MHB Show all invariant subspaces are of the form

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The discussion revolves around demonstrating that all invariant subspaces are of a specific form. A participant expresses confusion about how to start the proof, referencing a subspace spanned by a vector x. The solution is confirmed, linking it to a previous question about the dimension of V being equal to the degree of the minimal polynomial, which leads to a contradiction. The conversation highlights the importance of understanding the relationship between invariant subspaces and minimal polynomials in linear algebra. Ultimately, the problem is resolved with these insights.
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[solved] show all invariant subspaces are of the form

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i don't even know how to begin (Angry)

C_x is a subspace spanned by x that belongs to V

C_x = {x, L(x), L^2(x),...}

edit: SOLVED
 

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The hint refers to a question you asked in another thread. Do you remember it?
 
dim of v needs to be equal to degree of minimal poly and hence that would be a contradiction?

i'll see what i can do with that. thanks.
 
Thread 'How to define a vector field?'

Thread 'How to define a vector field?'

Hello! In one book I saw that function ##V## of 3 variables ##V_x, V_y, V_z## (vector field in 3D) can be decomposed in a Taylor series without higher-order terms (partial derivative of second power and higher) at point ##(0,0,0)## such way: I think so: higher-order terms can be neglected because partial derivative of second power and higher are equal to 0. Is this true? And how to define vector field correctly for this case? (In the book I found nothing and my attempt was wrong...
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