# find $$\alpha , \beta$$

• May 3rd 2009, 11:01 PM
zorro
find $$\alpha , \beta$$
If A= $\begin{pmatrix}
0 & 1 \\
-1 & 0
\end{pmatrix}$
, find $\alpha$, $\beta$ so that $(\alpha l + \beta A)^2 = A
$
• May 4th 2009, 02:23 AM
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 $\alpha (+/- \alpha)=1/2 \Rightarrow \alpha = \frac{1}{\sqrt{2}}=\beta$
• Dec 23rd 2009, 01:14 AM
zorro
I got this ?
Quote:

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 ????
• Dec 23rd 2009, 04:43 AM
HallsofIvy
Quote:

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?