Linear algebra: Prove the statement

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To prove the statement that \dim L(\mathbb F) + \dim Ker L = \dim(\mathbb F + Ker L), consider the linear transformation L: U → V, where U = \mathbb{F} + Ker L. By applying the theorem that states \dim U = \dim Ker L + \dim C(L^T), and recognizing that \dim C(L^T) equals \dim L(U), the relationship can be established. This leads to the conclusion that \dim L(\mathbb{F} + Ker L) is equivalent to \dim L(\mathbb{F}) when substituting the dimensions appropriately. Thus, the proof confirms the initial statement regarding the dimensions of the subspaces involved. The discussion effectively illustrates the application of fundamental concepts in linear algebra to validate the theorem.
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Homework Statement


Prove that \dim L(\mathbb F)+\dim Ker L=\dim(\mathbb F+Ker L) for every subspace \mathbb{F} and every linear transformation L of a vector space V of a finite dimension.

Homework Equations


-Fundamental subspaces
-Vector spaces

The Attempt at a Solution



Theorem: [/B]If L:U\rightarrow V is a linear transformation and \dim U=n, then \dim Ker L+\dim C(L^T)=n. Ker L is the null space, C(L^T) is the row space of L and n is the number of column vectors in [L].

How to use this theorem to prove the given statement?
 
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You may consider U=\mathbb{F}+\ker L and ##L: U → V##. Then according to your given formula \dim U = \dim \ker L + \dim C(L^T). With ##\dim C(L^T) = \dim L(U) = \dim L(\mathbb{F} + \ker L) = \dim L(\mathbb{F})## we have ##\dim(\mathbb{F}+\ker L) = \dim U = \dim \ker L + \dim L(\mathbb{F})##.
 
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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