- #1

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- Homework Statement
- True or false question? (Prove or give a counterexample)

Let ##V## be a real finite vector space and ##U_1, U_2## and ##U_3## be subspaces (of ##V##) with ##U_1 \cap U_2 = \{0\}##. Then the following statement holds

$$\dim (U_1 \cap U_3) + \dim (U_2 \cap U_3) \leq \dim (U_3)$$

- Relevant Equations
- N/A

My intuition tells me this is a true statements so let's try to prove it.

The dimension is defined as the number of elements of a basis. Hence, we can work in terms of basis to prove the statement.

Given that ##U_3## appears on both sides of the inequality, let's get a basis for it. How? Let's suppose that ##\{u_1, \dots, u_m \}## is a basis for ##U_1 \cap U_3##. By definition of intersection, ##\{u_1, \dots, u_m \} \in U_3##. The following is a well-known theorem in linear algebra: given a linearly independent list of vectors in a finite dimensional vector space (another one is that given a finite-dimensional vector space, any subspace of it is also finite dimensional), it is always possible to extend it to a basis of the vector space. Hence, a basis for ##U_3## is

$$\beta_{U_3} = \{u_1, \dots, u_m, w_1, \dots, w_j \}$$

But I do not really see how can we conclude the proof. I guess we still need to argue/prove why ##\{w_1, \dots, w_j \}## is a basis for ##U_2 \cap U_3##

Your guidance is appreciated, thanks!

The dimension is defined as the number of elements of a basis. Hence, we can work in terms of basis to prove the statement.

Given that ##U_3## appears on both sides of the inequality, let's get a basis for it. How? Let's suppose that ##\{u_1, \dots, u_m \}## is a basis for ##U_1 \cap U_3##. By definition of intersection, ##\{u_1, \dots, u_m \} \in U_3##. The following is a well-known theorem in linear algebra: given a linearly independent list of vectors in a finite dimensional vector space (another one is that given a finite-dimensional vector space, any subspace of it is also finite dimensional), it is always possible to extend it to a basis of the vector space. Hence, a basis for ##U_3## is

$$\beta_{U_3} = \{u_1, \dots, u_m, w_1, \dots, w_j \}$$

But I do not really see how can we conclude the proof. I guess we still need to argue/prove why ##\{w_1, \dots, w_j \}## is a basis for ##U_2 \cap U_3##

Your guidance is appreciated, thanks!