Please, help, nasty algebra got me (

[dextercioby]

Please, help, nasty algebra got me :((

I've been going nowhere with induction and possible favorable arrangements for A (see below) and I'm desperate for help...

Let

$$M_{3}=\{ A\in \mathcal{M}_{3}\left(\mathbb{Z}\right)| \forall i,j=\overline{1,3}, \ a_{ij}\in \{\pm 1 \} \}$$

Prove that

$$A^{2006} \neq 0_{3} , \forall A\in M_{3}$$.

I tried using complete induction, i.e. assuming A^k \neq 0_{3} and then trying to see whether A^(k+1) is 0_{3} or not. Of course, for this i have to assume that det A =0 (which fortunately is the case for a part of the 512 elements of M, for others, det A = \pm 4 and the proof is trivial). But this doesn't work, simply because A mixes plus & minus one and A^(k+1) could still be 0.

Then i tried using particular values for A. This lead me nowhere...The only relevant thing i found and hopefully is to be used is that if A\in M, then (-A) \in M, too.

This brought me to the following

$$-A^{2006}=(-A) A^{2005}$$

$$A^{2006}=(-A) (-A)^{2005}$$

But

$$A^{2005} \neq (-A)^{2005} \Leftrightarrow A^{2005} \neq 0_{3}$$

So i can assume that the latter is satisfied. But in proving that $A^{2006} \neq 0_{3}$ from the arguments above leads me to this issue

For matrices with det equal to 0

$$BA=C$$

$$EA=D$$

I know that $B\neq E$. However, det of A =0. Can i still prove that $C\neq D$ ??



Daniel.

 

[AKG]

Do you mean all the elements of A are plus or minus 1? It appears in your post that only when i and j are 1 or 3 does this need to happen, i.e. we're only guaranteed that the corners are plus or minus 1. But you say there are 512 elements in M, so I guess you mean all the elements are plus/minus 1. I clicked on your LaTeX, and where you have "1,3", you've written "overbar{1,3}". Do you somehow take "1,3" with a bar over it to mean 1,2,3? By the way, the overbar doesn't show. In fact, unless I'm mistaken, overbar isn't a command, but overline is.

Anyways, if everything is plus or minus 1, then it should be obvious. Look at a more general case. Let A and B be 3x3 matrices with all odd elements. Any element of AB is equal to a sum of three terms, each term being a product of two odd numbers. Well the product of two odd numbers is an odd number, and the sum of three odd numbers is an odd number, so AB will also be a matrix full of odd numbers. 0 is even. 1 and -1 are odd.

 

[dextercioby]

Yes, you were right about the Latex issue. I edited.

I see where you're coming from about the problem. Thankyou.


Daniel.

 

[George Jones]

I'm also confused about what the overbar means.

Here's another way to see it for the case where all entries are plus or minus 1.

For each such A, the trace of A is not equal to zero. But the trace of A equals the sum of the eigenvalues of A, so A has at least one non-zero eigenvalue, say $\alpha$. Since $\alpha^n$ is an eigenvalue of A^n, A^n has a non-zero eigenvalue.

Regards,
George

 

[AKG]

A either has 1 real eigenvalue or 3. Your argument doesn't hold if A only has 1, which is entirely possible, e.g. (1 1 1 // 1 -1 1 // -1 -1 1).

 

[George Jones]

A either has 1 real eigenvalue or 3. Your argument doesn't hold if A only has 1, which is entirely possible, e.g. (1 1 1 // 1 -1 1 // -1 -1 1).

You've lost me.

In this example, what part of my argument doesn't work?

In the example the trace = 1 , and the sum of the eigenvalues is 1, so A has at least one non-zero eigenvalue.

If $\alpha$ is any non-zero eigenvalue, then $\alpha^n$ is a non-zero eigenalue of A^n, and thus A^n is not zero.

Regards,
George

 

[AKG]

The matrix (cosX -sinX // sinX cosX) has trace of 2cosX but generally has no eigenvalues. So to say that 2cosX is the sum of the eigenvalues is wrong. Then again, we can always regard the underlying field to be a splitting field, in which case we can say that the matrix has a non-zero eigenvalue, so ultimately your argument does hold.

 

[George Jones]

The matrix (cosX -sinX // sinX cosX) has trace of 2cosX but generally has no eigenvalues. So to say that 2cosX is the sum of the eigenvalues is wrong. Then again, we can always regard the underlying field to be a splitting field, in which case we can say that the matrix has a non-zero eigenvalue, so ultimately your argument does hold.

Over the field C, this example has eigenvalues cosx + isinx and cosx - isinx, with corresponding eigenvectors (i , 1)^t and (-i , 1)^t.

The result that the trace of a matrix equals the sum of its eigenvalues is a general one.

Regards,
George