/sci/ - Science & Math » Thread #13122762

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Anonymous Thu 13 May 04:12:03 2021 No.13122762 View Reply Original Report

Quoted By: >>13122787 >>13123401 >>13123807 >>13123830 >>13123848

for matrix $\begin{bmatrix} \alpha & \beta\\ \beta & \gamma \end{bmatrix} $ where
$\alpha,\beta,\gamma \epsilon R$

How do I prove that there are always real eigenvalues

Anonymous

Anonymous Thu 13 May 2021 04:14:29 No.13122770 Report

Quoted By: >>13122776 >>13122919

A more important question. Why is apu so cute?

Anonymous

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Anonymous Thu 13 May 2021 04:15:35 No.13122776 Report

Quoted By:

>>13122770
I don't know anon, you tell me

Anonymous

Anonymous Thu 13 May 2021 04:19:49 No.13122787 Report

Quoted By:

>>13122762
idk i guess you can try solving for lambda of the determinant of the matrix minus lambda times identity matrix

Anonymous

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Anonymous Thu 13 May 2021 04:58:21 No.13122919 Report

Quoted By:

>>13122770
and intelligent

Anonymous

Anonymous Thu 13 May 2021 07:51:25 No.13123401 Report

Quoted By: >>13123871

>>13122762
develop the characteristic polinomial, which will be a quadratic, and find it's discriminant, you will see that it must always be positive. Thus, the polynomial will have 2 distinct real roots, which will be the matrix's eigenvalues.

Anonymous

Anonymous Thu 13 May 2021 11:08:12 No.13123807 Report

Quoted By: >>13123871

>>13122762
Post once you get it please, I got the discriminant but to me it seems like it can be negative.

Anonymous

Anonymous Thu 13 May 2021 11:18:59 No.13123830 Report

Quoted By:

>>13122762
delta of characteristic polynomial is square root of sum of 2 squares , so there are always 2 solutions for eigenvalue

Anonymous

Anonymous Thu 13 May 2021 11:24:53 No.13123848 Report

Quoted By:

>>13122762
Call the matrix $A$ . Let $v$ be an Eigenvector of $A$ belonging to the Eigenvalue $\lambda$ and let $\langle x,y \rangle = \bar y_1 x_1 + \bar y_2 x_2$ be the dot product of $\mathbb{C}^2$ then
$ \lambda \langle v,v \rangle = \langle \lambda v,v \rangle = \langle A v,v \rangle = \langle v, \bar A v \rangle = \langle v, A v \rangle = \langle v, \lambda v \rangle = \bar \lambda \langle v,v \rangle$
So
$\lambda = \bar \lambda$ which implies that $\lambda \in \mathbb{R}$ .

Anonymous

Anonymous Thu 13 May 2021 11:33:19 No.13123871 Report

Quoted By:

>>13123807
Nevermind, I was plain stupid. Do it as >>13123401 detailed it and you'll be done in 5 minutes.

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