# Thread: find $$\alpha , \beta$$

1. ## find $$\alpha , \beta$$

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

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

3. ## I got this ?

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 ????

4. 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?