#Linear Algebra - Eigenvectors and vector space
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smoky verge
You have $x_1$ which is free and $x_2 = 0$, so $\forall v \in E_1, v = (\alpha,0), \alpha \in R$. That means $E_1 = Vect( (1,0) )$. Is what you asked for ?
stray eagleBOT
Yojda
lavish jewel
You have $x_1$ which is free and $x_2 = 0$, so $\forall v \in E_1, v = (\alpha,0...
hmm
I am more or less trying to figure out
what I did wrong because I was told in my feedback sections aside from the red markings
that the problem isn't correct but it didnt specify why its wrong as in if i did the steps procedure wrong or something else like calculation
For the eignenvectors spanning, you now need to see if the set of eignvectors you found
S={(1,0), (1,-2)} spans R^2
lavish jewel
For the eignenvectors spanning, you now need to see if the set of eignvectors yo...
this was in the comment section
i know but at the same time i dont know what i did wrong for me to merit this feedback
smoky verge
Show that they are non collinear
But I don't know why he wants you to say this.
Because you did all this to do the matrix diagonalization.
Ohhh I see maybe
Wasn't necessary to say that S spans R^2 in my university, but it's interesting
It makes you thinking about what you are trying to do. You want to find such a basis for using the properties of the eigenvalues.
So your professor wants to be sure that you didn't lost any vector of your basis when you apply this matrix. In other words that the 2 vectors spans R^2
Those videos could help you to understand.