1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Putnam problem on matrices + invertibility

  1. Feb 28, 2017 #1
    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. jcsd
  3. Feb 28, 2017 #2

    fresh_42

    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?
     
  4. Feb 28, 2017 #3
    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##?
     
  5. Feb 28, 2017 #4

    fresh_42

    Staff: Mentor

    Oh, I missed this little word different. You should have written it ##A\neq B## in the first place. :wink:
    And this is the road to the solution, a contradiction to this condition.
     
  6. Feb 28, 2017 #5
    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?
     
  7. Feb 28, 2017 #6

    fresh_42

    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## ?
     
  8. Feb 28, 2017 #7
    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##
     
  9. Feb 28, 2017 #8

    fresh_42

    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.
     
  10. Feb 28, 2017 #9
    So... we have CA = CB. If C were invertible, then A = B, which is a contradiction. Hence C is not invertible. Is that right?
     
  11. Feb 28, 2017 #10

    fresh_42

    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##.
     
  12. Feb 28, 2017 #11
    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.
     
  13. Feb 28, 2017 #12

    fresh_42

    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.
     
  14. Feb 28, 2017 #13

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    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.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Putnam problem on matrices + invertibility
  1. Invertible Matrices (Replies: 2)

  2. Invertible Matrices (Replies: 3)

  3. Invertible matrices (Replies: 3)

Loading...