TITLE: positive definite and transpose
QUESTION [7 upvotes]: When a matrix A has m rows and n columns (m>n), explain why $AA^{T}$ can't be positive definite. For the same matrix A, is $A^{T}A$ always
  positive definite? If so, explain. If not, what is the condition for A
  so that $A^{T}A$ is positive definite?

Now $A^{T}$ is the transpose of A. This means the columns of $A^{T}$ are formed with the corresponding rows of A.
Positive definite means that $x^{T}Ax$ >0 for all$x\neq 0$.
Also with square symmetric matrices, the quadratic form $x^{T}Ax$ is positive definite if and only if the eigenvalues of A are all positive.
But how does this show that $AA^{T}$ is not positive definite?

REPLY [5 votes]: $AA^T$ is always positive-semidefinite, that is $x^TAA^Tx \geq 0$ and the eigenvalues of $AA^T$ are real and non-negative.
The only condition for $AA^T$ being positive-definite, that is $x^TAA^Tx > 0$ and the eigenvalues of $AA^T$ are real and positive, is in the case of $A$ being full column rank in your case. This is because, I assume that $A$ is a real matrix, $\operatorname{rank}\{A^TA\} = \operatorname{rank}\{A\}$. If $A$ is not full column rank, then $A^TA$ has a non-trivial kernel, and therefore $0$ is an eigenvalue of $A^TA$, thus $A^TA$ is not positive-definite in that case.