Math Is Fun Forum

Neela

Guest

Matrix powers..

Hey Everyone,

I'm not sure how to write in matrixes on this forum, but I'll try..

My task was to consider different matrixes and powers to find a general expression in terms of k and n (the power).

I was asked to consider the matrix:

k+1         k-1
k-1          k+1

After trying different values of k, I derived a pattern and got this general equation:

Matrix^n = 2^n-1     (k^n + 1           k^n - 1
                               k^n - 1           k^n + 1)  (the 2^n-1 is outside the bracket of the matrix)

My next task was to 'use technology' to see what happens with further values of k and n and then state the scope or limitation of k and n.

I'm not sure what 'use technology' refers to - but I'm guessing it's my TI calculator? Anyway, I tried testing the validity of the generalization by considering many different values. Identity matrixes, negative values for k etc. and all of them seem to fit with the generalization. I cannot find any limitations (what is scope?) of k and n. I considered negative values of n as you cannot put a matrix to a negative power - would this be right? Are there any other limitations? Something I'm missing?

I also need to explain why my results hold true in general - I'm guessing this is referring to the expression I found. I'm not sure how to do this, anyone?

Thanks a lot guys!
Neela.

Dross

Member

Registered: 2006-08-24

Posts: 325

Re: Matrix powers..

I also need to explain why my results hold true in general - I'm guessing this is referring to the expression I found. I'm not sure how to do this, anyone?

I would interpret a matrix raised to the power -3 to be the inverse of that matrix, raised to the power 3 - but whatever you've been taught, I guess.

As for showing your results in general, have you tried proof by induction? You can only have integer powers for a matrix, so it would seem that is the way to go...

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Bad speling makes me [sic]

Neela

Guest

Re: Matrix powers..

I also need to explain why my results hold true in general - I'm guessing this is referring to the expression I found. I'm not sure how to do this, anyone?

I would interpret a matrix raised to the power -3 to be the inverse of that matrix, raised to the power 3 - but whatever you've been taught, I guess.

As for showing your results in general, have you tried proof by induction? You can only have integer powers for a matrix, so it would seem that is the way to go...

Hey Dross,

Thanks for the help.

lol you're probably right about the inverse thing. Math is a weak side of mine, and I reached that conclusion by trying different numbers on my calculator and found that the calculator said 'error' everytime I tried using a negative power. Do you have any suggestions for any limitation?

Um, haha, as I said, my math is weak. Proof by induction? Integer powers? Could you perhaps explain the procedure?

Thanks!!

Dross

Member

Registered: 2006-08-24

Posts: 325

Re: Matrix powers..

Right - proof by induction is possibly something you've been taught under a different name. Have a look at wikipedia for a more thorough look.
Basically it involves:

1) formulate a statement, call it P(n), that you think is true
2) see if your statement, P(n), is true for n = some whole number (aka integer), usually 0 or 1 (but it could be any number)
3) show that if your statement is true for n = some arbitrary number x, then your statement is true for n = (x + 1)

(2) together with (3) above create a sort of "dominoes" effect. Since your statement is true for (say, for example...) n = 1, then P(1) is true. Then, by point (3), because P(1) is true, P(2) must be true. Similarly, P(3) must be true, and so on and so forth, meaning that your statement is true for all the whole numbers from 1 (for example) up.

For this specific case, you might proceed as follows:

1) formulate our statement that we think is true - in this case, formulate the statement:

and call it "P(n)".

2) have a look at P(1) - that statement would be:

...which is clearly true.

3) this is usually the tricky part, but here it's not so bad (just a little messy with the algebra!)

so, now we assume that P(n) is true for some random choice of n (that we don't know about), and see if we can use this fact to prove that P(n+1) is true. First step - have a look at what P(n+1) looks like. P(n+1) says:

So, we start off by wanting to find what

is:

(I missed out some calculations in the middle there - they were messy and nothing inspired)

So, we do indeed have our identity for P(n+1), assuming P(n) is true. Thus, by induction, P(n) is true for n = 1, 2, 3...

Note you could also have P(0) true if you held that a matrix raised to the power 0 be the identity matrix - I've never seen it, but it makes sense "I guess".

Did all that make any sense?

Now, you could possibly extend this to P(n) for negative values of n, but you have to be sure your matrix is invertible. As it happens, the determinant of

is 4k, which means that this matrix is invertible whenever k is not equal to zero. The rest I'll leave up to you.

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Bad speling makes me [sic]

Zhylliolom

Real Member

Registered: 2005-09-05

Posts: 412

Re: Matrix powers..

You can only have integer powers for a matrix...

This is slightly off topic from Neela's exact question, but there are ways to calculate square roots, logarithms, sines, etc. of matrices, based on how these operations act on the real numbers.

The easiest way to do this is when the matrix we wish to operate on is diagonalizable. Consider a matrix A such that A is diagonalizable, so that A = PDP[sup]-1[/sup] for appropriate matrices P and D (which I will assume you know how to determine from linear algebra... if not, feel free to ask how). We wish to find the square root of the matrix A, that is, a matrix A[sup]1/2[/sup] such that A[sup]1/2[/sup]A[sup]1/2[/sup] = A. Let D[sup]1/2[/sup] be the diagonal matrix consisting of the square roots of the eigenvalues of A. We claim that A[sup]1/2[/sup] = PD[sup]1/2[/sup]P[sup]-1[/sup]. To check that A[sup]1/2[/sup] has the desired property A[sup]1/2[/sup]A[sup]1/2[/sup] = A, we take A[sup]1/2[/sup]A[sup]1/2[/sup] = PD[sup]1/2[/sup]P[sup]-1[/sup]PD[sup]1/2[/sup]P[sup]-1[/sup] = PD[sup]1/2[/sup]D[sup]1/2[/sup]P[sup]-1[/sup] = PDP[sup]-1[/sup] = A (we have used the fact that P[sup]-1[/sup]P = I, XI = IX = X for any matrix X, and D[sup]1/2[/sup]D[sup]1/2[/sup] = diag(d[sub]i[/sub][sup]1/2[/sup]d[sub]i[/sub][sup]1/2[/sup]) = diag(d[sub]i[/sub], D). Then A[sup]1/2[/sup] = PD[sup]1/2[/sup]P[sup]-1[/sup] has the desired property, and we conclude A[sup]1/2[/sup] is the matrix version of a square root. Similarly, if A is diagonalizable, we have ln A = P ln D P[sup]-1[/sup], where ln D = diag(ln d[sub]i[/sub]).

If A is not diagonalizable, then one finds the Jordan canonical form of A and takes the square root/logarithm/sine/etc. of the Jordan blocks. This can be done for every square matrix, since every square matrix over the field C has a Jordan canonical form. We may write an r x r Jordan block J(λ) as λ(I + N(r)), where N(r) is the r x r matrix with 1's on the superdiagonal and 0's elsewhere. To calculate the square root/logarithm/sine/etc. of this block, we use a series expansion. Here is the logarithm for example:

Now you may argue that this series does not necessarily converge. However, N(r) is a nilpotent matrix of index r (this is easily verified by the reader ), so that (N(r))[sup]mr[/sup] = 0 for integers m. We are then left with a finite number of terms.

Anyway, I just wanted to show a little interesting side of matrices. Chances are you probably will never use it; I have only seen such a concept on a Putnam exam, where it was asked whether the sine of a certain form of a matrix ever exists. It's fun to know about this though.

Dross

Member

Registered: 2006-08-24

Posts: 325

Re: Matrix powers..

Cool - thanks very much for that, Zhylliolom. Though I'll have to read up on Jordan canonical form, is there much to it?

It is quite interesting to see how you can do these operations with matrices.

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Bad speling makes me [sic]

Zhylliolom

Real Member

Registered: 2005-09-05

Posts: 412

Re: Matrix powers..

If you've learned enough linear algebra to understand eigenvalues, then understanding Jordan canonical form should not be a problem. If it is, your source is being too rough with their explaination. The first part of this Wikipedia entry should be a good basic explaination:

http://en.wikipedia.org/wiki/Jordan_canonical_form

maria22

Guest

Re: Matrix powers..

I dont think the formula 2^(n-1)MATRIX works for all numbers.... do you know another formula?