# 4.4 Subspaces

When we talk about a vector space over a field $\mathbb{F}$, the word scalar refers to an element of $\mathbb{F}$.

###### Definition 4.4.1.

A subspace of a vector space $V$ is a subset $U$ of $V$ which

1. 1.

contains the zero vector $\textbf{0}_{V}$,

2. 2.

is closed under addition, meaning that for all $\mathbf{v},\mathbf{w}\in U$ we have $\mathbf{v}+\mathbf{w}\in U$, and

3. 3.

is closed under scalar multiplication, meaning that for all scalars $\lambda$ and all $\mathbf{u}\in U$ we have $\lambda\mathbf{u}\in U$.

We write $U\leqslant V$ to mean that $U$ is a subspace of $V$.

The idea this definition captures is that a subspace of $V$ is a nonempty subset which is itself a vector space under the same addition and scalar multiplication as $V$.

If $U\leqslant V$ and $\textbf{u}_{1},\ldots,\textbf{u}_{n}\in U$ and $\lambda_{1},\ldots,\lambda_{n}$ are scalars then $\sum_{i=1}^{n}\lambda_{i}\textbf{u}_{i}\in U$. This follows by using closure under scalar multiplication and closure under addition lots of times.

## 4.4.1 Subspace examples

###### Example 4.4.1.

If $V$ is any vector space, $V\leqslant V$. This is because, as a vector space, $V$ contains the zero vector, is closed under addition, and is closed under scalar multiplication.

A subspace of $V$ other than $V$ is called a proper subspace.

###### Example 4.4.2.

For any vector space $V$ we have $\{\textbf{0}_{V}\}\leqslant V$. Certainly this set contains the zero vector. It is closed under addition because $\mathbf{0}_{V}+\mathbf{0}_{V}=\mathbf{0}_{V}$, and it is closed under scalar multiplication by Lemma 4.3.3. This is called the zero subspace.

###### Example 4.4.3.

Let $U$ be the set of vectors in $\mathbb{R}^{2}$ whose first entry is zero. Then $U\leqslant\mathbb{R}^{2}$. We check the three conditions in the definition of subspace.

1. 1.

The zero vector in $\mathbb{R}^{2}$ is $\begin{pmatrix}0\\ 0\end{pmatrix}$. This has first coordinate 0, so it is an element of $U$.

2. 2.

Let $\textbf{v},\textbf{w}\in U$, so that $\textbf{v}=\begin{pmatrix}0\\ x\end{pmatrix}$ and $\textbf{w}=\begin{pmatrix}0\\ y\end{pmatrix}$ for some real numbers $x$ and $y$. Then $\textbf{v}+\textbf{w}=\begin{pmatrix}0\\ x+y\end{pmatrix}$ has first coordinate 0, so it is an element of $U$.

3. 3.

Let v be as above and $\lambda\in\mathbb{R}$. Then $\lambda\textbf{v}=\begin{pmatrix}0\\ \lambda x\end{pmatrix}$ which has first coordinate 0, so $\lambda\textbf{v}\in U$.

All three conditions hold, so $U\leqslant\mathbb{R}^{2}$. Of course, a similar argument shows the vectors in $\mathbb{F}^{n}$ with first entry 0 are a subspace of $\mathbb{F}^{n}$ for any field $\mathbb{F}$ and any $n$.

To every matrix $A$ we associate two important subspaces. The nullspace $N(A)$ (Definition 3.6.5) is the set of all vectors $\mathbf{x}$ such that $A\mathbf{x}=\mathbf{0}$, and the column space $C(A)$ is the set of all linear combinations of the columns of $A$.

###### Example 4.4.4.

Let $A$ be an $m\times n$ matrix with entries from the field $\mathbb{F}$. The nullspace $N(A)$ contains the zero vector as $A\mathbf{0}_{n}=\mathbf{0}_{m}$. It is closed under addition as if $\mathbf{u},\mathbf{v}\in N(A)$ then $A\mathbf{v}=\mathbf{0}_{m}$ and $A\mathbf{u}=\mathbf{0}_{m}$ so

 $\displaystyle A(\mathbf{u}+\mathbf{v})$ $\displaystyle=A\mathbf{u}+A\mathbf{v}$ $\displaystyle=\mathbf{0}_{m}+\mathbf{0}_{m}$ $\displaystyle=\mathbf{0}_{m}$

and therefore $\mathbf{u}+\mathbf{v}\in N(A)$. It is closed under scalar multiplication because if $\lambda$ is any scalar then $A(\lambda\mathbf{u})=\lambda A\mathbf{u}=\lambda\mathbf{0}_{m}=\mathbf{0}_{m}$ so $\lambda\mathbf{u}\in N(A)$. Therefore $N(A)\leqslant\mathbb{F}^{n}$.

The column space $C(A)$, defined to be the set of all linear combinations of the columns of $A$, is a subspace of $\mathbb{F}^{m}$. We won’t prove that here, because it is a special case of Proposition 4.7.1 which we prove later.

###### Example 4.4.5.

The set $U$ of all vectors in $\mathbb{R}^{3}$ with first entry 1 is not a subspace of $\mathbb{R}^{3}$. It doesn’t contain the zero vector (and it doesn’t meet the other two conditions either).

###### Example 4.4.6.

$\mathbb{Z}$ is not a subspace of $\mathbb{R}$. It contains the zero vector 0, it is closed under addition because if you add two integers you get another integer. But it is not closed under scalar multiplication: $\sqrt{2}$ is a scalar, $1\in\mathbb{Z}$, but $\sqrt{2}\times 1$ is not in $\mathbb{Z}$.

###### Example 4.4.7.

Let $U$ be the set of all functions $f:\mathbb{R}\to\mathbb{R}$ with $f(1)=0$. This is a subspace of the vector space $\mathcal{F}$ of all functions $\mathbb{R}\to\mathbb{R}$. The zero vector in $\mathcal{F}$ is the constant function that always takes the value zero, so certainly it belongs to $U$. If $f,g\in U$ then $(f+g)(1)=f(1)+g(1)=0+0=0$, so $f+g\in U$. If $\lambda\in\mathbb{R}$ and $f\in U$ then $(\lambda f)(1)=\lambda f(1)=\lambda\times 0=0$ so $\lambda f\in U$.

###### Example 4.4.8.

$\{A\in M_{n\times n}(\mathbb{R}):A^{T}=A\}\leqslant M_{n\times n}(\mathbb{R})$. The transpose operation satisfies $(A+B)^{T}=A^{T}+B^{T}$ and $(\lambda A)^{T}=\lambda A^{T}$, which you should check. This makes the three conditions straightforward to check.

###### Example 4.4.9.

$U=\{A\in M_{n\times n}(\mathbb{R}):A^{2}=\mathbf{0}_{m\times n}\}$ is not a subspace of $M_{n\times n}(\mathbb{R})$. For example, $U$ contains $E_{12}=\begin{pmatrix}0&1\\ 0&0\end{pmatrix}$ and $E_{21}=\begin{pmatrix}0&0\\ 1&0\end{pmatrix}$ but you can check that $E_{12}+E_{21}\notin U$.