Homework Help: Determining Matrix Powers without Eigenvectors ? (Worked out but inelegant soln)

1. Apr 23, 2012

sid9221

http://dl.dropbox.com/u/33103477/Untitled.png [Broken]

My solution:
$$M=\begin{bmatrix} t+15 & -12 \\ 24 & t-19 \end{bmatrix}$$

The eigen values are 1,3.

Hence as the matrix has real and distinct eigenvalues it is diagonalisable.

Now the characteristic equation is $$t^2 - 4t +3 =0$$

So $$M^2 - 4M +3I = 0$$

$$M^2 = 4M-3I$$

Hence $$M^7 = (4M-3I)(4M-3I)(4M-3I)(M)$$

This gives the correct answer but is a very inelegant solution. Is there a better way to determine the answer without Eingevectors ?

Last edited by a moderator: May 5, 2017
2. Apr 23, 2012

Robert1986

Why do you think that is inelegant? You used the Cayley Hamilton Theorem, right? I like it.

3. Apr 23, 2012

Staff: Mentor

I think that is exactly what you were supposed to do. The only thing left is to finish the calculation by substituting in the given matrix and doing the multiplications.

4. Apr 23, 2012

sid9221

Whats the point in having a theorem that just saves you 1 multiplication ?

5. Apr 23, 2012

Staff: Mentor

If you had a much larger matrix M, and it was being raised to a higher power, this could save you some work.

For this problem, the real goal is to get you comfortable working with the characteristic equation of a matrix.

6. Apr 23, 2012

sid9221

I see. Thank you.

7. Apr 23, 2012

micromass

No, no, you're not done yet!!

Work out the brackets, you'll see terms like M4 and M3. Substitute them also by using Caley-Hamilton.

The eventual goal is to write

$$M^7=aM^2+bM+cI$$

This only needs one matrix multiplication.

8. Apr 23, 2012

sid9221

$$27 IM-108 M^2-144 IM^3+64 M^4$$

Than sub in value for M^2 and multiply ? I don't see how I can get the answer with 1 multiplication ?

9. Apr 23, 2012

sharks

I think that's incorrect. The proper explanation should be:
Since the matrix does not have zero as one of its eigenvalues, it is invertible, hence it is diagonalisable.

But, replacing M^2 again would give M^7 in terms of M and I only, right?

10. Apr 23, 2012

micromass

No, that is totally incorrect. Invertibility and diagonalisable have nothing to do with eachother.

OP:

$M^2=4M-3I$

So

$M^3=4M^2-3M$

So you can replace all occurences of $M^3$ with occurences of [/itex]M^2[/itex].

Same way: you can replace $M^4$ with something and $M^2$ with something.

11. Apr 23, 2012

micromass

Right!! So this problem requires no matrix multiplications at all!!

12. Apr 23, 2012

sharks

But, do the eigenvalues need to be distinct in order for the matrix to be diagonalisable? What if the eigenvalues are real but equal?

13. Apr 23, 2012

micromass

Then it might or might not be diagonalisable.

A matrix is diagonalisable if and only if it has a basis of eigenvectors. So if the eigenvalues are distinct, then all eigenspaces are distinct and thus a basis of eigenvectors exist.

Another equivalence is: a matrix is diagonalisable if and only if the algebraic multiplicity is the geometric multiplicity.

However, if not all the eigenvalues are distinct, then it might be harder to check diagonalisability. Typically, one has to know the eigenvectors at that point.

Yet another equivalence states that a matrix is diagonalisable iff the minimal polynomial has distinct factors. But then we have to know the minimal polynomial.

14. Apr 23, 2012

sid9221

Apparently works out to:

$$M^7 = 27 - 108[4M-3I] - 144[16M-15I] + 64[64M-63I]$$

15. Apr 23, 2012

Staff: Mentor

That first term would have to be 27I.

16. Apr 23, 2012

sharks

I'll need to look into that a bit more. Thank you for the clarification, micromass.

17. Apr 23, 2012

sid9221

Yeah of course !

So extending this, you could potentially calculate the 27th power of a 6x6 matrix without a single multiplication(or a few) wow !!

Cayley and Hamilton were pretty smart dude's eh !