Show that the $3\times 3$ matrix $A^TA$ is symmetric but singular — what vectors are in its null space?

Question

Given a $3\times 3$ matrix $A$, it is obvious that $A^TA$ is symmetric, $$\because \left(A^TA\right)^T = A^TA.$$

However, I am unsure how to show that said matrix is in fact singular. I know that a matrix is singular if and only if $\det(A)=0$, but I am unsure how to show that this is the case.

Answer 1

To show that $A^T A$ is singular if and only if $\det A= 0$, you could use the results $$\det(A B ) = \det A \det B \qquad\text{and}\qquad \det A^T =\det A.$$

Answer 2

As for null space. We know that $KerA \subset Ker(A^TA)$ and also $dimKerA = 3 - rank(A) = 3 - rank(A^TA) = dimKer(A^TA)$ hence $Ker(A^TA) = Ker(A)$