I am confused about the bi-conditional statement I outlined in the title.
I'll write it again here: (This was part of a solution to an assignment question)
$A$ has full column rank if and only if the symmetric matrix $B=A^TA$ is positive definite.
The definition of column rank that I am aware of states that a $m \times n$ Matrix $A$ has full column rank if each of the columns are linearly independent. So it would be full rank if $rank(A) = n$ in this case.
And I guess, the dimension of $B = A^TA$ is $n \times n$. So it would be a square matrix that is full column rank. Now going back to the statement, why can we say that this matrix $B$ is symmetric, let alone a positive definite?
Best Answer
The definition of column rank is the dimension of the image of $A$, so 'full column rank' means that $A$ is surjective. The columns are linearly independent iff $A$ is injective. $B := A^{\mathsf T} A$ is symmetric: $$ B^{\mathsf T} = (A^{\mathsf T} A)^{\mathsf T} = A^{\mathsf T} A = B $$ $A$ is injective iff $B$ is positive definite: Let $A$ be injective. Then $v^{\mathsf T} B v = v^{\mathsf T} A^{\mathsf T} A v = \|Av\|^2 > 0$ for all $v \neq 0$. If $B$ is positive definite, then it is injective. Since $B = A^{\mathsf T} A$ we have $\ker(A) \subset \ker(B) = 0$, which implies that $A$ is injective.