Relationship between dimensions of subspaces

Question

Let $V$ and $W$ be subspaces with $V \subseteq W$. Show that $dimV \le dimW$ with equality iff $V = W$.

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My first problem is with understanding the question.

What does it mean by "with equality" when already using an $\le$ symbol? Am I correct in assuming that $\subseteq$ means, V is equal/equivalent to W, OR is a subset of W (conceptually similar to $\lt$ vs $\le$)?

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Apart from this... the answer seems intuitively simple. If V is a subset of W, then how could it have a higher dimension?

But I don't even know where to start on solving this mathematically. Which definitions should I be using here?

Answer 1

Start with a basis for $V$. If necessary, extend that basis to get a basis for $W$. Since the dimension of a subspace is the number of vectors in a basis for it, we have $\mathrm{dim}(V) \leq \mathrm{dim}(W)$.

Edited to add some details: Sticking with the finite-dimensional case for now. Suppose $A = \{a_1, \dots a_i\}$ is a basis for $V$. Let $b_1 \in W$ be a vector such that $A \cup \{b_1\}$ is a linearly independent set of vectors. (If no such $b_1$ exists, then $V = W$ and we're done.) Let $V_1 = \mathrm{span}(A \cup \{b_1\})$.

Now let $b_2 \in W$ be a vector such that $A \cup \{b_1, b_2\}$ is a linearly independent set of vectors. (If no such $b_2$ exists, then $V_1 = W$ and we're done.) Let $V_2 = \mathrm{span}(A \cup \{b_1, b_2\}$.

Etc.

Eventually, this process stops because $W$ is finite dimensional, and it produces the basis $\{a_1, \dots, a_i, b_1, \dots, b_j\}$ for $W$. Then $\mathrm{dim}(V) = i \leq \mathrm{dim}(W) = i + j$.

The statement that $\mathrm{dim}(V) = \mathrm{dim}(W)$ iff $V = W$ is only true for finite dimensional vector spaces. Infinite dimensional vector spaces have proper subspaces of the same dimension.