Introduction to Vectors and Subspaces

Not familiar 100% with what a vector space is? In this article, I cover vectors spaces and some basic theorems. I also emphasize linear combinations, the unions and intersections of subspaces, and provide several exercises.

This article is an introduction to vectors and subspaces. We use the terms vector space and linear space as synonyms.

Introduction to Vectors

Very precisely, here is what we mean by vector space.

Definition. Let $\mathbb{F}$ be a field (whose elements are called scalars) and let $\mathbb{V}$ be a nonempty set (whose elements are called vectors) on which two operations, called addition and scalar multiplication, have been defined. The addition operation (denoted by $+$), assigns to each pair $(u,v)\in \mathbb{V}\times \mathbb{V}$, a unique vector $u+v$ in $\mathbb{V}.$ The scalar multiplication operation (denoted by juxtaposition), assigns to each pair $(a,v)\in \mathbb{F}\times \mathbb{V}$ a unique vector $a v$ in $\mathbb{V}.$ We call $\mathbb{V}$ a linear space if the following axioms (A1)-(A8) are also satisfied.

(1) For all $u, v\in \mathbb{V}$, $u+v=v+u.$

(2) For all $u, v, w \in \mathbb{V}$, $(u+v)+w=u+(v+w).$

(3) There exists ${0}\in \mathbb{V}$ such that $v+{0}=v$ for all $v\in \mathbb{V}.$

(4) For every $v\in \mathbb{V}$, there exists $w\in \mathbb{V}$ such that $v+w={0}.$

(5) For all $v\in \mathbb{V}$, $1 v=v.$

(6) For all $a, b\in \mathbb{F}$ and $u\in \mathbb{V}$, $(a b) v=a (b v).$

(7) For all $a \in \mathbb{F}$ and $u, v\in \mathbb{V}$, $a(u+v)=a u+av.$

(8) For all $a, b \in \mathbb{F}$ and $u\in \mathbb{V}$, $(a+b)u=a u+ b u.$

If $\mathbb{F}=\mathbb{R}$ then $\mathbb{V}$ is a called a real linear space. If $\mathbb{F}=\mathbb{C}$ then $\mathbb{V}$ is called a complex linear space. We denote the zero vector as $0$, to distinguish between the zero vector and the zero $0$ in the field of scalars. For a short history.

Immediate Theorems when Introducing Vectors

The abstract definition above allows us to take a formal introduction to vectors that is very productive.

Theorem. Every linear space $\mathbb{V}$ has a unique additive identity (denoted by $0$).

Proof. Let $u_1$ and $u_2$ be additive identities in $V$, then $v+u_1=v$ and $v+u_2=v$ for every $v\in V.$ Thus, $u_1=u_1+u_2=u_2+u_1=u_2$ as desired.

In our introduction, we are emphasizing the importance of the special elements in any vector space: the additive identity.

Theorem. Every $v\in V$ has a unique additive inverse, (denoted by $-v$).

Proof. Let $v_1$ and $v_2$ be additive inverses of $w$ in $V$, then $w+v_1=\mathbf{0}$ and $w+v_2=\mathbf{0}.$ Thus,
\begin{align} v_1 & = v_1+\mathbf{0} = v_1+(w+v_2) = (v_1+w)+v_2 =(w+v_1)+v_2\\ & =\mathbf{0}+v_2=v_2 \end{align} as desired.

Theorem. If $v\in V$, then $0\, v={0}.$

Proof. Let $v\in V$, then $v=1 v=(1+0) v= 1 v+0 v= v+0v$ which shows that $0 v$ is the additive identity of $V$, namely $0 v=\mathbf{0}.$

Theorem. If $a\in \mathbb{F}$, then $a\, 0=0.$

Proof. Let $a\in \mathbb{F}$, then $ a \mathbf{0}=a(\mathbf{0}+\mathbf{0})=a\mathbf{0}+a\mathbf{0} $ which shows that $a \mathbf{0}$ is the additive identity of $V$, namely $a \mathbf{0}=\mathbf{0}.$

Theorem. If $v\in V$, then $-(-v)=v.$

Proof. Let $v\in V$, then $ v+(-1)v=1 v+(-1) v=(1+(-1)) v=0 v= \mathbf{0} $ which shows that $(-1)v$ is the unique additive inverse of $v$ namely, $(-1)v=-v.$

Theorem. If $v\in V$, then $(-1)\, v=-v.$

Proof. Since $-v$ is the unique additive inverse of $v$, $v+(-v)=\mathbf{0}.$ Then $(-v)+v=\mathbf{0}$ shows that $v$ is the unique additive inverse of $-v$, namely, $v=-(-v)$ as desired.

Theorem. If $a\,v=0$, then $a=0$ or $v=0.$

Proof. Suppose $a\neq 0.$ If $a v =\mathbf{0}$ then $v=1 v=(a^{-1} a) v=a^{-1} (a v)=a^{-1} \mathbf{0}=\mathbf{0}.$ Otherwise $a=0$ as desired.

Let $\mathbb{V}$ be a linear space and $U$ a nonempty subset of $\mathbb{V}.$ If $U$ is a linear space with respect to the operations on $\mathbb{V}$, then $U$ is called a subspace of $\mathbb{V}.$ To fully understand this introduction to vectors, you will need to master the idea of a subspace.

Theorem. A subset $U$ of $\mathbb{V}$ is a linear subspace of $\mathbb{V}$ if and only if $U$ has the following properties:

(1) $U$ contains the zero vector of $\mathbb{V}$,

(2) $U$ is closed under the addition defined on $\mathbb{V}$, and

(3) $U$ is closed under the scalar multiplication defined on $\mathbb{V}.$

More generally, a subset $U$ of $\mathbb{V}$ is called a subspace of $\mathbb{V}$ if $U$ is also a vector space using the same addition and scalar multiplication as on $\mathbb{V}.$ Any vector space is a subspace of itself. The set containing just the $0$ vector is also a subspace of any vector space. Given any vector space with a nonzero vector $v$, the scalar multiples of $v$ is a vector subspace of $\mathbb{V}$ and is denoted by $\langle v \rangle.$ Because any linear space $\mathbb{V}$ has $\mathbb{V}$ and $0$ as subspaces, these subspaces are called the trivial subspaces of $\mathbb{V}.$ All other subspaces are called proper subspaces of $\mathbb{V}.$

Example. Give an example of a real linear space $\mathbb{V}$ and a nonempty set $S$ of $\mathbb{V}$ such that, whenever $u$ and $v$ are in $S$, $u+v$ is in $S$ but $S$ is not a subspace of $\mathbb{V}.$

Example. Give an example of a real linear space $\mathbb{V}$ and a nonempty set $S$ of $\mathbb{V}$ such that, whenever $u$ and $v$ are in $S$, $c u$ is in $S$ for every scalar $c$ but $S$ is not a subspace of $\mathbb{V}.$

Example. Show that $P_n[0,1]$ is a proper subspace of $C[0,1].$

Example. Show that $C'[0,1]$ (continuous first derivative) is a proper subspace of $C[0,1].$

Example. Show that $R[0,1]$ (Riemann integrable) is a proper subspace of $C[0,1].$

Example. Show that $D[0,1]$ (Differentiable functions) is a proper subspace of $C[0,1].$

Linear Combinations

In any good introduction to vectors is the concept of linear combinations.

Definition. A linear combination of a list of vectors $(v_1, v_2, \ldots, v_m)$ in $\mathbb{V}$ is a vector of the form $a_1 v_1 + a_2 v_2 + \cdots + a_m v_m $ where $a_1, a_2, \ldots, a_m \in k.$

Theorem. Let $U$ be a nonempty subset of a vector space $\mathbb{V}.$ Then $U$ is a subspace of $\mathbb{V}$ if and only if every linear combination of vectors in $U$ is also in $U.$

Proof. If $U$ is a subspace of $\mathbb{V}$, then $U$ is a vector space and so is closed under linear combinations by definition of vector space. Conversely, suppose every linear combination of vectors in $U$ is also in $U.$ Thus for any $a, b \in k$, $a u+b v \in U$ for every $u, v\in U.$ In particular, when $a=b=1$ then $u+v \in U$ and so $U$ is closed with respect to addition. Notice when $b=0$ and $a=-1$, then $-u\in U$ for every $u\in U$ and so $U$ is closed under inverses. Notice when $u=v$, $a=1$, and $b=-1$ then $u+(-u)=0\in U$ so $U$ contains the identity element. The rest of the axioms in the definition of a vector space hold by containment.

Definition. The intersection and union of subspaces is just the intersection and union of the subspaces as sets. The sum of subspaces $U_1, U_2, \ldots, U_m$ of a vector space $U$ is defined by $$ U_1+ U_2+\cdots +U_m = \{ u_1 + u_2+\cdots + u_m \mid u_i \in U_i \text{ for } 1\leq i \leq m \}. $$

Unions and Intersections of Subspaces

Now that we have a good introduction to vectors going, we can discuss unions and intersections of subspaces. When do these operations on subspaces generate new subspaces?

Theorem. Let $\mathbb{V}$ be a linear space over a field $k.$ The intersection of any collection of subspaces of $\mathbb{V}$ is a subspace of $\mathbb{V}.$

Proof. Let $\{U_i\, |\, i \in I\}$ be a collection of subspaces where $I$ is some indexed set. Let $a,b\in k$ and $u,v\in \cap_{i\in I} U_i.$ Since each $U_i$ is a subspace of $\mathbb{V}$, $a u +b v\in U_i$ for every $i\in I.$ Thus $a u+b v\in \cap_{i\in I} U_i$ and therefore $\cap_{i\in I} U_i$ is a subspace of $\mathbb{V}.$

For example, the $x$-axis and the $y$-axis are subspaces on $\mathbb{R}^2$, yet the union of these axis is not.

Theorem. Let $\mathbb{V}$ be a linear space over a field $k.$ The union of two subspaces of $\mathbb{V}$ is a subspace of $\mathbb{V}$ if and only if one of the subspaces is contained in the other.

Proof. Suppose $U$ and $W$ are subspaces of $\mathbb{V}$ with $U\subseteq W.$ Then $U\cup W=W$ and so $U\cup W$ is also a subspace of $\mathbb{V}.$ Conversely, suppose $U$, $W$, $U\cup W$ are subspaces of $\mathbb{V}$ and suppose $u\in U.$ If $u\in W$ then $U$ is contained in $W$ as desired. Thus we assume, $u\not\in W.$ If $w\in W$, then $u+w \in U\cup W$ and either $u+w\in U$ or $u+w\in W.$ Notice $u+w\in W$ and $w\in W$ together yield $u\in W$ which is a contradiction. Thus $u+w\in U$ and so $w\in U$ which yields $W\subseteq U$ as desired.

Theorem. Let $\mathbb{V}$ be a linear space over a field $k.$ The sum $U_{1}+ U_2+\cdots +U_{m}$ is the smallest subspace containing each of the subspaces $U_1, U_2, \ldots, U_m.$

Proof. The sum of two subspaces is a subspace since the sum of two subspaces is closed under linear combinations. Thus $U_1, U_2, \ldots, U_m$ is a subspace containing $U_i$ for each $1\leq i \leq m.$ Let $U$ be another subspace containing $U_{i}$ for each $1\leq i \leq m.$ If $u\in U_{1}+ \cdots +U_{m}$, then $u$ has the form $u=u_1+\cdots + u_m$ where each $u_i\in U_i\subseteq U.$ Since $U$ is a subspace $u\in U$ and so $U_{1}+U_{2}+ \cdots +U_{m}$ is the smallest such subspace.

Definition. If $\{v_1, \ldots, v_m\}$ is a subset of a linear space $\mathbb{V}$, then the subspace of all linear combinations of these vectors is called the subspace generated (spanned) by $\{v_1, \ldots, v_m\}.$ The spanning set of the list of vectors $(\{v_1, \ldots, v_m\})$ in $\mathbb{V}$ is denoted by $$\mathop{span}(\{v_1, \ldots, v_m\}) = \{a_1 v_1 + a_2 v_2 + \cdots + a_m v_m \mid a_1, a_2, \ldots, a_m\in \mathbb{F} \}.$$

Theorem. The span of a list of vectors in $\mathbb{V}$ is the smallest subspace of $\mathbb{V}$ containing all the vectors in the list.

Proof. Let $(v_1, v_2, \ldots, v_n)$ be a list of vectors in $\mathbb{V}$ and let $S$ denote $\mathop{span}(v_1, v_2, \ldots, v_n).$ Clearly, $S$ contains $v_i$ for each $1\leq i \leq n.$ Let $u,v \in S$ and $a,b\in k.$ Then there exists $a_1, a_2, \ldots, a_n$ in $k$ and $b_1, b_2, \ldots, b_n$ in $k$ such that $u=a_1 v_1+\cdots a_n v_n$ and $v=b_1 v_1+ \cdots b_n v_n.$ Then $$a u+b v=(a a_1 +b b_1) v_1+ \cdots +(a a_n+b b_n) v_n$$ which shows $a u+b v\in S$ since $a a_i+b b_i \in k$ for each $1\leq i \leq n.$ Thus $S$ is a subspace containing each of the $v_i.$ Let $T$ be a subspace containing $v_i$ for $1 \leq i \leq n.$ If $s\in S$, then there exists $c_1, c_2, \ldots, c_n \in k$ such that $s=c_1 v_1+\cdots + c_n v_n.$ Since $v_i\in T$ for each $i$ and $T$ is closed under linear combinations (since $T$ is a subspace), $s\in T.$ Meaning $S \subseteq T$, so indeed $S$ is the smallest subspace of $\mathbb{V}$ containing all the vectors $v_i.$

Definition. Let $\emptyset$ denote the empty set. Then $\mathop{span}(\emptyset)={0}.$

Exercises on Introduction to Vectors

Example. Let $A=\begin{matrix}1 & 1 \ 0 & 0 \end{matrix}.$ Show that $S=\{X\in M_{2\times 2} \mid AX=XA\}$ is a subspace of $M_{2\times 2}$ under the standard operations.

Example. Let $f_1=x^2+1, f_2=3x-1, f_3=2.$ Determine the subspace generated by $f_1, f_2, f_3$ in $P_4.$

Example. Let $A_1=\begin{bmatrix} 1 & 0 & 0 & 3 \\ 0 & 0 & 2 & 0 \end{bmatrix} $ and $A_2= \begin{bmatrix} 0 & 2 & 0 & 1 \\ 0 & 0 & 0 & 1 \end{bmatrix}.$ Determine the subspace generated by $A_1$ and $A_2$ in $R_{2\times 4}.$

Example. Describe $\mathop{span}(0).$

Example. Consider the subset $$S=\{x^3-2x^2+x-3, 2x^3-3x^2+2x+5, 4x^3-7x^2+4x-1, 4x^2+x-3\}$$ of $P.$ Show that $3x^3-8x^2+2x+16$ is in $\mathop{span} (S)$ by expressing it as a linear combination of the elements of $S.$

Example. Determine if the matrices $\begin{bmatrix} 2 & -1 \\ 0 & 2 \end{bmatrix}, \begin{bmatrix}-4 & 2 \\ 3 & 0 \end{bmatrix}, \begin{bmatrix} -1 & 0 \\ 2 & 1 \end{bmatrix}, \begin{bmatrix}0 & 0 \\ 0 & 3\end{bmatrix}$ span $M_{2\times 2}.$

Exercises on Introduction to Vectors

Exercise. Show that the set of solution to a homogenous system form a linear space with standard operations.

Exercise. Show that the set of vectors for which a particular linear system has a solution is a linear space.

Exercise. Let $V=\{(x,y)\mid y=mx\}$, where $m$ is a fixed real number and $x$ is an arbitrary real number. Show that $\mathbb{V}$ is a linear space.

Exercise. Let $V=\{(x,y,x)\mid ax+by+cz=0\}$ where $a, b$ and $c$ are fixed real numbers. Show that $\mathbb{V}$ is a linear space with the standard operations.

Exercise. (Matrix Space) Show that the set $M_{m\times n}$ of all $m\times n$ matrices, with ordinary addition of matrices and scalar multiplication, forms a linear space.

Exercise. (Polynomial Space) Show that the set $P(t)$ of all polynomials with real coefficients, under the ordinary operations of addition of polynomials and multiplication of a polynomial by a scalar, forms a linear space. Show that the set of all polynomials with real coefficients of degree less than or equal to $n$, under the ordinary operations of addition of polynomials and multiplication of a polynomial by a scalar, forms a linear space.

Exercise. (Function Space) Show that the set $F(x)$ of all functions that map the real numbers into itself is a linear space. Show that the set $F[a,b]$ of all functions on the interval $[a,b]$ using the standard operations is a linear space.

Exercise. (The Space of Infinite Sequences) Show that the set of all infinite sequences of real numbers is a linear space, where addition and scale multiplication are defined term by term.

Exercise. (The Space of Linear Equations) Show that the set $L_n$ of all linear equations with $n$ variables, forms a linear space.

Exercise. Let $\mathbb{V}$ be a linear space with $u\in V$ and let $a$ and $b$ be scalars. Prove that if $a u=bu$ and $u\neq 0$, then $a=b.$

David A. Smith at Dave4Math

David Smith (Dave) has a B.S. and M.S. in Mathematics and has enjoyed teaching precalculus, calculus, linear algebra, and number theory at both the junior college and university levels for over 20 years. David is the founder and CEO of Dave4Math.

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