know there is something wrong here because from the above chain we can conclude that a symmetric matrix is always invertible, which is definitely not true.

Could someone please point out where I am going wrong?

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No, just cause it has linearly independent eigenvectors doesn't mean it's invertible / full rank. This tells you nothing about if any of those eigenvectors has eigenvalue zero.

In the diagonalizable case here, the subspace spanned by all the eigenvectors that have eigenvalue zero is the kernel. If the kernel is non trivial then the matrix is not invertible and is less than full rank.

If none of them have eigenvalue zero, then yes, it's invertible.

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  • $\begingroup$ So does having $n$ LI eigenvectors tell us anything at all then? $\endgroup$
    – information_interchange
    Mar 26, 2020 at 3:21
  • $\begingroup$ @information_interchange Having $n$ LI eigenvectors is the same thing as being diagonalizable. $\endgroup$
    – spaceisdarkgreen
    Mar 26, 2020 at 4:42
  • $\begingroup$ Not according to the question I asked: math.stackexchange.com/questions/3595431/… $\endgroup$
    – information_interchange
    Mar 26, 2020 at 5:07
  • $\begingroup$ @information_interchange Huh? Where in there do they say those are not equivalent? They say that diagonalizability doesn't imply invertability, which is true. (nor does invertability imply diagonalizability.) In fact you are asking pretty much the same question as this post and receiving pretty much the same answer. $\endgroup$
    – spaceisdarkgreen
    Mar 26, 2020 at 5:20
  • $\begingroup$ @information_interchange A quick rebuttal of what I think from your post is your reasoning: A diagonal (diagonal, not diagonalizable) matrix is not always invertible. In fact a diagonal matrix is invertible if and only if there are no zeros on the diagonal. $\endgroup$
    – spaceisdarkgreen
    Mar 26, 2020 at 5:29

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