# Show that if $S$ is a subspace of a vector space $V$, then $\dim (S) \leq \dim (V)$

Show that if $S$ is a subspace of a vector space $V$, then $\dim(S) \leq \dim(V)$. Furthermore, if $\dim(S)= \dim(V) < \infty$ then $S=V$. Give an example to show that the finiteness is required in the second statement.

Suppose, for contradiction, that $\dim(S) > \dim(V)$. Suppose that the basis of $S$ is $s_1, ..., s_n$ and the basis of $V$ is $v_1, ..., v_m$ where $n>m$.

$$a_1s_1 + ... + a_ns_n = 0 \iff a_1 = a_2 = ... = a_n = 0$$

This means that $s_1 , ... , s_n$ is linearly independent in $V$. We know that a spanning set containing $s_1, ... , s_n$ contains a basis for $V$ with cardinality greater than or equal to $n$. But $v_1, ... , v_m$ is also a basis, so this is a contradiction by the theorem that says "If $V$ is a vector space, then any two bases for $V$ have the same cardinality."

Now suppose that $\dim(S) = \dim(V)$. Suppose that $s_1, ... , s_n$ is a basis for $S$ and $v_1, ... , v_n$ is a basis for $V$.

Again, since $a_1s_1 + ... + a_ns_n = 0 \iff a_1 = a_2 = ... = a_n = 0$, $s_1, ... , s_n$ is linearly independent in $V$. Since $\dim(V)=n$ and any basis must be of the same cardinality, we know that $S$ is a basis for $V$ as well. So $V=S$.

Finiteness is required, because two infinite sets may not have the same number of elements. For example $\Bbb{R}$ and $\Bbb{Q}$ are both infinite but $\Bbb{R}$ contains more elements than $\Bbb{Q}$. So we wouldn't be able to use the same argument. In other words, in the argument we said that $s_1, ... , s_n$ is linearly independent and the basis of $V$ is of cardinality $n$...so it must be a basis as well. But if the dimensions were infinity, $s_1, s_2, ...$ has an infinite number of elements; but that doesn't necessarily mean that it contains the same number of elements as the $v_1, v_2, ...$.

For the example, we can think about the vector space $\Bbb{R}[X]$ and $\Bbb{R}$ as a subspace of that. Both have an infinite basis but they are not equal.

Do you think my answer are correct?

• You have the right idea about finiteness but I'm not seeing a physical example of two infinite vector spaces with the same dimension but $S\neq V$. – Alex R. Jul 27 '16 at 22:38