Subspace of Full Dimension Equals the Whole Space

The fact that a linearly independent list of the right length is a basis gives us a useful consequence.

Subspace of full dimension equals the whole space

Suppose that $V$ is finite-dimensional and $U$ is a subspace of $V$, such that $\dim U=\dim V$. Then $U=V$.

Proof. Let $u_1,\dots ,u_n$ be a basis of $U$. Thus $n=\dim U$. Since we have $\dim U=\dim V$, we also have $n=\dim V$.

Thus $u_1,\dots ,u_n$ is a linearly independent list of vectors in $V$ (because it is a basis of $U$) of length $\dim V$. From Linearly Independent List of Right Length is a Basis, we see that $u_1,\dots ,u_n$ is a basis of $V$, because it has length $\dim V$. In particular, every vector in $V$ is a linear combination of $u_1,\dots ,u_n$. Thus, $U=V$.

Examples

Example 1: A basis of ${\mathbb{F}}^2$.

Consider the list $(5,7),(4,3)$ of vectors in ${\mathbb{F}}^3$. This list of length two is linearly independent in ${\mathbb{F}}^2$, because neither vector is a scalar multiple of the other. Note that ${\mathbb{F}}^2$ has dimension 2.

Thus, Linearly Independent List of the Right Length is a Basis implies that the linearly independent list $(5,7),(4,3)$ of length two is a basis of ${\mathbb{F}}^2$.

• No need to bother with checking that it spans ${\mathbb{F}}^2$.

Example 2: A basis of a subspace of ${\mathcal{P}}_3(\mathbb{R})$.

Let $U$ be the subspace of ${\mathcal{P}}_3(\mathbb{R})$ defined by

$$U=\{p\in {\mathcal{P}}_3(\mathbb{R}):p^{\prime }(5)=0\}$$

To find a basis of $U$, first note that each of the polynomials $1,(x-5{)}^2,(x-5{)}^3$ are in $U$.

Suppose $a,b,c\in \mathbb{R}$ and

$$a+b(x-5{)}^2+c(x-5{)}^3=0$$

for every $x\in \mathbb{R}$.

• Without explicitly expanding the left side of the equation above, we can see that the left side has a $c{x}^3$ term. Because the right side has no $x^3$ term, this implies that $c=0$.
• Because $c=0$, we can see that the left side has a $b{x}^2$ term, which implies that $b=0$.
• Because $b=c=0$, we can conclude $a=0$.

Thus, the equation above implies that $a=b=c=0$. Hence the list $1,(x-5{)}^2,(x-5{)}^3$ is linearly independent in $U$. This means that $3\le \dim U$. Hence

$$3\le \dim U\le \dim {\mathcal{P}}_3(\mathbb{R})=4$$

where we used Dimension of a Subspace.

The polynomial $x$ is not in $U$ because its derivative is the constant function $1$. Thus $U\ne {\mathcal{P}}_3(\mathbb{R})$. Hence, $\dim U\ne 4$, by our previous result that “subspace of full dimension equals the whole space”, if $\dim U=4$, we would have $U={\mathcal{P}}_3(\mathbb{R})$.

The inequality above now implies that $\dim U=3$. Thus, the linearly independent list $1,(x-5{)}^2,(x-5{)}^3$ in $U$ has length $\dim U$ and hence is a basis of $U$.