Covariance of a nxm matrix with n<m is not positive semidefinite? [duplicate]

Is it true that, for a matrix $$A \in \mathbb{R}^{n \times m}$$ with $$n (so with more features than samples), its covariance matrix is (or might be?) not positive semidefinite? If that's the case, can someone explain and prove it?

• I've already seen that answer but I'm not able to generalize it to my case Oct 24 '18 at 20:47
• I'm pretty sure this question has been asked and answered, so I hope we can identify the duplicate. But in the meantime, note that a basic fact of linear algebra is that every matrix $B$ with more columns than rows has a nontrivial kernel, defined as the subspace of vectors $x$ it sends to zero: that is, $Bx=0$ for some nonzero $x.$ Consequently $x^\prime B^\prime B x=0,$ proving $B^\prime B$ is not definite. When $B$ is the centered version of $A,$ $B^\prime B$ is its covariance matrix.
– whuber
Oct 24 '18 at 21:02
• Thanks whuber, that makes sense. It's not so obvious to me that "every matrix B with more columns than rows has a nontrivial kernel", do you have some additional info/references regarding this statement? Oct 24 '18 at 21:18
• I think the OP has a point in that the proposed duplicate is not explicit about rank-deficient matrices. Someone who knows more about linear algebra than me could expand on @whuber comment so we do have an answered question. Oct 25 '18 at 9:20
• There are many ways to demonstrate the basic premise I asserted. One follows easily by counting dimensions, because the dimension of the kernel cannot be any less than the dimension of the domain $(m)$ minus that of the image $(n).$ See en.wikipedia.org/wiki/Rank%E2%80%93nullity_theorem.
– whuber
Oct 25 '18 at 12:05