# Linear Algebra: Linear Combinations

## Dimension

The **dimension** of a vector space is the cardinality of any basis of the vector space. The dimension of a vector space $V$ is written as $\text{dim}(V)$. Recall that all bases for a given vector space have the same cardinality, so the dimension of a vector space is invariant of the dimension chosen.

A vector space is **finite dimensional** if its dimension is finite. If $\text{dim}(V) = n$, we say that $V$ is $n$**-dimensional**. This definition of dimension corresponds exactly with our geometric intuition about dimensions Euclidean space: $\mathbb{R}^n$ is $n$-dimensional. Vector spaces that are not finite dimensional are **infinite dimensional**.

## Problems

Show that if $U$ is a subspace of $V$, then $\text{dim}(U) \leq \text{dim}(V)$.

Let $B_1$ be a basis for $U$. Then $B_1$ is a linearly independent subset of $V$. Let $B_2$ be a basis of $V$. Then $B_2$ spans $V$. Since the cardinality of a linearly independent set of vectors is at most the cardinality of a spanning set of vectors, $|B_1| \leq |B_2|$, so $\text{dim}(U) \leq \text{dim}(V)$.

Show that if a set of linearly independent vectors in a finite-dimensional vector space $V$ has $\text{dim}(V)$ vectors in it, it is a basis.

Let $\text{dim}(V) = n$, and let $A = \{a_1, \ldots, a_n\}$ be linearly independent. By definition of dimension, each basis of $V$ has $n$ vectors in it. Because a linearly independent set of vectors with the same cardinality as a basis is itself a basis, it follows that $A$ is a basis.

Let $U$ be a subspace of a finite-dimensional vector space $V$. Show that if $\text{dim}(U) = \text{dim}(V)$, then $U = V$.

Let $B_1$ be a basis for $U$, and let $B_2$ be a basis for $V$. If $\text{dim}(U) = \text{dim}(V)$, then $B_1$ is a set of linearly independent vectors in $V$ with the same number of vectors as $B_2$. This implies that $B_1$ is a basis of $V$. Therefore $U = \text{span}(B_1) = \text{span}(B_2) = V$.