If A=$\displaystyle \begin{pmatrix}

0 & 1 \\

-1 & 0

\end{pmatrix} $, find $\displaystyle \alpha$,$\displaystyle \beta$ so that $\displaystyle (\alpha l + \beta A)^2 = A

$

Printable View

- May 3rd 2009, 10:01 PMzorrofind [MATH]\alpha , \beta[/MATH]
If A=$\displaystyle \begin{pmatrix}

0 & 1 \\

-1 & 0

\end{pmatrix} $, find $\displaystyle \alpha$,$\displaystyle \beta$ so that $\displaystyle (\alpha l + \beta A)^2 = A

$ - May 4th 2009, 01:23 AMGamma
I think that is supposed to be the Identity matrix?

So you need

$\displaystyle \begin{pmatrix}0 & 1 \\-1 & 0\end{pmatrix}=\begin{pmatrix}\alpha & \beta \\-\beta & \alpha\end{pmatrix}^2$

Well multiply it out and see what you get. What I got was:

$\displaystyle \alpha^2=\beta^2 \Rightarrow \alpha = +/- \beta$

and

$\displaystyle 2\alpha \beta = 1 \Rightarrow \alpha \beta = 1/2$

But then use 1 and we see we need $\displaystyle \alpha (+/- \alpha)=1/2 \Rightarrow \alpha = \frac{1}{\sqrt{2}}=\beta$ - Dec 23rd 2009, 12:14 AMzorroI got this ?
- Dec 23rd 2009, 03:43 AMHallsofIvy