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Let $A$ be a square invertible matrix which its members are real numbers.

Prove/disprove:

There cannot be a matrix $A$ that satisfies: $A^3+A=0$

I did that:

$$A^3=-A$$ $$A^{-1}A^3=-AA^{-1}$$ $$A^2=-I$$

$$|A|^2=-|I|$$ $$|A|^2\ne-1$$

That is how a prove this, is this right?

because I'm not quite sure about my solution. should I consider a different approach of solving this?

Some help/tips will be highly appreciated :)

Thanks

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    $\begingroup$ If $A=I$ then you'll get $2I=\textbf 0$. clearly false $\endgroup$
    – David P
    Dec 27, 2014 at 19:51
  • $\begingroup$ Is my solution right? $\endgroup$
    – FigureItOut
    Dec 27, 2014 at 19:52
  • $\begingroup$ Your solution is fine. $\endgroup$
    – EuYu
    Dec 27, 2014 at 19:53
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    $\begingroup$ @EuYu Unless I'm delusional, $\det(-I)\neq -\det(I)$ in general. $\endgroup$
    – Gabriel Romon
    Dec 27, 2014 at 19:55
  • $\begingroup$ @LeGrandDODOM Hm, you're right. This at least works for odd sized matrices. $\endgroup$
    – EuYu
    Dec 27, 2014 at 19:56

5 Answers 5

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The key is to look at the polynomial $x^3+x = x(x^2+1) = 0$. If a matrix satisfies this polynomial, the eigenvalues must be in the set $\{0, \pm i\}$. Since we want the matrix to be invertible, this eliminates $0$. Since the matrix is real, the eigenvalues must be in conjugate pairs. Hence we know that the eigenvalues are $\pm i.$

Any matrix that satisfies $A^2+I = 0$ will suffice.

Let $A=\begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix}$.

Then $A$ is real, invertible and $A^3+A = 0$.

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  • $\begingroup$ Can you show me the approach of finding this matrix? I would be able to just come up with it right? Thanks :) $\endgroup$
    – FigureItOut
    Dec 27, 2014 at 20:04
  • $\begingroup$ Well, start with the polynomial $x^2+1 = 0$, this gives the eigenvalues. You want real values, so that adds an additional constraint. The plane rotations are always a good source of examples for this sort of thing. Here a rotation of $\pm {\pi \over 2}$ works. $\endgroup$
    – copper.hat
    Dec 27, 2014 at 20:05
  • $\begingroup$ Can you please elaborate a bit on the process of finding the matrix? Cause I haven't understood yes what I can do with $x^2+1$, how did you come up with it, how do you find eigenvalues with it and how do you find the rest of the matrix is quite not clear to me :| $\endgroup$
    – FigureItOut
    Dec 27, 2014 at 20:10
  • $\begingroup$ I added a comment showing how I know the eigenvalues. As an aside, you can represent complex numbers by a $2 \times 2$ matrix ('Google' something like '2x2 matrix complex numbers'). Then look at the representation for $i$ (since $i^2 = -1$). $\endgroup$
    – copper.hat
    Dec 27, 2014 at 20:10
  • $\begingroup$ @FigureItOut If you have seen companion matrices (which play an important role in the rational canonical form), then you can see that copper.hat's answer is just the (negative of) the companion matrix for the polynomial $x^2 + 1$. $\endgroup$
    – Viktor Vaughn
    Dec 27, 2014 at 21:14
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I always suggest to try the trivial example of $A=I$ to see what happens.

In this question, what do you get ?

More generally, try a few examples to get a feel for the question before pursuing a proof (this can be beneficial even if the claim is true, seeing how it works out in a real example)

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You changed the question!

Answer to old question: You are asked to prove or disprove that your formula hold for all real matrices. A good first step is to look for examples where the formula doesn't hold. If you can find one matrix where the formula doesn't hold, then you have disproved the formula. You have proved that if a matrix $A$ satisfies the equation, then $A^2$ must be $-I$. But not all matrices satisfy this. So you can just pick your favourite invertible matrix $A$ such that $A^2 \neq -I$.

How about $$ A = \pmatrix{1 & 0 \\ 0 & 1}? $$

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  • $\begingroup$ The above matrix is not invertiable as assumed $\endgroup$
    – Belgi
    Dec 27, 2014 at 19:57
  • $\begingroup$ @Belgi: Yes, I just realized that and updated my answer. $\endgroup$
    – Thomas
    Dec 27, 2014 at 19:58
  • $\begingroup$ OK. But wouldn't it be more educational to write $A=I$ so it won't look like a bunch of numbers the someone who just started learning about matrices ? $\endgroup$
    – Belgi
    Dec 27, 2014 at 20:01
  • $\begingroup$ @Belgi: Maybe true. I will leave it as is. $\endgroup$
    – Thomas
    Dec 27, 2014 at 20:01
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You see that your matrix must satisfy $A^2 + I = 0$ so its minimal polynomial is $x^2 + 1$. Can you find a matrix with this property? Hint: Try the matrix corresponding to a rotation of $\pi/2$ counterclockwise.

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One hint is that every matrix satisfies its own characteristic equation (Cayley-Hamilton Theorem). So, if you have a matrix $A$ whose characteristic polynomial is $p(\lambda)$, then $p(A) =0$.

In this case, you have $A^3+A = A (A^2+I) =0$. So, if you start with an invertible matrix whose characteristic polynomial is $p(\lambda) = \lambda^2+1$ (which is easy to find -- $A=\begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix}$ is the easy example), then you see that $A^2+I=0$ and thus $A (A^2+I) = A^3+A =0$.

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