Linear Algebra - four "true or false" questions about matrices and linear systems

Question

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)

Answer 1

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 $\det(A)=0 \iff \det(AA^T) = 0$

Answer 2

Here is an incomplete answer.

For 4), you need to prove both:

(i) $A$ is invertible implies $AA^{T}$ is invertible

(ii) $AA^{T}$ is invertible implies $A$ is invertible.

To prove (i), suppose $A$ is invertible. Then $A^{-1}$ exists. Then $(A^{-1})^{T}$, the transpose of the inverse matrix, exists. Now, recall that for any matrices $A$ and $B$, $(AB)^{T} = B^{T}A^{T}$. Then multiplying $AA^{T}$ by $(A^{-1})^{T}$, we get $AA^{T}(A^{-1})^{T} = A(A^{-1}A)^{T} = A(I^{T}) = AI = A$, and so $AA^{T}(A^{-1})^{T} = A$. Thus, multiplying on the right of each side of the equation by $A^{-1}$ gives $AA^{T}(A^{-1})^{T}A^{-1} = AA^{-1} = I$. So, the inverse of $AA^{T}$ is $(A^{-1})^{T}A^{-1}$.

The proof for (ii) is not exactly like the proof above, but I think it can be worked out using only as much thought as the proof above. But it looks like some other users have given a much shorter and simpler proof for 4.

Answer 3

For number 4, $$\det(AA^t) = \det(A)\det(A^t) = \det(A)^2.$$ Hence $\det(A) = 0 \iff \det(AA^t) = 0$.

Answer 4

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.