Matrix Proof

**lanierms** · September 6th 2011, 04:32 AM

Let A be a 2x2 matrix.

Then, prove the following:

$A^n=0 \Longrightarrow A^2=0$

**FernandoRevilla** · September 6th 2011, 04:53 AM

Originally Posted by lanierms

Let A be a 2x2 matrix. Then, prove the following: $A^n=0 \Longrightarrow A^2=0$

Hint: $p(x)=x^n$ is an annihilator polynomial of $A$ , this means that the minimal polynomial of $A$ is $\mu(x)=x$ or $\mu(x)=x^2$ so the canonical form of $A$ is $J=\begin{bmatrix}{0}&{0}\\{0}&{0} \end{bmatrix}$ or $J=\begin{bmatrix}{0}&{1}\\{0}&{0} \end{bmatrix}$

**alexmahone** · September 6th 2011, 04:53 AM

Originally Posted by lanierms

Let A be a 2x2 matrix.

Then, prove the following:

$A^n=0 \Longrightarrow A^2=0$

It is known that the degree of an n × n nilpotent matrix is always less than or equal to n. (I'm searching for a proof online...)

**lanierms** · September 6th 2011, 07:08 AM

what is annihilator polynomial? I don't get what you're saying.

**LoblawsLawBlog** · September 6th 2011, 07:29 AM

This might work, but check my computations. Since $A^n=0$ , $\mbox{det}A^n=0\Rightarrow \mbox{ det}A=0$ , so if we call the entries in A a,b,c,d, we get ad=bc. Now this is the part you need to check- it looks like using this gives a nice expression for any power of A, with entries involving powers of (a+d). I think this should do it.

**FernandoRevilla** · September 6th 2011, 08:17 PM

Originally Posted by lanierms

what is annihilator polynomial? I don't get what you're saying.

Have you covered canonical forms of a matrix?.

P.S. By the way, surely you had a little lapse posting this question in Pre-Calculus.

**HallsofIvy** · September 7th 2011, 08:10 AM

An "annihilator polynomial" for an object, A, which might be a number, matrix, function, or anything for which multiplication and addition are defined, is a polynomial, p(x), such that p(A)= 0.

**Drexel28** · September 7th 2011, 09:07 PM

Originally Posted by lanierms

Let A be a 2x2 matrix.

Then, prove the following:

$A^n=0 \Longrightarrow A^2=0$

If you don't know Canonical forms as Dr. Revilla suggested, you could just triangularize it and get the same conclusion.

**FernandoRevilla** · September 7th 2011, 11:16 PM

In a more auto content way we can reason in the following way: if $A^n=0$ then $\det A=0$ . This means that $A=0$ (and the result is trivial) or $\textrm{rank}A=1$ i.e. $A=\begin{bmatrix}{a}&{\lambda a}\\{b}&{\lambda b} \end{bmatrix}=\begin{bmatrix}{a}\\{b} \end{bmatrix}\begin{bmatrix}{1}&{\lambda} \end{bmatrix}$ with $\begin{bmatrix}{a}&{b} \end{bmatrix}\neq \begin{bmatrix}{0}&{0} \end{bmatrix}$ . So:

$A^n=\ldots=(a+\lambda b)^{n-1}A=0\Leftrightarrow (A=0) \vee (a=-\lambda b)$

If $A=0$ , then $A^2=0$ . If $a=-\lambda b$ then, $A=\begin{bmatrix}{-\lambda b}&{-\lambda^2b}\\{b}&{\lambda b} \end{bmatrix}$ and also $A^2=0$ .

**Unbeatable0** · September 8th 2011, 02:48 AM

This can be done with quite simple linear algebra.

Since $A^n = 0$ , there is a minimal $m$ such that $A^m=0$ .

$A^{m-1} \ne 0$ and therefore there is a vector $v$ for which $A^{m-1}v \ne 0$ .

It's easy to prove that the set $\{v,Av,A^2v,\cdots,A^{m-1}v\}$

is linearly independent, and contains $m$ elements.

Can you finish now?

**FernandoRevilla** · September 8th 2011, 09:55 AM

Originally Posted by Unbeatable0

This can be done with quite simple linear algebra.

In my previous post we can avoid the concept of rank : if $A=\begin{bmatrix}{a}&{c}\\{b}&{d} \end{bmatrix}$ with $\det A=0$ and $(a,b)\neq (0,0)$ then, $(c,d)=\lambda (a,b)$ etc. So, the problem can be solved using only the concepts of determinant and product of matrices, obtaining more simple algebra . I don't mean exactly linear algebra (some say Linear Algebra ends when Cramer's Rule starts) .

**ymar** · September 8th 2011, 06:58 PM

Originally Posted by FernandoRevilla

(some say Linear Algebra ends when Cramer's Rule starts) .

Isn't it "where determinants start"? I read an interesting paper lately: http://www.axler.net/DwD.pdf It was a revelation to me, a mediocre student who instantly falls asleep on seeing formulas with many indices.

As for the problem, do you think it's tractable by means of finding a formula for $A^n$ dependent on a, b, c and d? I've just tried it and I can't but I'm not very good.

**FernandoRevilla** · September 8th 2011, 08:11 PM

Originally Posted by ymar

Isn't it "where determinants start"?

Well, that depends on the way we categorize.

As for the problem, do you think it's tractable by means of finding a formula for $A^n$ dependent on a, b, c and d? I've just tried it and I can't but I'm not very good.

The characteristic polynomial for a $2\times 2$ matrix is $\chi (\lambda)=\lambda ^2-\textrm{tr}(A)\lambda +\det A$ . If we know that $\det A=0$ then, $\chi (\lambda)=\lambda(\lambda-\textrm{tr}A)$ . Now, we can use the Cayley-Hamilton theorem. For example for $\textrm{tr}A\neq 0$ you'll obtain $A^n=(\textrm{tr}A)^nA$ .

**ymar** · September 9th 2011, 03:53 AM

Originally Posted by FernandoRevilla

Well, that depends on the way we categorize.

The characteristic polynomial for a $2\times 2$ matrix is $\chi (\lambda)=\lambda ^2-\textrm{tr}(A)\lambda +\det A$ . If we know that $\det A=0$ then, $\chi (\lambda)=\lambda(\lambda-\textrm{tr}A)$ . Now, we can use the Cayley-Hamilton theorem. For example for $\textrm{tr}A\neq 0$ you'll obtain $A^n=(\textrm{tr}A)^nA$ .

Never heard of the theorem! And I'm defending my BA thesis in algebra on Thursday...

Thank you.

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