TITLE: Linear Algebra - four "true or false" questions about matrices and linear systems
QUESTION [6 upvotes]: I'm reviewing for my linear algebra course, and have four "true or false" questions that I'm struggling to prove. I've included my approach to the solutions in brackets below them:
1) If $A^2 = B^2$, then A = B or A = -B, where A and B are nxn matrices 
(Not sure how to approach this one at all)
2) Every 3x3 skew symmetric matrix is singular 
(Pretty sure I have this one correct: Because this is a skew symmetric matrix, $\det(A) = \det(A^T) = \det(-A) = (-1)^n\det(A)$, and when n is odd $\det(A) = -\det(A)$, so $2\det(A) = 0$ and therefore $\det(A) = 0$. As such, the answer is "False" because it is only singular when n is odd)
3) Any system of n linear equations in n variables has at most n solutions 
(A system can have infinitely many solutions if the determinant is zero, right? I just don't know how to prove it)
4) For a square matrix A, A is invertible if and only if $AA^T$ is 
(Not sure how to approach this one, either)

REPLY [2 votes]: I think that it is not necessarily best to prove that invertibility of $A$ is equivalent to that of $A^T$ using the determinant.  Indeed, as user46944 pointed out at https://math.stackexchange.com/a/930345/87579, one can explicitly construct an inverse for $A A^T$ from one for $A$.  This might not be slick, but I think that it gives a more immediate insight into what's going on.
For the converse, note that, if $B \mathrel{:=} A A^T$ is invertible, then $A(A^T B^{-1}) = I$, so that $A$ has a right inverse.  In general, this is all that one can say (consider the shift operator $A(x_0, x_1, x_2, \dotsc) = (x_1, x_2, \dotsc)$, for which $A A^T = I$).  In the finite-dimensional case, an operator is (two-sided) invertible if and only if it has either one-sided inverse, so we are done.