Let $A\in M_{n\times n}(\mathbb{R})$ be a matrix. Is it true that $AA^{T}$ is positive-definite?
Clearly $AA^{T}$ is symmetric. I have shown that a symmetric matrix $S\in M_{n\times n}(\mathbb{R})$ is positive-definite if and only if $S$ has only positive eigenvalues. Can this be helpful?
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$\begingroup$Hint: let $v$ be a non zero vector; then, setting $B=A^T$ for simplicity, $$ v^TB^TBv=(Bv)^T(Bv) $$ is positive if and only if $Bv\ne 0$. How can you ensure that $Bv\ne0$ if and only if $v\ne0$?
Conversely, if $AA^T$ is positive definite, what can you say about the rank of $A$?
So, what's a necessary and sufficient condition so that $AA^T$ is positive definite?
$\endgroup$ 3 $\begingroup$$AA^T$ is not necessarily positive definite, but it is positive semi-definite, meaning that $\langle x, AA^Tx \rangle \ge 0$ for all vectors $x$. To see this, note that $\langle x, AA^Tx \rangle = \langle A^Tx, A^Tx \rangle = \Vert A^Tx \Vert^2 \ge 0$. A counter example to positive definiteness is provided, when $n = 2$, by taking
$A= \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}; \tag{1}$
then
$AA^T = A, \tag{2}$
so if $x = (0, 1)^T$,
$\langle x, AA^Tx \rangle = 0. \tag{3}$
It is easy to generalize this example by taking $A$ to be a diagonal matrix in $M_{n \times n}(\Bbb R)$ with at least one zero on the diagonal; many other generalizations are also possible.
Hope this helps. Cheerio,
and as always,
Fiat Lux!!!
$\endgroup$ 4 $\begingroup$For a matrix to be positive definite we need $x^TMx \geq0$. Consider the case where the matrix A is not full rank so therefore has more rows than columns. It should be clear there exist a vector $x$ where $ x \neq 0 $, such that $x^TA = 0$. Thus we have $x^TAA^Tx = 0$ Therefore $AA^T$ cannot be strictly positive definite. (But you can show it is always semi-positive definite.)
$\endgroup$ 1 $\begingroup$No, it is not a positive definite but clearly it is positive semidefinite. Consider the general complex case, and consider $AA^*$ which is Hermitian.$<AA^*x,x>=<A^*x,A^*x>=||A^*x|| \geq 0 \forall x$. Therefore, $AA^*$ is positive semidefinite. Clearly, if $A$ is real, then A is symmetric i.e. $A^*=A^T$ since $\overline {a_{ij}}=a_{ij}$.
However, if you want it to be positive look at $V\backslash\{0\}$ and $A$ injective.
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