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Thread: matrices to the power of 100

  1. April 6th 2008, 07:00 PM #1
    smplease
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    Question matrices to the power of 100

    My question is as follows,
    Let B = ( 1/sq.rt.2 -1/sq.rt.2)
    ( 1/sq.rt.2 1/sq.rt.2)
    (sq.rt obviously meaning square root)

    Calculate B to the power of 100.

    Is there a formula for this or is it a very time consuming question?

    Thanks for your help
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  2. April 6th 2008, 07:05 PM #2
    ThePerfectHacker
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    Quote Originally Posted by smplease View Post
    My question is as follows,
    Let B = ( 1/sq.rt.2 -1/sq.rt.2)
    ( 1/sq.rt.2 1/sq.rt.2)
    (sq.rt obviously meaning square root)

    Calculate B to the power of 100.

    Is there a formula for this or is it a very time consuming question?

    Thanks for your help
    First factor out the radical. The way to do this is to use matrix diagnolization. Find the eigenvectors and form a matrix out of them. So you want to write B = MDM^{-1} where D is a diagnol matrix. Then it means B^n = MD^nM^{-1} but D^n is easy to compute because it is a diagnol matrix.
    Last edited by ThePerfectHacker; April 6th 2008 at 07:55 PM.
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  3. April 6th 2008, 07:06 PM #3
    Jhevon
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    Quote Originally Posted by smplease View Post
    My question is as follows,
    Let B = ( 1/sq.rt.2 -1/sq.rt.2)
    ( 1/sq.rt.2 1/sq.rt.2)
    (sq.rt obviously meaning square root)

    Calculate B to the power of 100.

    Is there a formula for this or is it a very time consuming question?

    Thanks for your help
    formula? not really. there are methods though.

    check out this

    and this
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  4. April 6th 2008, 08:24 PM #4
    Soroban
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    Hello, smplease!

    \text{Let }\,B \;= \;\begin{bmatrix}\frac{1}{\sqrt{2}} & \text{-}\frac{1}{\sqrt{2}} \\ \\ \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \end{bmatrix}

    Calculate B^{100}
    There is no formula, but we can crank it out . . .

    \text{We have: }\;B \;=\;\frac{1}{\sqrt{2}}\begin{bmatrix}1 & \text{-}1 \\ 1 & 1 \end{bmatrix}

    . . \text{Then: }\;B^2 \;=\;\frac{1}{2}\begin{bmatrix}1 & \text{-}1 \\ 1 & 1 \end{bmatrix}\begin{bmatrix}1 & \text{-}1 \\ 1 & 1 \end{bmatrix} \;=\;\frac{1}{2}\begin{bmatrix}0 & \text{-}2 \\ 2 & 0\end{bmatrix} \;=\;\begin{bmatrix}0 & \text{-}1 \\ 1 & 0\end{bmatrix}

    . . \text{And: }\;B^4\;=\;\begin{bmatrix}0 & \text{-}1 \\ 1 & 0 \end{bmatrix}\begin{bmatrix}0 & \text{-}1 \\ 1 & 0 \end{bmatrix} \;=\;\begin{bmatrix}\text{-}1 & 0 \\ 0 & \text{-}1\end{bmatrix} \;=\;-I

    . . \text{Hence: }\;B^8 \;=\;B^4\cdot B^4 \;=\;(-I)(-I) \;=\;I


    Therefore: . B^{100} \;=\;B^{96}\cdot B^4 \;=\;(B^8)^{12}\cdot B^4 \;=\;(I^{12})\cdot B^4 \;=\;B^4 \;=\;\begin{bmatrix}\text{-}1 & 0 \\ 0 & \text{-}1\end{bmatrix}
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  5. April 7th 2008, 04:47 PM #5
    smplease
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    Quote Originally Posted by Soroban View Post
    Hello, smplease!

    There is no formula, but we can crank it out . . .

    \text{We have: }\;B \;=\;\frac{1}{\sqrt{2}}\begin{bmatrix}1 & \text{-}1 \\ 1 & 1 \end{bmatrix}

    . . \text{Then: }\;B^2 \;=\;\frac{1}{2}\begin{bmatrix}1 & \text{-}1 \\ 1 & 1 \end{bmatrix}\begin{bmatrix}1 & \text{-}1 \\ 1 & 1 \end{bmatrix} \;=\;\frac{1}{2}\begin{bmatrix}0 & \text{-}2 \\ 2 & 0\end{bmatrix} \;=\;\begin{bmatrix}0 & \text{-}1 \\ 1 & 0\end{bmatrix}

    . . \text{And: }\;B^4\;=\;\begin{bmatrix}0 & \text{-}1 \\ 1 & 0 \end{bmatrix}\begin{bmatrix}0 & \text{-}1 \\ 1 & 0 \end{bmatrix} \;=\;\begin{bmatrix}\text{-}1 & 0 \\ 0 & \text{-}1\end{bmatrix} \;=\;-I

    . . \text{Hence: }\;B^8 \;=\;B^4\cdot B^4 \;=\;(-I)(-I) \;=\;I


    Therefore: . B^{100} \;=\;B^{96}\cdot B^4 \;=\;(B^8)^{12}\cdot B^4 \;=\;(I^{12})\cdot B^4 \;=\;B^4 \;=\;\begin{bmatrix}\text{-}1 & 0 \\ 0 & \text{-}1\end{bmatrix}



    Wow thank you so much that makes it a lot more easy to understand and work out

    thanks again
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  6. April 7th 2008, 06:17 PM #6
    mr fantastic
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    Quote Originally Posted by Soroban View Post
    Hello, smplease!

    There is no formula, but we can crank it out . . .

    \text{We have: }\;B \;=\;\frac{1}{\sqrt{2}}\begin{bmatrix}1 & \text{-}1 \\ 1 & 1 \end{bmatrix}

    . . \text{Then: }\;B^2 \;=\;\frac{1}{2}\begin{bmatrix}1 & \text{-}1 \\ 1 & 1 \end{bmatrix}\begin{bmatrix}1 & \text{-}1 \\ 1 & 1 \end{bmatrix} \;=\;\frac{1}{2}\begin{bmatrix}0 & \text{-}2 \\ 2 & 0\end{bmatrix} \;=\;\begin{bmatrix}0 & \text{-}1 \\ 1 & 0\end{bmatrix}

    . . \text{And: }\;B^4\;=\;\begin{bmatrix}0 & \text{-}1 \\ 1 & 0 \end{bmatrix}\begin{bmatrix}0 & \text{-}1 \\ 1 & 0 \end{bmatrix} \;=\;\begin{bmatrix}\text{-}1 & 0 \\ 0 & \text{-}1\end{bmatrix} \;=\;-I

    . . \text{Hence: }\;B^8 \;=\;B^4\cdot B^4 \;=\;(-I)(-I) \;=\;I


    Therefore: . B^{100} \;=\;B^{96}\cdot B^4 \;=\;(B^8)^{12}\cdot B^4 \;=\;(I^{12})\cdot B^4 \;=\;B^4 \;=\;\begin{bmatrix}\text{-}1 & 0 \\ 0 & \text{-}1\end{bmatrix}
    And of course if you recognise the original matrix \begin{bmatrix}\frac{1}{\sqrt{2}} & \text{-}\frac{1}{\sqrt{2}} \\ \\ \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \end{bmatrix} as a rotation matrix, where the angle of rotation is \frac{\pi}{4} in the anti-clockwise direction, you don't even need to crank ........
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