# Proving facts about matrices without determinants

1. Apr 2, 2017

### Mr Davis 97

1. The problem statement, all variables and given/known data
Let $A$ and $B$ be $n \times n$ matrices

1) Suppose $A^2 = 0$. Prove that $A$ is not invertible.
2) Suppose $AB=0$. Could $A$ be invertible.
3) If $AB$ is invertible, then $A$ and $B$ are invertible

2. Relevant equations

3. The attempt at a solution
1) Suppose that $A$ were invertible. Then $A^2 = 0 \Rightarrow A = 0$. This is a contradiction, because we know that the zero matrix is not invertible.

So I think this is one way of doing it. What are some others?

2) It seems that we can proceed in a similar way as 1). Assume that $A$ is invertible, then $B=0$. Thus, since $A$ is can be an arbitrary matrix, it can be invertible.

Is this the correct way of doing this?

3) This is the one that I am having the most trouble with. Assume that $AB$ is invertible. So we know that there exists a $n \times n$ matrix $C$ such that $C(AB) = (AB)C = I_n$. So $(CA)B = A(BC) = I_n$. So we can conclude that $B$ has a left-inverse and that $A$ has a right-inverse. But I don't see how we can conclude that $B$ has a right-inverse and $A$ has a left-inverse.

2. Apr 2, 2017

### vela

Staff Emeritus
How do you know $A^2=0$ implies $A=0$? There are non-zero matrices that square to 0.

Same problem. Completely misread the second question.

You should add one more step to each the first proof.

Last edited: Apr 2, 2017
3. Apr 2, 2017

### Staff: Mentor

It doesn't follow that if $A^2 = 0$ then A must necessarily be the zero matrix. Consider $A = \begin{bmatrix} 0 & 1 \\ 0 & 1 \end{bmatrix}$.
All you are given is that AB is invertible, which means it is a square matrix. Does this necessarily mean that both A and B have to be square matrices as well?

4. Apr 2, 2017

### Mr Davis 97

So for the first problem, I am assuming that $A$ is invertible, and then deriving a contradiction. I am not just assuming that the zero matrix is the only matrix that squares to zero.
If I assume that $A$ is invertible, then $A^2 = 0 \Rightarrow A^{-1}A^2 = A^{-1} 0 \Rightarrow A = 0$, but zero can't be invertible.

And for the third problem, we are given that $A$ and $B$ are square matrices.

5. Apr 2, 2017

### Staff: Mentor

Right.
Of course the determinant. Another way is to assume $A \neq 0$ which implies, there is a vector $v$ with $Av \neq 0$ and $Av \in \operatorname{ker} A$.

6. Apr 2, 2017

### Staff: Mentor

No. You don't have an arbitrary matrix. You have certain matrices $A,B$ with $AB=0$.
Of course $A$ can be invertible, namely if $B=0$ as you have said. So the general answer is "yes" by your own reasoning.

7. Apr 2, 2017

### Mr Davis 97

So I am actually confused by my own reasoning. To show that $A$ can be invertible, don't I need to also find a sufficient condition? I showed that $A$ being invertible implies that $B=0$, which means that $B$ being $0$ is necessary. But this is not sufficient to show that there is an instance where $A$ can be invertible, right? Don't I need a sufficient condition?

8. Apr 2, 2017

### vela

Staff Emeritus
You could just give an example. For example, A=I and B=0 satisfied AB=0 but A is invertible.

9. Apr 2, 2017

### Staff: Mentor

The question is: $(\; \exists A \; \exists B \, : \, A \textrm{ invertible } \;\wedge \; AB=0\;)$ true or false?
There is no statement about an equivalence or an implication.

10. Apr 2, 2017

### Mr Davis 97

So to completely answer the question, do I need to supply an example of an invertible matrix $A$, such as $I_n$, as vela said?

11. Apr 2, 2017

### Staff: Mentor

If the statement is false, all that's needed is a single counterexample. If the statement is true, no amount of examples would be sufficient. Also, even though the questions are about n x n matrices, you can usually think about things using 2 x 2 or 3 x 3 matrices, especially when you're looking for counterexamples.

12. Apr 2, 2017

### Mr Davis 97

So what about the last problem?

13. Apr 2, 2017

### Staff: Mentor

That's probably a trick I can't remember. Try to show (in general for any (non Abelian) group) that left- and right-inverses cannot be different by using associativity.

14. Apr 2, 2017

### vela

Staff Emeritus
Of course, answering the question posed depends on determining what the correct statement is. In this case, it's probably a bit easier and clearer to step back and look at the question as a whole. The question asks: if AB=0, is it possible for A to be invertible? Supplying an example definitively shows the answer to the question is yes.

15. Apr 2, 2017

### vela

Staff Emeritus
The complication here is showing that if a left (right) inverse exists, that the right (left) inverse also exists.

16. Apr 2, 2017

### Staff: Mentor

Yes, and it's one of those "Why didn't I see this?" questions, once you know the right equation.

17. Apr 2, 2017

### nuuskur

Argh, without determinants. Must learn how to read :(

If we go by definition: $X$ singular if and only if $X$ not invertible.
Alternatively, could argue assuming for a contradiction $B$ was singular, but then there would exist a vector $x\neq 0$ s.t $Bx = 0$, therefore $(AB)x = 0$ (why?) making $AB$ singular, a contradiction. Therefore $B$ must be invertible, therefore $A$ is invertible.

I can also tell you that using only semigroup properties is not enough to tackle this problem.
Specifically, the existence of L-inverse does not, in general, guarantee the existence of R-inverse.

Since determinants are evil, you should use vectors or perhaps you can think of something ingenious!

What do you think of this proposition?
$AB$ invertible only if $BA$ invertible

Last edited: Apr 2, 2017
18. Apr 2, 2017

### Mr Davis 97

Doesn't this argument only show that $B$ is injective, not necessarily surjective too?

19. Apr 2, 2017

### nuuskur

Since you haven't specified any exotic properties of the ring of matrices you are dealing with, I'm assuming we are talking about matrices over Fields not division rings or god knows what else.

The implications (rather straightforward) are the following
$$B \mbox{ invertible} \iff\ \left (Bx=0\Longrightarrow x=0\right )$$
You should prove it, if you are not convinced, but the proof is obvious.
This will solve the problem at hand.

20. Apr 2, 2017

### Staff: Mentor

One also needs a dimension argument to conclude that injectivity is sufficient.