For subspace $U$ of $V$ where $V$ is finite-dimensional, is $U = V$ if and only if $\dim U = \dim V$?

Question

For subspace $U$ of $V$ where $V$ is finite-dimensional, $U = V$ if and only if $\dim U = \dim V$.

A. Is the statement above true?

B. If true, is the following proof correct?

Suppose $U = V$. Then $\dim U = \dim V$ by definition of dimension.

Conversely, suppose $\dim U = \dim V = n$. Let $u_1, \dotsc, u_n$ be a basis of $U$. Since $U$ is a subspace of $V$, $u_1, \dotsc, u_n$ is linearly independent in $V$ and thus can be extended to a basis of $V$. However, since $u_1, \dotsc, u_n$ already contains $n$ vectors, there is nothing to add, hence it is already a basis of $V$. Since $U$ and $V$ share the same basis, $U = V$.

Answer 1

The statement is true in finite dimensions. Your proof is essentially correct but just slightly unclear. A somewhat clearer way to say it: $u_1,\dots,u_n$ can be extended to a basis of $V$, but all bases of $V$ have cardinality exactly $n$, so $u_1,\dots,u_n$ is already a basis of $V$. Thus $U$ and $V$ have a common basis and are thus the same vector space.

The statement is false in infinite dimensions, because in this case the basis of $U$ may be a proper subset of a basis of $V$ even though both have the same cardinality.

Answer 2

Yes, it is true, assuming that we are dealing with finite-dimensional spaces here.

Your proof is correct. You don't need to write that if $U=V$, them $\dim U=\dim V$ by definition of dimension. If two things are equal, everything that we can build from them will also be equal.