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 3×3 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)
Best Answer
1: Nope. Consider $\left(\begin{smallmatrix}0&1\\0&0\end{smallmatrix}\right)$ and $\left(\begin{smallmatrix}0&0\\0&0\end{smallmatrix}\right)$, or $\left(\begin{smallmatrix}1&0\\0&1\end{smallmatrix}\right)$ and $\left(\begin{smallmatrix}1&0\\0&-1\end{smallmatrix}\right)$.
2: By your reasoning it's true since we're only considering $n=3$.
3: You're right; this is false. It's enough to have a counter-example such as $$ x+y = 0\\ 2x + 2y =0 $$ 4: This is true; note $\det(AA^T)=\det(A) \det(A^T) = [\det(A)]^2$. Conclude $\det(A)=0 \iff \det(AA^T) = 0$