I am trying to show that if $A$ is a $2 \times 3$ matrix of rank $2$, then $AA^{T}$ is symmetric and positive-definite.
Showing that it's symmetric seems simple. Take $A = \left[\begin{array}{lll} a & c & f \\ b & d & g \end{array}\right]$. Then $AA^{T} = \left[\begin{array}{ll} a^{2}+c^{2}+f^{2} & a b+c d+f g \\ a b+c d+f g & b^{2}+d^{2}+g^{2} \end{array}\right]$, which is clearly symmetric.
However, this seems a bit long-winded. Is there a more general way to show this ? What is the usual interpretation of the matrix $AA^{T}$ ?
Now, to show that it's positive definite I understand that I need to show that $q(x) = x^{T}Ax \geq 0 \space \forall x$ and $q(x)=0 \iff x=\vec{0}$.
What is the best route to show this here ? Can I use the fact that the matrix has rank 2? Calcualting the eigenvalues, and determining if they're both positive, for general entries, also seems long-winded. I'm sure there a straight-forward way to show this.
A matrix $M$ is symmetric iff $M^T = M$. You can therefore prove the claim for any matrix $A$ as follows: $$ (AA^T)^T = (A^T)^T(A)^T = AA^T $$ by using the transpose property that $(AB)^T = B^TA^T$. To show positive-definiteness, notice that $$ u^T (AA^T) u = (A^T u)^T(A^Tu) = ||A^Tu||^2 \geq 0$$ for any $u \in \mathbb{R}^2$. Moreover, $\text{rank}(A)=2$ implies that $A^T : \mathbb{R}^2 \to \mathbb{R}^3$ is also of rank 2 (since column rank is equal to row rank), which in turn implies by the rank-nullity theorem that $A^T$ has trivial kernel, i.e. $A^Tu = 0$ implies $u = 0$. We conclude that $AA^T$ is positive-definite.