Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $A^2 = 0$?

share|improve this question

4 Answers 4

up vote 20 down vote accepted

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 $A^2=0$.

share|improve this answer
    
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 2 '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 2 '12 at 12:57
1  
See also math.stackexchange.com/questions/36651/… –  lhf Feb 2 '12 at 12:58
    
Cheers.. thank you.. –  meinzlein Feb 2 '12 at 13:13
    
@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 2 '12 at 16:00

The $(A^nK^2)_{n\ge0}$ form a sequence of subspaces of $K^2$ which is strictly decreasing until it reaches its minimum.

share|improve this answer

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 $A^2=0$.

share|improve this answer
2  
Your last sentence doesn't follow (and indeed isn't true for larger matrices!) –  Cam McLeman Feb 2 '12 at 12:48
1  
@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 2 '12 at 12:49
2  
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 2 '12 at 13:33

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 $\mu$.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.