If we have $B$ is an $(n,n+1)$ matrix and $A$ is a positive definite $(n,n)$ matrix is the resulting quantity positive definite?


It seems like it should be since we have both a $B^T$ and a $B$ in there and we know that $B^TB$ is positive definite, but I am getting stuck on an explicit proof.



It is never positive definite. $B$ is a fat matrix. Thus $Bx=0$ has a nontrivial solution $x$ and $x^TB^TABx=0$.

  • $\begingroup$ Hmm what is a fat matrix? $\endgroup$ Jul 13 '19 at 15:47
  • $\begingroup$ Do all fat matrices have a guaranteed solution? $\endgroup$ Jul 13 '19 at 15:50
  • 1
    $\begingroup$ @user2879934 When the number of columns exceeds the dimension of the vector space, the columns must be linearly dependent. $\endgroup$
    – user1551
    Jul 13 '19 at 15:51
  • $\begingroup$ What about semi-positive definite? $\endgroup$ Jul 13 '19 at 16:05
  • $\begingroup$ Interesting, so if both A and B are semi-positive definite, does that rearrangement guarantee that quantity is >= 0. Sorry B can't be PSD $\endgroup$ Jul 13 '19 at 16:12

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