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The problem is

The matrix $$ A = \begin{bmatrix} 2 & 2 & -2 \\ 5 & 1 & -3 \\ 1 & 5 & -3 \end{bmatrix} $$ has the property $A^3 = 0$. How many distinct eigenvalues does $A$ have?

I can, of course, compute all the eigenvalues, but I am wondering what does $A^3 = 0$ tell us here about the number of distinct eigenvalues?

user5965026
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    If $Av=\lambda v$, what about $A^3v$? – Hagen von Eitzen Aug 09 '21 at 19:38
  • @HagenvonEitzen Ahh then it is $\lambda^3 v$. And the eigenvalues of the zero matrix is all 0, so there is only 1 distinct eigenvalue? – user5965026 Aug 09 '21 at 19:39
  • yes indeed, although you still have to prove $0$ is an eigenvalue I guess. – Asinomás Aug 09 '21 at 19:40
  • @Yorch What "proof" is sufficient? I think my inclination would be to say that the eigenvalues of an upper or lower triangular matrix (which a zero matrix is) are just the elements along the main diagonal, which in this case is all zeros? Is that sufficient proof, or do I need to prove why the eigenvalues of a triangular matrix are the diagonal elements? – user5965026 Aug 09 '21 at 19:41
  • that shows $0$ is an eigenvalue of $A^3$ I think. – Asinomás Aug 09 '21 at 19:42
  • https://en.wikipedia.org/wiki/Nilpotent_matrix

    It's funny that you are given the values of the matrix, when they are irrelevant, the property $A^3 = 0$ suffices.

     – leonbloy Aug 09 '21 at 19:44
  • @leonbloy It really is more funny that red herrings like the explicit entries of $A$ aren't more prevalent in math exercises. They are everywhere in the real world, and most people who will use their math education in their professional lives are going to work there. And even in abstract academics, the ability to filter out unnecessary information is crucial. Yet it is rare that students get a chance to practice it. – Arthur Aug 09 '21 at 19:57
  • @leonbloy It could be that students that had missed the abstract thinking of $A^3=0$ would still have a chance to get points by a direct computation (by the price of spending more time for the test, of course). – A.Γ. Aug 09 '21 at 20:08

2 Answers2

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$A^3=0$ is the characteristic equation applied to $A$:

$$p(A)=0$$

The characteristic equation ($p(\lambda)=\lambda^3=0$) has three roots, all roots are zero, so there is a unique (triple) eigenvalue $\lambda =0.$

mjw
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  • @TravisWillse I thought only eigenvectors could be "generalised"? $0$ is definitely an eigenvalue in the traditional sense, so I guess it depends on how you define its "multiplicity". I'm curious: do you have a reference to an author who refers to generalised eigenvalues? – Theo Bendit Aug 09 '21 at 21:28
  • Please ignore my previous, now-deleted comment, apparently written in a state of sleep deprivation! – Travis Willse Aug 10 '21 at 09:14
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For this particular matrix I don't have anything else to add. So I will look at the general case that you talk about.

Let's suppose that $A$ is a (square) matrix. Then eigenvalues are solutions to the equation $A\vec v = \lambda \vec v$, where $\vec v$ is the corresponding eigenvector. Now it is a relatively simple exercise to show that if $\lambda$ is one such eigenvalue, with corresponding eigenvector $\vec v$, then for each $n\geq 1$, $\lambda ^n$ is an eigenvalue of $A^n$ with eigenvector $\vec v$. (Just multiply each side of the defining equation on the left by $A$ repeatedly.)

Now if $A^3 = 0$, then $\lambda^3$ should be an eigenvalue of $A^3$ and so must be zero. (Note that $A^3\vec v = 0\vec v = \vec 0$ so the only possible eigenvalue is zero.) But then $\lambda^3 = 0$ has only a single solution $\lambda = 0$.

Here are some other questions you might want to think about: what do you think would happen if instead $A^3 = I$ or $A^3 = A$?

SamM
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