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Math Help - find [MATH]\alpha , \beta[/MATH]

  1. #1
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    Exclamation find [MATH]\alpha , \beta[/MATH]

    If A= \begin{pmatrix}<br />
0 & 1 \\<br />
 -1 & 0<br />
\end{pmatrix} , find \alpha, \beta so that (\alpha l + \beta A)^2 = A<br />
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  2. #2
    Super Member Gamma's Avatar
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    I think that is supposed to be the Identity matrix?

    So you need
    \begin{pmatrix}0 & 1 \\-1 & 0\end{pmatrix}=\begin{pmatrix}\alpha & \beta \\-\beta & \alpha\end{pmatrix}^2

    Well multiply it out and see what you get. What I got was:
    \alpha^2=\beta^2 \Rightarrow \alpha = +/- \beta
    and
    2\alpha \beta = 1 \Rightarrow \alpha \beta = 1/2

    But then use 1 and we see we need \alpha (+/- \alpha)=1/2 \Rightarrow \alpha = \frac{1}{\sqrt{2}}=\beta
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  3. #3
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    I got this ?

    Quote Originally Posted by Gamma View Post
    I think that is supposed to be the Identity matrix?

    So you need
    \begin{pmatrix}0 & 1 \\-1 & 0\end{pmatrix}=\begin{pmatrix}\alpha & \beta \\-\beta & \alpha\end{pmatrix}^2

    Well multiply it out and see what you get. What I got was:
    \alpha^2=\beta^2 \Rightarrow \alpha = +/- \beta
    and
    2\alpha \beta = 1 \Rightarrow \alpha \beta = 1/2

    But then use 1 and we see we need \alpha (+/- \alpha)=1/2 \Rightarrow \alpha = \frac{1}{\sqrt{2}}=\beta

    ( \alpha I + \beta A)^2 = \begin{bmatrix} \alpha^2 + \beta^2 & 0 \\ 0 & \alpha^2 - \beta^2 \end{bmatrix}


    Is this right and what should i do next ????
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  4. #4
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    Quote Originally Posted by zorro View Post
    ( \alpha I + \beta A)^2 = \begin{bmatrix} \alpha^2 + \beta^2 & 0 \\ 0 & \alpha^2 - \beta^2 \end{bmatrix}


    Is this right and what should i do next ????
    According to the time stamp, gamma gave you the answer 8 months ago. Why are you asking now?
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