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How to show the following claim

$A^{100} = 0 \implies A^2 = 0$ with $A \in Mat(2 \times 2, K)$

If A is the matrix of a linear map $\phi$ then for all $v \in K^2$ the following identity should be true

$\phi^{99}(v) = A^{99}\cdot v = A^{99} \cdot col_i(A) = 0$

But how to show that $A^2 = 0$?

Arturo Magidin
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meinzlein
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4 Answers4

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In general, if a square matrix $A$ satisfies $p(A)=0$ where $p$ is a polynomial, then the minimal polynomial of $A$ divides $p$. Thus, since your $A$ satisfies $p(A)=0$ with $p(x)=x^{100}$, it follows that the minimal polynomial of $A$ divides $x^{100}$. But the degree of the minimal polynomial is $\le$ the size of $A$ which is $2$. Hence, the minimal polynomial is either $x$ or $x^2$, so that $A=0$ or $A^2=0$.

the L
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  • Thanks for your answer. I can follow so far. The only thing I didn't know is that the degree of the minimal polynimial is $\leq$ the size of A. How do I know this? Does this fact imply that there are no more eigenvalues then the size of a matrix? – meinzlein Feb 02 '12 at 12:53
  • By the Cayley Hamilton theorem, it follows that the size of $A$= the degree of the characteristic polynomial of $A$ is greater or equal of the degree of the minimal polynomial. – the L Feb 02 '12 at 12:57
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    See also http://math.stackexchange.com/questions/36651/interpreting-the-cayley-hamilton-theorem/36662#36662 – lhf Feb 02 '12 at 12:58
  • @anonymous: One does not absolutely need the Cayley-Hamilton theorem to see why the minimal polynomial cannot have degree more than the size of $A$; just careful consideration of the minimal polynomial suffices, as I have indicated in my answer. (This is more a comment on the other question linked to by lhf, actually.) Of course if you know Cayley-Hamilton, it immediately gives the result as well. – Marc van Leeuwen Feb 02 '12 at 16:00
  • Why exactly do I need $A^100 = 0$? The degree of the minimal polynomial is still $\leq$ 2 and therefore $A = 0$ or $A^2 = 0$. – meinzlein Feb 02 '12 at 16:12
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The $(A^nK^2)_{n\ge0}$ form a sequence of subspaces of $K^2$ which is strictly decreasing until it reaches its minimum.

Pierre-Yves Gaillard
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Pick $\lambda$ an eigenvalue of $A$. Then $Ax=\lambda x$ for $x$ a non-zero eigenvector. Inductively you find that $A^{100}x=\lambda^{100}x=0$. Therefore $\lambda^{100}=0$. Hence $A$ has only zero eigenvalues. Since $A$ is a $2\times 2$ matrix, from Cayley Hamilton theorem it follows that $A^2=0$.

Beni Bogosel
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    Your last sentence doesn't follow (and indeed isn't true for larger matrices!) – Cam McLeman Feb 02 '12 at 12:48
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    @CamMcLeman Why it doesn't follow? The characteristic polynomial is $A^2-Tr(A)A+det(A)$, and since all eigenvalues are zero, $Tr(A)=0,det(A)=0$. Of course it is not true for larger matrices, but we are talking about $2\times 2$ matrices. – Beni Bogosel Feb 02 '12 at 12:49
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    Okay, sure. My objection was that your answer makes it appear that having all-zero eigenvalues implies $A^2=0$ without referencing the crucial hypothesis that $A$ be $2\times 2$. – Cam McLeman Feb 02 '12 at 13:33
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The minimal polynomial of $A$ is of degree at most the size $2$ of $A$, and divides $X^{100}$; this means it divides $X^2$, so $A^2=0$.

For an elementary proof that the minimal polynomial $\mu$ of any $n\times n$ matrix $A$ has degree at most $n$ (without using the Cayley-Hamilton theorem) one may argue that for any irreducible factor $P$ of the minimal polynomial the dimension of the kernel of $P(A)$ is at least the degree $d$ of $P$, and so its image has dimension at most $n-d$; induction and the fact that $\mu/P$ is the minimal polynomial of the restriction of $A$ to the image of $P(A)$ complete the proof. But in this concrete situation you don't even need that, as it is clear that the factor $P$ must be $X$, so $P(A)=A$ and $\dim(\ker A)\geq1$ is obvious; therefore it boils down to the argument given by Pierre-Yves Gaillard.

For the record I'll add why $\dim(\ker(P(A)))\geq\deg P$: because $P$ is a nontrivial divisor of $\mu$ it kills some nonzero vector $v$, and then $A^i(v)$ for $0\leq i<\deg P$ are linearly independent (by the irreducibility of $P$) vectors, and they all lie in $\ker(P(A))$ since $A$ commutes with $P(A)$. And that $\mu/P$ is the minimal polynomial of the restriction of $A$ to the image of $P(A)$ is because $(\mu/P)(A)$ must kill that image, and no lower degree polynomial could do that by minimality of $\mu$.

Marc van Leeuwen
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