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

[sid9221]

http://dl.dropbox.com/u/33103477/Untitled.png


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 ?

 

[Robert1986]

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

 

[Mark44]

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.

 

[sid9221]

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.

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

 

[Mark44]

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.

 

[sid9221]

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.

I see. Thank you.

 

[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.

 

[sid9221]

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.

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

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

 

[DryRun]

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

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.

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.

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

 

[micromass]

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.

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

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.

 

[micromass]

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

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

 

[DryRun]

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

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

 

[micromass]

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

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.

 

[sid9221]

Apparently works out to:

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

 

[Mark44]

Apparently works out to:

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

That first term would have to be 27I.

 

[DryRun]

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.

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

 

[sid9221]

That first term would have to be 27I.

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 !