# Thread: matrices to the power of 100

1. ## 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?

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?

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 $\displaystyle B = MDM^{-1}$ where $\displaystyle D$ is a diagnol matrix. Then it means $\displaystyle B^n = MD^nM^{-1}$ but $\displaystyle D^n$ is easy to compute because it is a diagnol matrix.

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?

formula? not really. there are methods though.

check out this

and this

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

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

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

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

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

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

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

5. Originally Posted by Soroban

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

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

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

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

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

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

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

thanks again

6. Originally Posted by Soroban

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

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

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

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

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

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

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# matrix power leading to 100

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