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)

For 1) a counter example is $\begin{pmatrix} 1 & 0 \ 0 & -1 \end{pmatrix}$ and the identity matrix. — James, Sep 14 '14 at 00:42
I'm not sure if everyone would agree, but I personally think these would be more useful as separate questions. It would be hard for anyone in the future searching for your problems to find them if they are in a question called Linear Algebra — Ben Millwood, Sep 14 '14 at 00:54
@BenMillwood I understand, I just didn't want to flood the new question area with four of my own. I'll try to group things together so they're easier to search in the future! — Alex, Sep 14 '14 at 01:02
The first question is duplicate of this one: http://math.stackexchange.com/questions/871591/if-a2-b2-then-a-b-or-a-b — Martin Sleziak, Sep 14 '14 at 07:11
For the second question see http://math.stackexchange.com/questions/200594/a-problem-on-skew-symmetric-matrix and http://math.stackexchange.com/questions/795160/odd-dimensional-skew-symmetric-matrix-is-singular-even-in-a-field-of-characteri — Martin Sleziak, Sep 14 '14 at 07:15
The fourth question is closely related to http://math.stackexchange.com/questions/327666/if-operatornameranka-m-can-we-say-anything-about-operatornamerankaa — Martin Sleziak, Sep 14 '14 at 07:45
Now I think that some parts of your question should be closed at duplicates. This serves several purposes. It is better to have information about a problem collected in one place, rather then in several posts. Although it is better not to waste other users time by answering questions, which already have answers on this site. But it is not very probable that we will find an exact duplicate of your question having for unrelated parts. (cont...) — Martin Sleziak, Sep 14 '14 at 07:49
(cont...) The consensus from this discussion on meta seems to be that posts consisting of several unrelated questions should be closed. — Martin Sleziak, Sep 14 '14 at 07:50

score 6 · Accepted Answer · edited Sep 14 '14 at 11:46

6

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

edited Sep 14 '14 at 11:46

David Richerby

2,450

answered Sep 14 '14 at 00:45

Ben Grossmann

225,327

Alright, I see. I never even thought of using SPECIFIC counter-examples for any of these, I was trying to work out some generalized ones. Is there a way to answer this question while giving a generalized answer?
Oh right, thanks for pointing that out!
Just as in question 1, is there any way I could use a generalized solution to disprove this?
I'm afraid I'm not following you on this one.

Alex

Sep 14 '14 at 00:54

A general statement can be proven with a general proof or disproven with a specific counter-example. See my update for 4. – Ben Grossmann Sep 14 '14 at 01:02

1

How does your example refute 1? $;$ – Sep 14 '14 at 06:50

@RickyDemer If you let $A$ be the $0$-matrix and $B$ be the matrix $\begin{pmatrix} 0 & 1 \ 0 & 0 \ \end{pmatrix}$ then $A^{2} = B^{2}$, but $A \neq B$ and $A \neq -B$, which is a counter example to the statement in $1$. – layman Sep 14 '14 at 18:37

score 2 · Answer 2 · answered Sep 14 '14 at 01:46

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.

layman · Answer 3 · 2014-09-14T00:57:28.880

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.

score 1 · Answer 4 · answered Sep 14 '14 at 00:56

1

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

answered Sep 14 '14 at 00:56

ncmathsadist

49,383

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

4 Answers4