Dimension of a Vector Space Examples 1

Recall from the Dimension of a Vector Space page that if $V$ is a finite-dimensional vector space then the dimension of $V$ is equal to the length of any basis of $V$. If the length of any basis of a finite-dimensional vector space $V$ is $n$ then we write $\mathrm{dim} (V) = n$.

We will now look at some problems regarding the dimension vector spaces.

Example 1

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

We know that since $V$ is a finite-dimensional vector space, then any subspace $U$ of $V$ will also be finite-dimensional from the Infinite / Finite-Dimensional Vector Space Comparison Theorem page.

Now suppose that $\mathrm{dim} (U) > \mathrm{dim} (V)$. Let $\{ u_1, u_2, ..., u_n \}$ be a basis of $U$ and let $\{ v_1, v_2, ..., v_m \}$ be a basis of $V$ where $n > m$. Since $U$ is a subspace of $V$ we have that since every vector $v \in V$ can be written as a linear combination of the vectors in $\{ v_1, v_2, ..., v_m \}$, and so every vector $u \in U$ can be written as a linear combination of the vectors in $\{ v_1, v_2, ..., v_m \}$. Therefore, any basis of $U$ must have $m$ vectors or less, that is, $n ≤ m$ which contradicts our assumption that $n > m$.

Therefore $\mathrm{dim} (U) ≤ \mathrm{dim} (V)$.

Example 2

Prove that if $V$ is a finite-dimensional vector space with $\mathrm{dim} (V) = n$ then there exists one-dimensional subspaces $U_1$, $U_2$, …, $U_n$ such that $V = U_1 \oplus U_2 \oplus ... \oplus U_n$.

Let $V$ be a finite-dimensional vector space with $\mathrm{dim} (V) = n$. Then there exists a basis of $V$ containing $n$ vectors, say $\{ v_1, v_2, ..., v_n \}$ so that for every vector $v \in V$ we have that for scalars $a_1, a_2, ..., a_n \in \mathbb{F}$ that:

(1)

\begin{align} \quad v = a_1v_1 + a_2v_2 + ... + a_nv_n \end{align}

Let $U_1 = \mathrm{span} (v_1)$, $U_2 = \mathrm{span} (v_2)$, …, $U_n = \mathrm{span} (v_n)$. Each of these subspaces are clearly one-dimensional.

Now we note that for every vector $v \in V$ we have that $v = u_1 + u_2 + ... + u_n$ where $u_j \in U_j$ for each $j = 1, 2, ..., n$. More specifically, $u_j = a_jv_j \in \mathrm{span} (v_j)$. Therefore:

(2)

\begin{align} \quad V = U_1 + U_2 + ... + U_n \end{align}

Furthermore, we note that every vector $v \in V$ can be written uniquely as $v = u_1 + u_2 + ... + u_n = a_1v_1 + a_2v_2 + ... + a_nv_n$ since $\{ v_1, v_2, ..., v_n \}$ is a linearly independent set of vectors, and so we conclude that

(3)

\begin{align} \quad V = U_1 \oplus U_2 \oplus ... \oplus U_n \end{align}

Example 3

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

Let $V$ be a finite-dimensional vector space and suppose that $\mathrm{dim} (V) = n$. Let $U$ be a subspace of $V$ and assume that $\mathrm{dim} (U) = n$.

Since $V$ and $U$ are both finite-dimensional vector spaces whose dimension is $n$ then there exists a basis $\{ v_1, v_2, ..., v_n \}$ of $V$ and a basis $\{ u_1, u_2, ..., u_n \}$ of $U$. Since $U$ is a subspace of $V$, we have that every vector in the basis of $U$ is also in $V$. Furthermore, since $\{ u_1, u_2, ..., u_n \}$ is linearly independent in $U$, we must have that $\{ u_1, u_2, ..., u_n \}$ is linearly independent in $V$.

Now $\{ u_1, u_2, ..., u_n \}$ is linearly independent in $V$ and contains $\mathrm{dim} (V) = n$ vectors which implies that $\{ u_1, u_2, ..., u_n \}$ is a basis of $V$. Therefore, $\mathrm{span} (v_1, v_2, ..., v_n) = V = \mathrm{span} (u_1, u_2, ..., u_n)$. By definition, we have that $U = \mathrm{span} (u_1, u_2, ..., u_n)$.

Thus $U = V$.