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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?

Thank you in advance

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  • 2
    $\begingroup$ A small remark: It makes no sense to talk of "the basis" of a vector space. Typically, there are infinitely many possible bases. $\endgroup$ – Andrés E. Caicedo Jun 22 '13 at 5:32
  • $\begingroup$ 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 $S\neq V$. $\endgroup$ – Alex R. Jul 27 '16 at 22:38
  • $\begingroup$ @AndrésE.Caicedo: This seems to be a very picky statement if I am understanding it correctly. Once someone says dimension of vector space it is automatically implied that, that value comes from the smallest set of linearly independent vectors which span, at least that's how I was taught. $\endgroup$ – Faraad Armwood Jul 27 '16 at 22:40
  • $\begingroup$ It is a theorem that every spanning linearly independent set of a given vector space has the same cardinality. Therefore there's no ambiguity in saying "the dimension" of a vector space. $\endgroup$ – Greg Martin Jul 27 '16 at 22:47

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