The dimension of a vector space is the length of any basis in the vector space. It is denoted as \(\dim V\).
additional information
See also finite-dimensional vector space and infinite-demensional vector space
dimension of subspace is smaller or equal to that of its parent
If we have a finite-dimensional \(V\) and a subspace thereof \(U\), then \(\dim U \leq \dim V\).
Firstly, the every subspace of a finite-dimensional vector space is a finite-dimensional vector space is itself a finite-dimensional vector space. Therefore, it has a finite dimension.
Then, we will simply think of the basis of \(U\) as an linearly independent list in \(V\); and of course, the basis of \(V\) spans \(V\). As length of linearly-independent list \(\leq\) length of spanning list, we have that length of basis of \(U \leq\) length of basis of \(V\).
This makes \(\dim U \leq \dim V\), as desired. \(\blacksquare\)
lists of right length are a basis
These are two results that tell us if you are given a list of list of right length, one condition (spanning or linear independence) can tell you that they are a basis. It’s also known (as a John McHugh special:tm:) as the Half Is Good Enough theorems.
linearly independent list of length dim V are a basis of V
Begin with an linearly independent list in \(V\) of length \(\dim V\). We aim to extend this list into a basis of \(V\).
As we know all basis in \(V\) must have length \(\dim V\), and the list is already length \(\dim V\), no extension is needed to form a basis.
As every linearly independent list expends to a basis, we conclude that the list is already a basis of \(V\), as desired \(\blacksquare\).
spanning list of length of dim V are a basis of V
Begin with a spanning list in \(V\) of length \(\dim V\). We aim to reduce this list into a basis of \(V\).
As we know all basis in \(V\) must have length \(\dim V\), and the list is already length \(\dim V\), no reduction is needed to form a basis.
As all spanning lists contains a basis of which you are spanning, we conclude that the list is a basis of \(V\), as desired \(\blacksquare\).