## rsst123 one year ago **Will Medal**

2. anonymous

without calculator

4. zzr0ck3r

do you know how to find eigen values and eigen vectors? Do you know what diagonalization is ?

5. zzr0ck3r

there are other ways, so I need to know what you are doing in class if not.

yes isn't it A=SDS^-1?

7. zzr0ck3r

Correct. But you first need to find the Eigen values and vectors so that you can construct $$S$$

ok so after finding eigen values and vectors what would i do to find the matrix?

9. zzr0ck3r

Ok if we call your 2x2 matrix given above A and the matrix of eigen vectors S. Then you will find D by D = $$SAS^{-1}$$. Now once you have D you just take the square root of the entries (they will only be non 0 on the diagonal). You with me?

oh ok i see, I would get a diagonal matrix and then I can just square root the diagonal is that correct?

11. zzr0ck3r

then $$\sqrt{A} = S^{-1}\sqrt{D}S$$

12. zzr0ck3r

yes

ok perfect thank you!

14. zzr0ck3r

np

15. amoodarya

$\left[\begin{matrix}4 & 3 \\ 2 & 3\end{matrix}\right]=\left[\begin{matrix}-1 & \frac{3}{2} \\ 1 & 1\end{matrix}\right]\times \left[\begin{matrix}1 & 0 \\ 0 & 6 \end{matrix}\right] \times \left[\begin{matrix}\frac{-2}{5} & \frac{3}{5} \\ \frac{2}{5} & \frac{2}{5} \end{matrix}\right]=\\=S D S^{-1}\\\sqrt{A}==S \left[\begin{matrix}\sqrt{1} & 0 \\ 0 & \sqrt{6}\end{matrix}\right] S^{-1}$