Invariant subspace and linear transformation

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The discussion centers on proving that if U is an invariant subspace of V under every linear transformation, then U must equal V. The approach begins by assuming U is non-trivial and extending a nonzero vector from U to a basis for V. A linear map is constructed that maps the basis vectors in a specific sequence, demonstrating that all basis vectors must belong to U due to its invariance. This leads to the conclusion that the span of the basis vectors covers V, confirming that U equals V. The main challenge highlighted is establishing that the constructed map is indeed a linear transformation.
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Homework Statement


Let U be a subspace of V. Suppose that U is a invariant subspace of V for every linear
transformation from V to V. Show that U=V.


Homework Equations



no

The Attempt at a Solution


Assume U is not trivial: Now we only need to show that U = V. Let dimV = n: We can choose
a nonzero vector u in U and then extend u to a basis u1,u2,u3,...u{n-1} for V.
There is a linear map k such that
k(u) = u1;
k(u1) = u2;
k(u2) = u3;
...
k(u{n-1})=u
As U is invariant, u1 also in U,then as k(u1)=u2,u2 also in U,.....then u{n-1} also in U.

On the other hand, as u1,u2,u3,...u{n-1} is a basis for V, we have
span(u1,u2,u3,...u{n-1}) = V:
So it follows that U = V.

The difficulty is proving k is a linear transformation, and if it is, is the argument above correct? Thanks!
 
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Sounds reasonable to me, Apply k to two arbitrary vectors in V and linearity should follow
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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