Quick way of raising matrices to indicies

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In summary, the conversation discusses finding the term in the first row and second column of matrix A^5 when x = 18°, using trigonometric identities to simplify the computation, and using the pattern of multiplying A^2 by A to find A^5. It is also mentioned that A corresponds to rotation through an angle of x, and multiplying A by itself 5 times is equivalent to rotating by -5x degrees.
  • #1
thomas49th
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Homework Statement



A given matrix [tex]
A(x) =\left | \begin{array}{cc} cos(x)&sin(x)\\ -sin(x)&cos(x) \end{array}\right |
[/tex]

Find the term in the first row and second column A^{5}(x) when x = 18°

Homework Equations


The Attempt at a Solution



Instead of multiplying the matrix again and again by itself in a iterative procedure is there any quicker way?

Thanks
Thomas
 
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  • #2
For this particular matrix, why don't you find A^2 for general x, and use your knowledge of trigonometric identities...then you may see a pattern.
 
  • #3
A^2 = [ cos2x 2cosxsinx
-2cosxsinx cos2x ]

what I want is cos²x + sin²x so i get one, but what I got at the mooment doesn't seem to get me any where :( I wanted somthing like if the power is odd then you get -1 and even 1 in the first element, but can't see that happening :(
 
  • #4
Since 2 cos x sin x = sin 2x, a pattern is emerging, just not the one you thought.
 
  • #5
aha so the formula for that specific elemnt is [tex]\sin 2^{n-1}x[/tex]

Correct?

Thanks
Tom
 
  • #6
Maybe for [tex]A^{2^{n-1}}[/tex] but that doesn't help with [tex]A^5[/tex].

In other words, multiply [tex]A^2[/tex] by [tex]A[/tex], not by [tex]A^2[/tex], to find [tex]A^3[/tex].

(I know you know this -- your brain just went into overdrive and caused a blunder.)
 
  • #7
ahhh I am getting
A^3
[tex]

A(x) =\left | \begin{array}{cc} cos(x)cos(2x)-sin(x)sin(2x)& cos(x)cos(2x) +sin(x)sin(2x)\\ -cos(x)cos(2x)-sin(x)sin(2x)& cos(x)cos(2x)-sin(x)sin(2x) \end{array}\right |

[/tex]

where is this pattern? I can see they're all similar
I might as well just expand to A^5 now :(
 
  • #8
What is the trig identity for cos(A+B)? For sin(A+B)?

Also, your A^3 has two incorrect entries.
 
  • #9
cos(A+B) = cosACosB - sinAsinB
sin(A+B) = sinAcosB + cosAsinB

right?

I can see how we can use RCos(x+a) for A^3 but I want to get A^5. There must be some relationship!

Which of the terms are incorrect in A^3btw for A^2 i got [cos2x sin2x
-sin2x cos2x]

Thanks
Thomas
 
  • #10
I am sure BB means something much faster, but whenever I hear about anything raised to large integer power I think in terms of powers of 2, in this case it means A5=A*(A2)2 - that means just three multiplications instead of 5. The larger the exponent, the more you save.
 
  • #11
thomas49th said:
btw for A^2 i got [cos2x sin2x
-sin2x cos2x]
Can't you guess the correct pattern just from this! :smile:

Additional hint: A(x) corresponds to rotation through an angle of x.
 
  • #12
I can see A^2 = sin2x

but I cannot see why that is so argghh! must be something really simple!
 
  • #13
A=
[cos x sin x
-sin x cos x]

A^2=
[cos 2x sin 2x
-sin 2x cos 2x]

A^3=
what would you guess! Yes, it's very simple, sorry. :smile:

Now compute A^3, and if it doesn't match your guess, then your computation is probably wrong, not your guess!
 
  • #14
You might also note that
[tex]\begin{bmatrix}cos(\theta) & sin(\theta) \\ -sin(\theta) & sin(\theta)\end{bmatrix}[/tex]
corresponds to a rotation by -x degrees. Multiplying it by itself 5 times is the same as rotating by -5x degrees. In this case that is -5(18)= -90 degrees. The result is suprisingly simple.
 
  • #15
yeah I didn't think of that!

still need to expand it manually to see the maths but bed time now

Thanks for the help!
Thomas
 

Related to Quick way of raising matrices to indicies

1. What is the purpose of raising matrices to indicies?

The purpose of raising matrices to indicies is to quickly calculate large powers of a matrix without having to perform multiple matrix multiplications. This can be useful in various applications, such as solving systems of linear equations or finding the long-term behavior of a dynamical system.

2. How do you raise a matrix to an index?

To raise a matrix to an index, you can use the binomial theorem or the diagonalization method. The binomial theorem is used for matrices with repeated eigenvalues, while the diagonalization method works for matrices with distinct eigenvalues. Both methods involve finding the eigendecomposition of the matrix and using it to calculate the desired power.

3. What is the binomial theorem method for raising matrices to indicies?

The binomial theorem method involves expressing the matrix as a sum of powers of a diagonal matrix and using the binomial theorem to expand each term. The resulting diagonal matrix can then be raised to the desired power, and the original matrix can be reconstructed using the eigendecomposition.

4. How does the diagonalization method work for raising matrices to indicies?

The diagonalization method involves finding the eigendecomposition of the matrix, which expresses the matrix as a product of a diagonal matrix and an invertible matrix. The diagonal matrix can then be raised to the desired power, and the original matrix can be reconstructed using the eigendecomposition.

5. Are there any limitations to raising matrices to indicies?

Yes, there are limitations to raising matrices to indicies. This method only works for square matrices with distinct eigenvalues or repeated eigenvalues. Additionally, the resulting matrix may not always be a true power of the original matrix, but rather an approximation due to rounding errors in the calculation.

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