### Basis

##### 5. Properties of dimension

Since vector spaces of dimension n may be treated just like Rn, all properties of euclidean vectors concerning the dimension n can be extended to general vector spaces. For example, the property on the size of spanning sets and the property on the size of linearly independent sets may be extended.

Let dimV = n. Then

• v1, v2, ..., vk span Vkn.
• v1, v2, ..., vkV are linearly independent ⇒ kn.

Moreover, the property on the size of onto linear transformations and the property on the size of one-to-one linear transformations may also be extended.

Let T: VW be a linear transformation. Then

• T is onto ⇒ dimVdimW.
• T is one-to-one ⇒ dimVdimW.

In particular, isomorphic vector spaces have the same dimension.

The following extends the basic principal of linear algebra.

Let v1, v2, ..., vn be vectors in a space V of dimension n. Then the following are equivalent.

• v1, v2, ..., vn form a basis of V.
• v1, v2, ..., vn span V.
• v1, v2, ..., vn are linearly independent.

For linear transformations, the basic principal takes the following form (see this criterion for the invertibility of a linear transformation between euclidean spaces).

Let T: VW be a linear transformation between finite dimensional vector spaces. If dimV = dimW, then the following are equivalent.

• T is invertible.
• TS = id for some transformation S.
• ST = id for some transformation S.
• T is onto.
• T is on-to-one.

Finally, as a consequence of the properties above, we have

Let H be a subspace of V. Then

• dimHdimV.
• If dimH = dimV < ∞, then H = V.

The property is easy to understand if H and V are replaced by finite sets and dimension is replaced by number of elements.

Proof Let {v1, v2, ..., vk} be a basis of H. Then k = dimH and v1, v2, ..., vk are linearly independent as vectors of H. Clearly, v1, v2, ..., vk are also linearly independent as vectors of V. Thus by this property, we have kdimV. Since k = dimH, the first part is proved.

For the second part, note that v1, v2, ..., vk being linearly independent and k = dimH = dimV implies that v1, v2, ..., vk form a basis of V. In particular, the vectors span V. On the other hand, they also form a basis of H and hence span H. Thus we conclude

H = span{v1, v2, ..., vk} = V.