- #1

Hall

- 351

- 87

- Homework Statement
- If a linear space V is finite dimensional then S, a subspace of V, is also finite-dimensional and dim S is less than or equal to dim V.

- Relevant Equations
- Let A = {u_1, u_2, ... u_n} be a basis for V.

If a linear space ##V## is finite dimensional then ##S##, a subspace of ##V##, is also finite-dimensional and ##dim ~S \leq dim~V##.

$$

x = \sum_{i=1}^{n} c_i u_i

$$

As ## S \subset V## therefore, all the elements of ##S## can also be represented by linear combinations of ##u_i##s. As, ##u_i## s are finite, this implies that ##S## is finite dimensional.

Let's say ##dim~S \gt dim~V##. Then, a basis for ##S## would look like

$$

A' = \{ v_1, v_2, \cdots v_m\}

$$

where ##m \gt n##. But that would imply that ##A'## is independent thus contradicting that "any set of (n+1) elements in V would be dependent if ##L(A) = V## and number of elements in A is n". Hence, ##dim ~S \leq dim~V##.

Thats ends my proof.

It seems to me that my proof of the first part is quite verbal and needs to be a little more rigorous but the question is: Was the reasoning correct? And how to make it rigorous?

*Proof:*Let's assume that ##A = \{u_1, u_2, \cdots u_n\}## be a basis for ##V##. Well, then any element ##x## of ##V## can be represented as$$

x = \sum_{i=1}^{n} c_i u_i

$$

As ## S \subset V## therefore, all the elements of ##S## can also be represented by linear combinations of ##u_i##s. As, ##u_i## s are finite, this implies that ##S## is finite dimensional.

Let's say ##dim~S \gt dim~V##. Then, a basis for ##S## would look like

$$

A' = \{ v_1, v_2, \cdots v_m\}

$$

where ##m \gt n##. But that would imply that ##A'## is independent thus contradicting that "any set of (n+1) elements in V would be dependent if ##L(A) = V## and number of elements in A is n". Hence, ##dim ~S \leq dim~V##.

Thats ends my proof.

It seems to me that my proof of the first part is quite verbal and needs to be a little more rigorous but the question is: Was the reasoning correct? And how to make it rigorous?