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. dimension of sums See dimension of sums