1. ## Matrix Proof

Let A be a 2x2 matrix.

Then, prove the following:

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

2. ## Re: Matrix Proof

Originally Posted by lanierms
Let A be a 2x2 matrix. Then, prove the following: $\displaystyle A^n=0 \Longrightarrow A^2=0$
Hint: $\displaystyle p(x)=x^n$ is an annihilator polynomial of $\displaystyle A$ , this means that the minimal polynomial of $\displaystyle A$ is $\displaystyle \mu(x)=x$ or $\displaystyle \mu(x)=x^2$ so the canonical form of $\displaystyle A$ is $\displaystyle J=\begin{bmatrix}{0}&{0}\\{0}&{0} \end{bmatrix}$ or $\displaystyle J=\begin{bmatrix}{0}&{1}\\{0}&{0} \end{bmatrix}$

3. ## Re: Matrix Proof

Originally Posted by lanierms
Let A be a 2x2 matrix.

Then, prove the following:

$\displaystyle A^n=0 \Longrightarrow A^2=0$
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...)

4. ## Re: Matrix Proof

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

5. ## Re: Matrix Proof

This might work, but check my computations. Since $\displaystyle A^n=0$, $\displaystyle \mbox{det}A^n=0\Rightarrow \mbox{ det}A=0$, 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.

6. ## Re: Matrix Proof

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.

7. ## Re: Matrix Proof

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.

8. ## Re: Matrix Proof

Originally Posted by lanierms
Let A be a 2x2 matrix.

Then, prove the following:

$\displaystyle A^n=0 \Longrightarrow A^2=0$
If you don't know Canonical forms as Dr. Revilla suggested, you could just triangularize it and get the same conclusion.

9. ## Re: Matrix Proof

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

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

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

10. ## Re: Matrix Proof

This can be done with quite simple linear algebra.

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

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

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

is linearly independent, and contains $\displaystyle m$ elements.

Can you finish now?

11. ## Re: Matrix Proof

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 $\displaystyle A=\begin{bmatrix}{a}&{c}\\{b}&{d} \end{bmatrix}$ with $\displaystyle \det A=0$ and $\displaystyle (a,b)\neq (0,0)$ then, $\displaystyle (c,d)=\lambda (a,b)$ 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) .

12. ## Re: Matrix Proof

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 $\displaystyle A^n$ dependent on a, b, c and d? I've just tried it and I can't but I'm not very good.

13. ## Re: Matrix Proof

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

14. ## Re: Matrix Proof

Originally Posted by FernandoRevilla
Well, that depends on the way we categorize.

The characteristic polynomial for a $\displaystyle 2\times 2$ matrix is $\displaystyle \chi (\lambda)=\lambda ^2-\textrm{tr}(A)\lambda +\det A$ . If we know that $\displaystyle \det A=0$ then, $\displaystyle \chi (\lambda)=\lambda(\lambda-\textrm{tr}A)$ . Now, we can use the Cayley-Hamilton theorem. For example for $\displaystyle \textrm{tr}A\neq 0$ you'll obtain $\displaystyle A^n=(\textrm{tr}A)^nA$ .
Never heard of the theorem! And I'm defending my BA thesis in algebra on Thursday... Thank you.