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Let $A$ be a $ 2 \times 2 $ matrix and a positive integer $k \geq 2$. Prove that $A^k = 0 $ iff $A^2 = 0$.

I can make it to do this exercise if I have $ \det (A^k) = (\det A)^k $. But this question comes before this.

Thank you very much for your help!

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    $\begingroup$ $A^3=A^2\cdot A,\ A^4=A^2\cdot A^2\cdots$ $\endgroup$ – Samrat Mukhopadhyay May 23 '14 at 8:42
  • $\begingroup$ @SamratMukhopadhyay It was "iff", which means "if and only if". Both direction needs to be proved. $\endgroup$ – peterh May 23 '14 at 8:42
  • $\begingroup$ If $A^k = 0$ then ? thanks $\endgroup$ – daisy May 23 '14 at 8:43
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    $\begingroup$ @SamratMukhopadhyay No, it doesn't. These are matrices. $\endgroup$ – peterh May 23 '14 at 8:47
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    $\begingroup$ @SamratMukhopadhyay I think the problem is very clear even without any editing. $\endgroup$ – Tunococ May 23 '14 at 8:53
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The solution using the minimal polynomial and Cayley-Hamilton is a bit of an over-kill and somewhat of a magic solution. I prefer the following non-magic solution (for the non-trivial implication, and there is no need to assume the matrix is $2\times 2$).

Think of $A$ as a linear operator $A: V \to V$ with $V$ an $n$-dimensional vector space, and suppose $A^t=0$. We'll show that $A^n=0$. Now, for each $m\ge 1$ consider the space $K_m$, the kernel of $A^m$. It is immediate that $K_{m}\subseteq K_{m+1}$. Since $A^t=0$ it follows that $K_t=V$. It is also immediate that if $K_m=K_{m+1}$, then $K_m=K_{r}$ for all $r>m$.

Thus, the sequence of kernels is an increasing sequence of subspaces, that stabilizes as soon as the next step is equal to the previous one, and it is eventually all of $V$. Now we will use the fact that $V$ is of dimension $n$. Considering the dimensions of the kernels, the above implies that the sequence of dimensions is strictly increasing until it stabilizes. Since it reaches $V$, the dimensions reach $n$ and since the dimension of $K_1$ is not zero, the sequence of dimensions starts at $1$ or more. That means it has to stabilize after no more than $n$ steps, and thus $K_t=V$ for some $t\le n$. But then $K_n=V$, and thus $A^n=0$.

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    $\begingroup$ Very nice answer! But just to spare future readers the confusion that I experienced for a while, let me comment here that $A^t$ means the $t$th power of $A$, not the transpose... $\endgroup$ – Hans Lundmark Apr 6 '16 at 9:53
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    $\begingroup$ Good answer, +1. I'll gripe very slightly about the last sentence of the second paragraph; this is not quite as immediate as the previous stuff. There is an easy one-line proof, but it does require some imagination and is not a matter of just writing down what the statement means. $\endgroup$ – Marc van Leeuwen Apr 8 '16 at 8:31
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The standard and more general approach to solve this kind of problems is to consider the minimal polynomial of $A$ (see the other answers here). However, in this particular problem, as $A$ is only $2\times2$, one may solve the problem as follows. As the forward implication is trivial, we consider only the backward implication

Proof 1. If $A^k=0$ for some $k\ge2$, then $A$ is singular and it has at most rank 1. Therefore, $A=uv^T$ for some vectors $u$ and $v$. Note that for any $n\ge1$, $$ A^n=\underbrace{(uv^T)(uv^T)\cdots(uv^T)}_{n \text{ times}} =u\underbrace{(v^Tu)}_{\text{scalar}}\cdots(v^Tu)v^T=(v^Tu)^{n-1}A. $$ In particular, $A^k=(v^Tu)^{k-1}A$ and $A^2=(v^Tu)A$. Now the result follows immediately.

Proof 2. If $A^k=0$ for some $k\ge2$, then $A$ is singular. Let $Ax=0$ for some $x\ne0$. Extend $x$ to a basis $\{x,y\}$ of the underlying vector space. Let $Ay=px+qy$. By mathematical induction, we have $A^ny=pq^{n-1}x + q^ny$ for any $n\ge1$. Now use $A^k=0$ to argue that $q=0$. Hence $A^2x=A^2y=0$.

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  • $\begingroup$ @Willie Wong Thanks for the edit. $\endgroup$ – user1551 May 23 '14 at 11:07
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    $\begingroup$ You're welcome. Nice solution, by the way. In spirit it is the same as Ittay's, but in this specific case the argument is much cleaner. $\endgroup$ – Willie Wong May 23 '14 at 11:10
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One direction is trivial, so let's consider the other.

I will give you some hints towards a solution.

Suppose $A^k = 0$. This means that $A$ satisfies the polynomial $x^k$. What does this say about the minimal polynomial of the matrix? What in turn does this say about the characteristic polynomial? Finally, consider the Cayley-Hamilton theorem.

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  • $\begingroup$ I known the minimal polynomical of the matrix .And $A$ satisfies $X^2 - (a + d)X +(ad - bc)E = 0 $ $\endgroup$ – daisy May 23 '14 at 8:56
  • $\begingroup$ @daisy You can narrow the minimal polynomial down to one of two forms. Do you know what form(s) the minimal polynomial must take? $\endgroup$ – EuYu May 23 '14 at 8:58
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    $\begingroup$ I think one only needs the facts that the minimal polynomial $f$ divides $x^k,$ and that $\text{deg}(f)\le 2.$ $\endgroup$ – awllower May 23 '14 at 8:59
  • $\begingroup$ @awllower The fact that $\deg(f) \le 2$ comes from the Cayley-Hamilton theorem though (although I guess you can prove it independently). $\endgroup$ – EuYu May 23 '14 at 9:03
  • $\begingroup$ I would like to comment that this solution, while classical, is, in my view, a magical over-kill. I used to love it the first time I saw it cause it is somewhat awe inspiring. But then I grew to dislike it since it does not explain what is going on. $\endgroup$ – Ittay Weiss May 23 '14 at 10:32
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The operator $A$ is nilpotent, therefore there exists a basis where $A$ is strictly upper triangular by Engel's Theorem. It follows that $A^2 = 0$.

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    $\begingroup$ talk about an over-kill :D $\endgroup$ – Ittay Weiss May 23 '14 at 11:10
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    $\begingroup$ @IttayWeiss Maybe. But Engel's Theorem can be proved using elementary linear algebra. $\endgroup$ – mez May 23 '14 at 11:14
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Here's a naive solution that uses only the definition of linear independence, the notion of dimension, and the fact that $\dim \Bbb F^2 = 2$.

One direction is trivial. For the other direction, for clarity of presentation we'll give an explicit argument for $k = 3$, but the general case is totally analogous:

Suppose $A^2 \neq 0$; then, there is a vector $\xi \in \Bbb F^2$ such that $A^2 \xi \neq 0$. We'll show that the triple $(\xi, A \xi, A^2 \xi)$ is linearly independent, which implies that $\dim \Bbb F^2 \geq 3$, a contradiction:

Suppose not; then, there are coefficients $a, b, c$, not all $0$, such that $$0 = a \xi + b A \xi + c A^2 \xi.$$ Multiplying both sides of this equation by $A^2$ gives $$0 = a A^2 \xi + b A^3 \xi + c A^4 \xi = a A^2 \xi$$ (the second equality uses that $A^3 = 0$). Since $A^2 \xi \neq 0$, we must have $a = 0$ and so $$0 = b A \xi + c A^2 \xi .$$ Multiplying both sides by $A$ similarly leads to $b = 0$, leaving $$c A^2 \xi = 0,$$ so $c = 0$, a contradiction.

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One direction is clear.

Suppose $A^k = 0$ . This means that the minimal polynomial of $A$, call it $g(x)$, satisfies $$g(x) \mid x^k$$ Thus $$g(x) = x^n \ \ \ n \leq k$$ But the minimal polynomial divides the characteristic polynomial, which has degree $2$, and so $n = 1$ or $n = 2$. In both cases $$A^2 = 0$$

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  • $\begingroup$ Unfortunately, using the characteristic polynomial requires the determinant, which is not an admissibile tool, according to the question. $\endgroup$ – egreg May 23 '14 at 12:15
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A^k=0 iff A^2=0 k positive interger >2 you can write k=2+n and n>=1. Using induction: A^2+n=0 Then A^2=0. Take A^3=A^2A=0. And prove A^2+n+1=0 =>A^2=0 and A^2+nA=0. Then A^2=0.

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  • $\begingroup$ If you assume $A^2=0$, then it's quite obvious that $A^2=0$, isn't it? If your aim is to prove generally that $A^k=0$ for some $k>0$ implies $A^2=0$, then you can't, because it's false. The hypothesis that the matrix $A$ is $2\times2$ is essential: without using it, any proof of the statement is wrong. $\endgroup$ – egreg Sep 26 '16 at 16:15
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A^k=0 iff A^2=0 k positive interger >2 you can write k=2+n and n>=1. Using induction: A^2+n=0 Then A^2=0. Take A^3=A^2A=0. And prove A^2+n+1=0 =>A^2=0 and A^2+nA=0. Then A^2=0.

First direction A^2=0 then A^k=0.

You can take A and B as a matrix.. Lets represent both matrix . We have A=[a b over c d] and B=[b1 b2 over b3 b4] then we have B times A=0. (multiplying) we have BA= (b1a+b2c and b1b+ b2d over b3a+b4c and b3b+b4d) It was the matrix. Then b1a+b2c=0 b1b+ b2d=0 b3a+b4c=0 b3b+b4d=0 Where we have (ad-bc)b1=0 (ad-bc)b2=0 (ad-bc)b3=0 and (ad-bc)b4=0 Then we have 2 possibilities: b1=b2=b3=b4=0 or ad-bc=0 this is B=0 or ad=bc Obviously if A^2=0 then A^k =0 for every k bigger than 2.

A second direction A^k=0 then A^2=0 :

Now let's prove A^k=0 for k>2 with k as positive integer. Then A^2=0. Watch first k>2, then you can write k=2+n with k positive integer n>=1. Let's prove using induction that: if A^2+n=0 then A^2=0 For n=1 we have A^3=A^2.A=0. Previously we worked on matrix we can say A^2=0 or ad=bc. Let's represent A^3 as matrix: A^3=[(a^2+bc)a+(ab+bd)c and (a^2+bc)b+(ab+bd)d over (ca+dc)a+(bc+d^2)c and (ca+dc)b+(bc+d^2)d)] It was the matrix. A^3=0 then (a^2+bc)(a+d)=0 (ab+bc)(a+d)=0 (ca+dc)(a+d)=0 (bc+d^2)(a+d)=0 Then let's represent the matrix as : A^2=[a^2+bc and ab+bd over ca+dc and bc+d^2]=[0 and 0 over 0. And 0] Then a=-d but we have ad=bc then a^2+bc=0 bc+d^2=0 ab+bd=0 and ca+dc=0 then A^2=0 And prove that A^2+n+1=0 then A^2=0 If A^2+n.A=0 then A^2+n.A=0. Because previous work A^2+n=0 or ad=bc (induction) we conclude A^2+n=0 and Because the induction hypothesis we have A^2=0 finally we have A^k=0 then A^2=0. Done

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  • $\begingroup$ Do you consider this procedure as trivial solution or trivial implications? I am learning to do it by myself and I would really like to know why. $\endgroup$ – Chucky Oz Sep 26 '16 at 19:03
  • $\begingroup$ Is Cayley-Hamilton the only way you know to solve this? Come on guys I am not an expert but I have your dislikes but not your arguments... $\endgroup$ – Chucky Oz Sep 26 '16 at 19:52

protected by Alex M. Sep 26 '16 at 17:23

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