5 Vector Spaces

A structured guide to vector spaces, subspaces, spans, independence, bases, dimension, and coordinates, with definitions, tests, examples, and a unified problem-solving workflow.

Foundations and Examples

A is an abstract setting in which vectors can be added and multiplied by scalars while obeying familiar algebraic laws. These objects need not look like geometric arrows. Common examples include:

$\mathbb{R}^n$ , the set of real $n$ -tuples;
$P_n$ , the real polynomials of degree at most $n$ ;
the set of all $m\times n$ real matrices;
real-valued functions on a fixed domain; and
solution sets of homogeneous linear differential equations.

For example, every element of $P_2$ can be written as

p(x)=a+bx+cx^2,

where $a,b,c\in\mathbb{R}$ . Adding two such polynomials or multiplying one by a scalar still produces a polynomial in $P_2$ .

The axioms ensure that there is a zero vector, every vector has an additive inverse, addition is commutative and associative, and scalar multiplication distributes over addition. The key perspective is that the nature of the objects is less important than the linear operations they support.

Takeaway: Vector-space methods apply to many kinds of mathematical objects, not only numerical tuples.

Testing Subspaces

A subset becomes a when it is itself a under the operations inherited from the larger space. A practical test is:

Confirm that the subset is nonempty.
For all $u,v$ in the subset and all scalars $c,d$ , verify that $cu+dv$ remains in the subset.

This single closure condition guarantees the zero vector and additive inverses as well as closure under addition and scalar multiplication.

Consider

W=\{(x,y,z)\in\mathbb{R}^3:x+y+z=0\}.

If $u$ and $v$ satisfy the equation, then

(cu+dv)_1+(cu+dv)_2+(cu+dv)_3 =c(u_1+u_2+u_3)+d(v_1+v_2+v_3)=0.

Thus $W$ is closed under linear combinations. Solving for $z$ gives

(x,y,z)=(x,y,-x-y)=x(1,0,-1)+y(0,1,-1),

so the set can also be described using two generating vectors.

By contrast,

\{(x,y,z)\in\mathbb{R}^3:x+y+z=1\}

is not a because it does not contain $(0,0,0)$ , and scalar multiplication does not preserve the defining equation.

Takeaway: Homogeneous linear conditions typically define subspaces; a nonzero constant on the right-hand side usually prevents a set from being a .

Building Spaces with Spans

A is formed by multiplying given vectors by scalars and adding the results. For vectors $v_1,\ldots,v_k$ , their is

\operatorname{span}\{v_1,\ldots,v_k\} =\{c_1v_1+\cdots+c_kv_k:c_1,\ldots,c_k\text{ are scalars}\}.

To determine whether a vector $w$ belongs to a , solve the coefficient equation

c_1v_1+\cdots+c_kv_k=w.

For example, let

v_1=(1,2,0),\qquad v_2=(0,1,1).

Then

a v_1+b v_2=(a,2a+b,b),

so every vector in the satisfies

x-2y+2z=0.

Indeed, $(3,8,2)=3v_1+2v_2$ , so it belongs to the . To test another vector, substitute its coordinates into the equation or solve directly for $a$ and $b$ .

The of any set is a because sums of linear combinations are still linear combinations, and scalar multiples of linear combinations are also linear combinations.

Takeaway: Spanning asks which vectors can be constructed from a given collection; membership is decided by solving a equation.

Recognizing Independence

A set is when the only way to produce the zero vector from its vectors is to use all zero coefficients:

c_1v_1+\cdots+c_kv_k=0 \quad\Longrightarrow\quad c_1=\cdots=c_k=0.

If a nonzero coefficient choice produces the zero vector, the set is dependent. Dependence means that at least one vector is redundant and can be written as a of the others.

For example,

v_1=(1,2,3),\qquad v_2=(2,4,6)

are dependent because

2v_1-v_2=0.

The standard vectors $e_1=(1,0)$ and $e_2=(0,1)$ are independent: if

c_1e_1+c_2e_2=(0,0),

then both coefficients must be zero.

Useful consequences include:

Any set containing the zero vector is dependent.
Any subset of an independent set is independent.
Any set with more than $n$ vectors in $\mathbb{R}^n$ is dependent.
In a matrix whose columns are the vectors, independence is equivalent to having a pivot in every column after row reduction.

Takeaway: Independence measures whether a collection contains redundancy.

Constructing Bases

A combines the two central requirements: it spans the space and is . Therefore, a generates every vector without redundancy, and every vector has exactly one representation as a of the vectors.

The standard of $\mathbb{R}^3$ is

\mathcal{E}=\{(1,0,0),(0,1,0),(0,0,1)\}.

Every vector satisfies

(x,y,z)=x(1,0,0)+y(0,1,0)+z(0,0,1).

The polynomial space $P_2$ has $\{1,x,x^2\}$ , because every polynomial $a+bx+cx^2$ is generated by these elements and the only polynomial identity

a+bx+cx^2=0

for every $x$ has $a=b=c=0$ .

To extract a from a spanning set, remove redundant vectors until the remaining vectors are independent. In a matrix calculation, row reduction identifies pivot columns. The corresponding columns of the original matrix, not merely the columns of the row-reduced matrix, form a for the column space.

For

A=\begin{bmatrix}1&0&1\\0&1&1\\1&1&2\end{bmatrix},

the third column is the sum of the first two, so the first and second columns form a for the column space.

Takeaway: A is an efficient generating system: enough vectors to , but no redundant vectors.

Measuring

counts the number of vectors in a . The fact that every of a finite-dimensional has the same number of vectors makes well-defined.

Important examples are

\dim(\mathbb{R}^n)=n, \qquad \dim(P_n)=n+1, \qquad \dim(M_{m\times n})=mn.

For instance, the four matrix units form a of $M_{2\times 2}$ , so

\dim(M_{2\times 2})=4.

If $V$ has $n$ , then:

every independent set in $V$ has at most $n$ vectors;
every spanning set for $V$ has at least $n$ vectors;
any set of exactly $n$ independent vectors is a ; and
any set of exactly $n$ vectors that spans $V$ is a .

The zero $\{0\}$ has $0$ , with the empty set as its .

Takeaway: Once the is known, the number of vectors in a candidate set can quickly reveal whether it could be a .

Coordinates in a Chosen

Let $\mathcal{B}=\{v_1,v_2,\ldots,v_n\}$ be an ordered . Every vector $v$ has a unique expression

v=c_1v_1+c_2v_2+\cdots+c_nv_n.

Its relative to $\mathcal{B}$ is

[v]_{\mathcal{B}}= \begin{bmatrix} c_1\\c_2\\\vdots\\c_n \end{bmatrix}.

For the

\mathcal{B}=\{(1,1),(1,-1)\}

and the vector $w=(5,1)$ , solve

c_1(1,1)+c_2(1,-1)=(5,1).

The resulting equations are

c_1+c_2=5, \qquad c_1-c_2=1.

Therefore, $c_1=3$ and $c_2=2$ , so

[w]_{\mathcal{B}}=\begin{bmatrix}3\\2\end{bmatrix}.

The is not the original vector; it is the list of coefficients used with the chosen ordered . If the vectors are placed as columns of

P_{\mathcal{B}}=\begin{bmatrix}v_1&v_2&\cdots&v_n\end{bmatrix},

then

v=P_{\mathcal{B}}[v]_{\mathcal{B}}.

For a of $\mathbb{R}^n$ , the matrix is invertible and

[v]_{\mathcal{B}}=P_{\mathcal{B}}^{-1}v.

Takeaway: To find coordinates, express the vector as a of the ordered and record the coefficients in the same order.

An Integrated Problem-Solving Workflow

The concepts fit together as a systematic workflow for analyzing a or a generated set:

Describe the vectors using equations or parameters.
Rewrite a general vector as a of parameter vectors.
Use those parameter vectors as candidate vectors.
Check that they are .
Count them to obtain the .
Solve a coefficient equation to find coordinates relative to the resulting .

For example, consider

W=\{(x,y,z,w)\in\mathbb{R}^4:x+y+z+w=0\}.

Solving for $w$ gives $w=-x-y-z$ , so

(x,y,z,w) =x(1,0,0,-1)+y(0,1,0,-1)+z(0,0,1,-1).

Thus

W=\operatorname{span}\{(1,0,0,-1),(0,1,0,-1),(0,0,1,-1)\}.

The three vectors are independent because the first three coordinates force all coefficients to be zero in any equal to the zero vector. They therefore form a , and

\dim W=3.

For $u=(2,-1,4,-5)$ , the coordinates relative to this are immediately visible:

[u]_{\mathcal{B}}=\begin{bmatrix}2\\-1\\4\end{bmatrix}.

Final checklist: distinguish the space from its subsets, test closure for subspaces, solve coefficient equations for spans and coordinates, test the zero relation for independence, and use spanning plus independence to establish a .