find [MATH]\alpha , \beta[/MATH]

**zorro** · May 3rd 2009, 10:01 PM

If A= $\begin{pmatrix}<br /> 0 & 1 \\<br /> -1 & 0<br /> \end{pmatrix}$ , find $\alpha$ , $\beta$ so that $(\alpha l + \beta A)^2 = A<br />$

**Gamma** · May 4th 2009, 01:23 AM

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 $\alpha (+/- \alpha)=1/2 \Rightarrow \alpha = \frac{1}{\sqrt{2}}=\beta$

**zorro** · December 23rd 2009, 12:14 AM

Originally Posted by Gamma

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 ????

**HallsofIvy** · December 23rd 2009, 03:43 AM

Originally Posted by zorro

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

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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