# $A$ has full column rank if and only if the symmetric matrix $B=A^TA$ is positive definite.

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?

• Note that $x^TBx=(Ax)^T(Ax)\geq0$, so $B$ is positive semidefinite no matter what $A$ is. When is $(Ax)^T(Ax)=0$? – Arthur Dec 20 '16 at 22:04
• @πr8 yes I know that. – h94 Dec 20 '16 at 22:05
• @Arthur never thought of it that way. That makes a lot of sense. But still don't understand the symmetry. – h94 Dec 20 '16 at 22:07
• @πr8 But where would the column rank come into play? – h94 Dec 20 '16 at 22:09
• It doesn't, for the symmetry - the column rank is only relevant (I think) for the positive-definiteness. – πr8 Dec 20 '16 at 22:09

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.