When we talk about a vector space over a field $\mathbb{F}$, the word scalar refers to an element of $\mathbb{F}$.
A subspace of a vector space $V$ is a subset $U$ of $V$ which
contains the zero vector ${\text{\U0001d7ce}}_{V}$,
is closed under addition, meaning that for all $\mathbf{v},\mathbf{w}\in U$ we have $\mathbf{v}+\mathbf{w}\in U$, and
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 $U\u2a7dV$ 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 $U\u2a7dV$ and ${\text{\mathbf{u}}}_{1},\mathrm{\dots},{\text{\mathbf{u}}}_{n}\in U$ and ${\lambda}_{1},\mathrm{\dots},{\lambda}_{n}$ are scalars then ${\sum}_{i=1}^{n}{\lambda}_{i}{\text{\mathbf{u}}}_{i}\in U$. This follows by using closure under scalar multiplication and closure under addition lots of times.
If $V$ is any vector space, $V\u2a7dV$. 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.
For any vector space $V$ we have $\{{\text{\U0001d7ce}}_{V}\}\u2a7dV$. Certainly this set contains the zero vector. It is closed under addition because ${\mathrm{\U0001d7ce}}_{V}+{\mathrm{\U0001d7ce}}_{V}={\mathrm{\U0001d7ce}}_{V}$, and it is closed under scalar multiplication by Lemma 4.3.3. This is called the zero subspace.
Let $U$ be the set of vectors in ${\mathbb{R}}^{2}$ whose first entry is zero. Then $U\u2a7d{\mathbb{R}}^{2}$. We check the three conditions in the definition of subspace.
The zero vector in ${\mathbb{R}}^{2}$ is $\left(\begin{array}{c}0\\ 0\end{array}\right)$. This has first coordinate 0, so it is an element of $U$.
Let $\text{\mathbf{v}},\text{\mathbf{w}}\in U$, so that $\text{\mathbf{v}}=\left(\begin{array}{c}0\\ x\end{array}\right)$ and $\text{\mathbf{w}}=\left(\begin{array}{c}0\\ y\end{array}\right)$ for some real numbers $x$ and $y$. Then $\text{\mathbf{v}}+\text{\mathbf{w}}=\left(\begin{array}{c}0\\ x+y\end{array}\right)$ has first coordinate 0, so it is an element of $U$.
Let v be as above and $\lambda \in \mathbb{R}$. Then $\lambda \text{\mathbf{v}}=\left(\begin{array}{c}0\\ \lambda x\end{array}\right)$ which has first coordinate 0, so $\lambda \text{\mathbf{v}}\in U$.
All three conditions hold, so $U\u2a7d{\mathbb{R}}^{2}$. 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 $A\mathbf{x}=\mathrm{\U0001d7ce}$, and the column space $C(A)$ is the set of all linear combinations of the columns of $A$.
Let $A$ be an $m\times n$ matrix with entries from the field $\mathbb{F}$. The nullspace $N(A)$ contains the zero vector as $A{\mathrm{\U0001d7ce}}_{n}={\mathrm{\U0001d7ce}}_{m}$. It is closed under addition as if $\mathbf{u},\mathbf{v}\in N(A)$ then $A\mathbf{v}={\mathrm{\U0001d7ce}}_{m}$ and $A\mathbf{u}={\mathrm{\U0001d7ce}}_{m}$ so
$A(\mathbf{u}+\mathbf{v})$ | $=A\mathbf{u}+A\mathbf{v}$ | ||
$={\mathrm{\U0001d7ce}}_{m}+{\mathrm{\U0001d7ce}}_{m}$ | |||
$={\mathrm{\U0001d7ce}}_{m}$ |
and therefore $\mathbf{u}+\mathbf{v}\in N(A)$. It is closed under scalar multiplication because if $\lambda $ is any scalar then $A(\lambda \mathbf{u})=\lambda A\mathbf{u}=\lambda {\mathrm{\U0001d7ce}}_{m}={\mathrm{\U0001d7ce}}_{m}$ so $\lambda \mathbf{u}\in N(A)$. Therefore $N(A)\u2a7d{\mathbb{F}}^{n}$.
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.
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).
$\mathbb{Z}$ 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 $\mathbb{Z}$.
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 $\lambda f\in U$.
$\{A\in {M}_{n\times n}(\mathbb{R}):{A}^{T}=A\}\u2a7d{M}_{n\times n}(\mathbb{R})$. 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.
$U=\{A\in {M}_{n\times n}(\mathbb{R}):{A}^{2}={\mathrm{\U0001d7ce}}_{m\times n}\}$ is not a subspace of ${M}_{n\times n}(\mathbb{R})$. For example, $U$ contains ${E}_{12}=\left(\begin{array}{cc}0& 1\\ 0& 0\end{array}\right)$ and ${E}_{21}=\left(\begin{array}{cc}0& 0\\ 1& 0\end{array}\right)$ but you can check that ${E}_{12}+{E}_{21}\notin U$.