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?
2 Answers
Maybe it helps you to distinguish between a matrix $M$ being positive definite (and allow for $x^TMx \geq 0$) and strictly positive definite (your definition).
If a matrix $A$ has more rows than columns, it cannot have full row rank and there is a vector $x\neq 0$, such that $x^TA=0$. And thus $x^TAA^Tx = 0$, so that $AA^T$ can be shown to be positive but not in the strict sense.
With the same arguments you can then conclude, that $A^TA$ is strictly positive definite, if, and only if, $A$ has full rank.
$AA^T$ 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.
