Critical points of matrix

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Homework Statement
Give their critical points of this system which can be written as a matrix:
##x^{\prime}=x - 2y, y^{\prime}=-2x+ 4y##
Relevant Equations
##x^{\prime}=x - 2y, y^{\prime}=-2x+ 4y##
My attempt is:

Condition for critical point is ##x' = y' = 0##,
##0 = x - 2y \implies 2y = x##
##-2x + dy = 0##
Then ##-4y + 4y = 0##

However, this means that critical points are ##(2y, y)## as system is linearly dependent (both equations are the same) where ##y \in \mathbb{R}##. However, that means there are infinitely many critical points which I have a doubt about.

I express gratitude to those who help.
 
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TanWu said:
However, that means there are infinitely many critical points which I have a doubt about.
Why so?
 
TanWu said:
Homework Statement: Give their critical points of this system which can be written as a matrix:
##x^{\prime}=x - 2y, y^{\prime}=-2x+ 4y##
Relevant Equations: ##x^{\prime}=x - 2y, y^{\prime}=-2x+ 4y##

My attempt is:

Condition for critical point is ##x' = y' = 0##,
##0 = x - 2y \implies 2y = x##
##-2x + dy = 0##
Do you mean ##-2x + 4y = 0## here?
TanWu said:
Then ##-4y + 4y = 0##

However, this means that critical points are ##(2y, y)## as system is linearly dependent (both equations are the same) where ##y \in \mathbb{R}##. However, that means there are infinitely many critical points which I have a doubt about.
Your work looks ok to me. If you have doubts, have you checked that you got the initial problem equations right?
 
docnet said:
Why so?
FactChecker said:
Do you mean ##-2x + 4y = 0## here?

Your work looks ok to me. If you have doubts, have you checked that you got the initial problem equations right?
Thank you Sirs. I apoglize, that is a typo of me. Yes, got the initial problem equations correct. I was only expecting one critical point.
 
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TanWu said:
Thank you Sirs. I apoglize, that is a typo of me.
Mathematics is very unforgiving in many ways. It's a learned skill to review your work very carefully.
TanWu said:
Yes, got the initial problem equations correct. I was only expecting one critical point.
You did a good job! The problems where you get a different answer than you expected are ones that really test you.
 
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There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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