# Putnam problem on matrices + invertibility

1. Feb 28, 2017

### Mr Davis 97

1. The problem statement, all variables and given/known data
Let $A$ and $B$ be different $n \times n$ with real entries. If $A^3 = B^3$ and $A^2 B = B^2 A$, can $A^2 + B^2$ be invertible?

2. Relevant equations

3. The attempt at a solution
So, first of all I am just trying to interpret the question correctly. Does "can $A^2 + B^2$ be invertible" mean "does there exist distinct matrices A and B such that $A^2+B^2$ is invertible" or does it mean "prove that $A^2 + B^2$ is invertible for all A and B"?

2. Feb 28, 2017

### Staff: Mentor

It means, prove or disprove that $C:=A^2+B^2$ is invertible for all matrices $A$ and $B$ which satisfy the conditions $A^3=B^3$ and $A^2B=B^2A$.

So either $C$ is never invertible, or we have to find a case, i.e. certain matrices $A$ and $B$ with these conditions, that make $C$ regular. Since this looks not quite easy, I assume the conditions on $A$ and $B$ force $C$ to be not invertible.

Correction: What if $A$ and $B$ are diagonal matrices?

3. Feb 28, 2017

### Mr Davis 97

If A and B were diagonal, and if $A^3 = B^3$, wouldn't that imply that $A = B$, which violates the condition that $A \neq B$?

4. Feb 28, 2017

### Staff: Mentor

Oh, I missed this little word different. You should have written it $A\neq B$ in the first place.
And this is the road to the solution, a contradiction to this condition.

5. Feb 28, 2017

### Mr Davis 97

Sorry, I just pasted the question word-for-word.

So as of right now should I be trying to show that $\det(A^2 + B^2) = 0$, given the conditions?

6. Feb 28, 2017

### Staff: Mentor

No sorry, that was a joke. I once said: "I'm a guy, you have to shout!"

Anyway, it's far easier than that. What is $(A^2+B^2)\cdot A$ ?

7. Feb 28, 2017

### Mr Davis 97

Well, $(A^2 + B^2)A = A^3 + B^2 A = A^3 + A^2 B = A^2 (A+B)$ or $A^3 + B^2 A = B^3 + A^2 B = (A^2 + B^2)B$, but I don't see where any of those two expressions get me. What exactly am I trying to show, if I am not trying to show that $\det (A^2 + B^2) = 0$? That $\det (A^2 + B^2) \neq 0$

8. Feb 28, 2017

### Staff: Mentor

You have $(A^2+B^2)A=(A^2+B^2)B$. Why do you stop here? Write it with $C=A^2+B^2$ if this is easier. I don't think that you have a chance to get your hands on the determinant and if, certainly not the easy way.

9. Feb 28, 2017

### Mr Davis 97

So... we have CA = CB. If C were invertible, then A = B, which is a contradiction. Hence C is not invertible. Is that right?

10. Feb 28, 2017

### Staff: Mentor

Yes. It's even shorter than mine. I wrote $(A^2+B^2)(A-B) = C(A-B)=0$ and with $A \neq B$ we have an element in the image of $A-B$ which is not zero, and thus has to be in the kernel of $C=A^2+B^2$.

11. Feb 28, 2017

### Mr Davis 97

So what got me stuck was that I was convinced that I should have been trying to prove something with determinants. How did you know that determinants wouldn't be involved? If I knew beforehand that using determinants would be futile, I could have gotten the answer quite easily.

12. Feb 28, 2017

### Staff: Mentor

We both used $A \neq B$ and as you have mentioned as answer to my suggestion of diagonal matrices, it's really necessary. But if we apply the determinant, e.g. on $C(A-B)=0$ we get $\det C = 0$ or $\det (A-B) = 0$. But how can we exclude the second case? The determinant can well be zero without the matrices being equal. The same goes for your equation $CA=CB$ and then $\det A = \det B$. The determinant "destroys" some properties in the sense that it makes some different matrices indistinguishable. As we now know that $A^2+B^2$ cannot be regular, there might be a way to operate with determinants, I just don't see one and the multiplication $(A^2+B^2)A$ was too tempting. I probably did too many calculations in groups in my life, so it became somehow natural to play with them as I saw the commutation conditions given.

13. Feb 28, 2017

### Ray Vickson

Usually we try to avoid determinants of sums, so try to stay away from things like $\det(A+B)$. If you look at the horrible formulas for the determinant of a sum, you will see why.