The Central Idea
A vector space is a setting in which two basic operations make sense:
- adding two vectors;
- multiplying a vector by a scalar.
The important requirement is that these operations stay inside the space and obey a consistent collection of algebraic rules.
In the familiar plane,
\[ (x_1,y_1)+(x_2,y_2)=(x_1+x_2,\,y_1+y_2) \]and
\[ c(x,y)=(cx,\,cy). \]But linear algebra is broader than arrows in the plane. Matrices, polynomials, functions, and other objects can also form vector spaces when they support the same linear structure.
Why Vector Spaces Matter
Many apparently different mathematical problems have the same underlying structure. A force vector, a polynomial, a column of data, a matrix, and a function may look unrelated, yet each can often be added and scaled in the same algebraic way.
Vector-space language lets us develop one theory that applies across all of these settings.
This is one of the central efficiencies of linear algebra: instead of solving each type of object from scratch, we identify the shared structure and reuse the same tools.
What Is a Vector?
In elementary geometry, a vector is often introduced as a directed arrow with magnitude and direction. That picture is useful, but linear algebra uses a more general meaning.
A vector is an element of a vector space.
Depending on the space, a vector might be:
- an ordered list of numbers;
- a geometric displacement;
- a matrix;
- a polynomial;
- a function;
- a signal or data record;
- another object with suitable addition and scalar multiplication.
The word vector therefore describes a structural role, not only a geometric shape.
Vectors and Scalars
A scalar is a number used to scale a vector.
If
\[ \mathbf{v}=(2,-1) \]and the scalar is
\[ c=3, \]then
\[ 3\mathbf{v}=3(2,-1)=(6,-3). \]Unless otherwise stated, this lesson uses real numbers as scalars. Such a space is called a real vector space.
Vector spaces can also be built over other scalar systems, especially the complex numbers.
The Familiar Example: \(\mathbb{R}^2\)
The symbol
\[ \mathbb{R}^2 \]denotes the set of all ordered pairs of real numbers:
\[ \mathbb{R}^2=\{(x,y):x,y\in\mathbb{R}\}. \]Vectors are added component by component:
\[ (a,b)+(c,d)=(a+c,\,b+d), \]and scaled component by component:
\[ k(a,b)=(ka,\,kb). \]These operations satisfy all of the vector-space rules, so \(\mathbb{R}^2\) is a vector space over \(\mathbb{R}\).
Three Dimensions: \(\mathbb{R}^3\)
The set
\[ \mathbb{R}^3=\{(x,y,z):x,y,z\in\mathbb{R}\} \]contains all ordered triples of real numbers.
Addition is
\[ (a,b,c)+(d,e,f)=(a+d,\,b+e,\,c+f), \]and scalar multiplication is
\[ k(a,b,c)=(ka,\,kb,\,kc). \]This is the standard algebraic model for vectors in three-dimensional space.
Higher Dimensions: \(\mathbb{R}^n\)
The pattern continues beyond dimensions we can directly visualize.
\[ \mathbb{R}^n =\{(x_1,x_2,\ldots,x_n):x_i\in\mathbb{R}\}. \]For vectors
\[ \mathbf{u}=(u_1,u_2,\ldots,u_n) \]and
\[ \mathbf{v}=(v_1,v_2,\ldots,v_n), \]addition is defined by
\[ \mathbf{u}+\mathbf{v} =(u_1+v_1,\ldots,u_n+v_n), \]while scalar multiplication is
\[ c\mathbf{u}=(cu_1,\ldots,cu_n). \]Higher-dimensional vector spaces let us represent systems with many interacting quantities without requiring a geometric picture of every dimension.
What Is a Vector Space?
A vector space over \(\mathbb{R}\) is a set \(V\) equipped with vector addition and scalar multiplication that obey the standard vector-space axioms.
If \(\mathbf{u},\mathbf{v},\mathbf{w}\in V\) and \(a,b\in\mathbb{R}\), the rules require the operations to behave like the familiar operations in \(\mathbb{R}^n\).
Closure Under Addition
If two vectors are in the space, their sum must also be in the space.
\[ \mathbf{u},\mathbf{v}\in V \quad\Longrightarrow\quad \mathbf{u}+\mathbf{v}\in V. \]This property is called closure under addition.
A set that fails this requirement cannot be a vector space under the stated operations.
Closure Under Scalar Multiplication
If a vector is in the space, multiplying it by any allowed scalar must keep the result in the space.
\[ \mathbf{v}\in V,\ c\in\mathbb{R} \quad\Longrightarrow\quad c\mathbf{v}\in V. \]This property is called closure under scalar multiplication.
Scaling by negative numbers, fractions, zero, or very large values must all remain valid.
The Vector-Space Axioms
A real vector space satisfies the following rules for all \(\mathbf{u},\mathbf{v},\mathbf{w}\in V\) and \(a,b\in\mathbb{R}\):
- \(\mathbf{u}+\mathbf{v}\in V\).
- \(\mathbf{u}+\mathbf{v}=\mathbf{v}+\mathbf{u}\).
- \((\mathbf{u}+\mathbf{v})+\mathbf{w}=\mathbf{u}+(\mathbf{v}+\mathbf{w})\).
- There is a zero vector \(\mathbf{0}\) such that \(\mathbf{v}+\mathbf{0}=\mathbf{v}\).
- Every \(\mathbf{v}\) has an additive inverse \(-\mathbf{v}\).
- \(a\mathbf{v}\in V\).
- \(a(\mathbf{u}+\mathbf{v})=a\mathbf{u}+a\mathbf{v}\).
- \((a+b)\mathbf{v}=a\mathbf{v}+b\mathbf{v}\).
- \(a(b\mathbf{v})=(ab)\mathbf{v}\).
- \(1\mathbf{v}=\mathbf{v}\).
The first and sixth items state closure. The remaining items ensure that the operations interact coherently.
The Zero Vector
Every vector space contains a unique zero vector, written \(\mathbf{0}\), satisfying
\[ \mathbf{v}+\mathbf{0}=\mathbf{v}. \]In \(\mathbb{R}^3\),
\[ \mathbf{0}=(0,0,0). \]In a polynomial vector space, the zero vector is the zero polynomial.
In a function vector space, the zero vector is the function whose value is zero at every input.
The notation is the same because the structural role is the same.
Additive Inverses
Every vector \(\mathbf{v}\) has an additive inverse \(-\mathbf{v}\) such that
\[ \mathbf{v}+(-\mathbf{v})=\mathbf{0}. \]For example, if
\[ \mathbf{v}=(4,-2,7), \]then
\[ -\mathbf{v}=(-4,2,-7). \]The existence of additive inverses is one reason a vector space cannot be restricted to only nonnegative coordinate vectors.
Two Useful Consequences of the Axioms
The axioms imply several familiar facts. Two especially useful ones are
\[ 0\mathbf{v}=\mathbf{0} \]and
\[ (-1)\mathbf{v}=-\mathbf{v}. \]These are not separate assumptions. They follow from the vector-space rules.
Polynomials Can Form a Vector Space
Let \(P_2\) denote the set of all real polynomials of degree at most two:
\[ P_2=\{a+bx+cx^2:a,b,c\in\mathbb{R}\}. \]Two such polynomials can be added, and the result still has degree at most two.
For example,
\[ (1+2x-x^2)+(3-x+4x^2)=4+x+3x^2. \]Scalar multiplication also stays inside the set:
\[ 5(1+2x-x^2)=5+10x-5x^2. \]With the usual polynomial operations, \(P_2\) is a vector space.
Functions Can Form a Vector Space
Consider all real-valued functions defined on the real numbers.
If \(f\) and \(g\) are two such functions, define
\[ (f+g)(x)=f(x)+g(x) \]and
\[ (cf)(x)=c f(x). \]These operations satisfy the vector-space axioms, so functions themselves can behave as vectors.
This perspective becomes important in differential equations, Fourier analysis, approximation, and many models of continuous systems.
Matrices Can Form a Vector Space
Let \(M_{2\times2}\) be the set of all real \(2\times2\) matrices.
\[ A= \begin{bmatrix} a&b\\ c&d \end{bmatrix}. \]Matrix addition and scalar multiplication are performed entry by entry.
The sum of two \(2\times2\) matrices is another \(2\times2\) matrix, and a scalar multiple of a \(2\times2\) matrix is another \(2\times2\) matrix.
Therefore, \(M_{2\times2}\) is a vector space over \(\mathbb{R}\).
The Same Structure in Different Objects
Compare these three objects:
\[ (a,b,c), \] \[ a+bx+cx^2, \]and
\[ \begin{bmatrix} a\\b\\c \end{bmatrix}. \]They are not literally the same object, but each is determined by three real coefficients and supports componentwise linear operations.
Linear algebra focuses on this reusable structure.
Nonexample: The First Quadrant
Consider the set
\[ S=\{(x,y):x\ge0,\ y\ge0\}. \]This set is closed under addition, but it is not closed under multiplication by every real scalar.
For example,
\[ (1,2)\in S, \]but
\[ -1(1,2)=(-1,-2)\notin S. \]Therefore, the first quadrant is not a vector space under the usual operations.
Nonexample: A Shifted Line
Consider the line
\[ y=2x+1. \]The point \((0,1)\) lies on the line, but the zero vector \((0,0)\) does not.
Since every vector space must contain the zero vector, this line is not a vector space under the usual operations.
Geometrically, a linear subspace must pass through the origin.
What Is a Subspace?
A subspace of a vector space \(V\) is a subset \(W\subseteq V\) that is itself a vector space under the same addition and scalar multiplication already used in \(V\).
A subspace therefore inherits its operations from the larger space.
The key question is not whether we can invent new operations that work. The question is whether the subset remains a vector space using the operations already present in \(V\).
The Subspace Test
Suppose \(W\) is a subset of a known vector space \(V\). To prove that \(W\) is a subspace, it is enough to verify three conditions:
- \(\mathbf{0}\in W\);
- if \(\mathbf{u},\mathbf{v}\in W\), then \(\mathbf{u}+\mathbf{v}\in W\);
- if \(\mathbf{u}\in W\) and \(c\in\mathbb{R}\), then \(c\mathbf{u}\in W\).
The remaining vector-space axioms are inherited automatically from the larger vector space.
Why the Subspace Test Is Efficient
Rechecking all ten vector-space axioms every time would be unnecessary.
If \(V\) is already known to be a vector space, properties such as commutativity, associativity, and distributivity already hold for every vector in \(V\). A subset only needs to preserve the required vectors under the operations.
This reduces most subspace problems to closure and the zero vector.
A Compact Subspace Test
There is a useful one-step version of the subspace test.
A nonempty subset \(W\) of \(V\) is a subspace if, for every \(\mathbf{u},\mathbf{v}\in W\) and every scalars \(a,b\),
\[ a\mathbf{u}+b\mathbf{v}\in W. \]In other words, a subspace is closed under every linear combination of two of its vectors.
Linear combinations will be developed systematically in the next lesson.
The Two Subspaces Every Vector Space Has
Every vector space \(V\) has at least two subspaces.
The first is the zero subspace:
\[ \{\mathbf{0}\}. \]The second is the entire space itself:
\[ V. \]These are sometimes called the trivial subspaces. Any other subspace is called a proper nonzero subspace.
Example: A Line Through the Origin
Consider
\[ W=\{(x,y)\in\mathbb{R}^2:y=2x\}. \]Every vector in \(W\) can be written
\[ (t,2t)=t(1,2). \]The zero vector occurs when \(t=0\).
Adding two vectors in the set gives
\[ (s,2s)+(t,2t)=(s+t,2(s+t)), \]which is still in the set.
Scaling gives
\[ c(t,2t)=(ct,2ct), \]which is also in the set. Therefore, \(W\) is a subspace of \(\mathbb{R}^2\).
Contrast: A Line Not Through the Origin
Now consider
\[ S=\{(x,y)\in\mathbb{R}^2:y=2x+3\}. \]The zero vector does not satisfy the defining equation because
\[ 0\ne2(0)+3. \]Therefore,
\[ \mathbf{0}\notin S, \]so \(S\) is not a subspace.
This distinction separates linear sets from translated or affine sets.
Example: A Plane Through the Origin
Consider the subset of \(\mathbb{R}^3\) defined by
\[ W=\{(x,y,z):x+2y-z=0\}. \]The zero vector satisfies the equation.
If \(\mathbf{u}=(x_1,y_1,z_1)\) and \(\mathbf{v}=(x_2,y_2,z_2)\) both satisfy the equation, then
\[ x_1+2y_1-z_1=0 \]and
\[ x_2+2y_2-z_2=0. \]Adding the equations gives
\[ (x_1+x_2)+2(y_1+y_2)-(z_1+z_2)=0, \]so the sum remains in \(W\).
Multiplying the defining equation by a scalar \(c\) shows that scalar multiples also remain in \(W\). Therefore, \(W\) is a subspace.
Homogeneous Linear Equations Naturally Produce Subspaces
An equation such as
\[ a_1x_1+a_2x_2+\cdots+a_nx_n=0 \]is called a homogeneous linear equation.
Its solution set contains the zero vector and is preserved by vector addition and scalar multiplication.
More generally, the solution set of a homogeneous linear system
\[ A\mathbf{x}=\mathbf{0} \]is a subspace. Later, this subspace will be called the null space of \(A\).
Nonhomogeneous Equations Usually Do Not Produce Subspaces
Compare
\[ x+2y-z=0 \]with
\[ x+2y-z=5. \]The first equation is homogeneous and its solution set is a subspace.
The second equation does not contain the zero vector because
\[ 0+2(0)-0\ne5. \]Its solution set is an affine plane, not a vector subspace.
Example: A Coordinate Restriction
Let
\[ W=\{(x,y,z)\in\mathbb{R}^3:z=0\}. \]This is the \(xy\)-plane inside \(\mathbb{R}^3\).
If
\[ \mathbf{u}=(a,b,0) \]and
\[ \mathbf{v}=(c,d,0), \]then
\[ \mathbf{u}+\mathbf{v}=(a+c,b+d,0) \]and
\[ k\mathbf{u}=(ka,kb,0). \]Therefore, \(W\) is a subspace of \(\mathbb{R}^3\).
Example: A Polynomial Subspace
Inside \(P_3\), consider the polynomials with zero constant term:
\[ W=\{ax+bx^2+cx^3:a,b,c\in\mathbb{R}\}. \]The zero polynomial is in \(W\).
Adding two polynomials with zero constant term produces another polynomial with zero constant term. Scaling such a polynomial also preserves the zero constant term.
Therefore, \(W\) is a subspace of \(P_3\).
Nonexample: Polynomials with Constant Term One
Consider
\[ S=\{1+ax+bx^2:a,b\in\mathbb{R}\}. \]Every polynomial in this set has constant term \(1\).
The zero polynomial is not in \(S\), so the set cannot be a subspace.
It also fails closure under scalar multiplication. For example,
\[ 2(1+x)=2+2x, \]whose constant term is \(2\), not \(1\).
Example: Symmetric Matrices
A square matrix \(A\) is symmetric when
\[ A^T=A. \]The set of all symmetric \(n\times n\) real matrices is a subspace of \(M_{n\times n}\).
If \(A^T=A\) and \(B^T=B\), then
\[ (A+B)^T=A^T+B^T=A+B. \]Also,
\[ (cA)^T=cA^T=cA. \]The zero matrix is symmetric, so all three subspace conditions hold.
Example: Functions Vanishing at a Point
Let \(V\) be a vector space of real-valued functions, and define
\[ W=\{f:f(0)=0\}. \]The zero function belongs to \(W\).
If \(f(0)=0\) and \(g(0)=0\), then
\[ (f+g)(0)=f(0)+g(0)=0. \]For any scalar \(c\),
\[ (cf)(0)=c f(0)=0. \]Therefore, the functions that vanish at zero form a subspace.
Nonexample: Functions with \(f(0)=1\)
Consider
\[ S=\{f:f(0)=1\}. \]The zero function is not in this set.
Also, if \(f(0)=1\), then
\[ (2f)(0)=2, \]so scalar multiplication leaves the set.
Therefore, \(S\) is not a subspace.
The Zero-Vector Check Is a Fast Filter
When testing whether a set is a subspace, check the zero vector first.
If the zero vector is missing, the problem is finished immediately: the set is not a subspace.
This quickly eliminates many translated lines, translated planes, fixed nonzero-value function sets, and other affine constraints.
How to Disprove Closure
To show that a set is not closed under an operation, one counterexample is enough.
Consider
\[ S=\{(x,y)\in\mathbb{R}^2:xy=0\}. \]This set contains vectors on the two coordinate axes.
Both
\[ (1,0)\in S \]and
\[ (0,1)\in S, \]but
\[ (1,0)+(0,1)=(1,1), \]and
\[ 1\cdot1\ne0. \]Therefore, \((1,1)\notin S\), so \(S\) is not a subspace.
Linear Constraints versus Nonlinear Constraints
Subspaces frequently arise from homogeneous linear constraints such as
\[ x-3y+2z=0. \]Nonlinear constraints such as
\[ x^2+y^2=1 \]usually do not define subspaces.
The unit circle, for example, does not contain the zero vector and is not closed under scalar multiplication.
Linear algebra is fundamentally concerned with structures preserved by linear combinations.
Subspaces of \(\mathbb{R}^2\)
Geometrically, the subspaces of \(\mathbb{R}^2\) have only three possible forms:
- the zero vector \(\{(0,0)\}\);
- a line through the origin;
- all of \(\mathbb{R}^2\).
A line not passing through the origin is not a subspace.
Subspaces of \(\mathbb{R}^3\)
The geometric possibilities in \(\mathbb{R}^3\) are:
- the zero vector;
- a line through the origin;
- a plane through the origin;
- all of \(\mathbb{R}^3\).
This geometric classification foreshadows the idea of dimension, which will later measure the number of independent directions available inside a vector space.
A Parametric View of a Subspace
Consider the plane
\[ x+y-z=0. \]Solve for \(z\):
\[ z=x+y. \]Let
\[ x=s,\qquad y=t. \]Then every vector in the plane has the form
\[ (s,t,s+t). \]Rewrite this as
\[ (s,t,s+t) =s(1,0,1)+t(0,1,1). \]This representation exposes the linear structure of the plane. The next lesson will formalize this idea using linear combinations and span.
Vector Spaces Can Sit Inside Other Vector Spaces
A vector space can contain a hierarchy of smaller vector spaces.
For example, inside \(\mathbb{R}^3\), a plane through the origin may contain many lines through the origin.
Symbolically, we might have
\[ L\subseteq P\subseteq\mathbb{R}^3, \]where \(L\) is a line subspace and \(P\) is a plane subspace.
This nested structure becomes important when comparing solution sets, invariant spaces, null spaces, and spaces generated by data.
Intersections of Subspaces
If \(U\) and \(W\) are subspaces of the same vector space \(V\), then their intersection
\[ U\cap W \]is also a subspace.
A vector in the intersection belongs to both subspaces. Since each subspace contains the zero vector and is closed under addition and scalar multiplication, the intersection inherits those properties.
Geometrically, two planes through the origin in \(\mathbb{R}^3\) may intersect in a line through the origin.
Unions of Subspaces Are Usually Not Subspaces
The union
\[ U\cup W \]of two subspaces is generally not a subspace.
For example, let \(U\) be the \(x\)-axis and \(W\) the \(y\)-axis in \(\mathbb{R}^2\).
Then
\[ (1,0)\in U\cup W \]and
\[ (0,1)\in U\cup W, \]but their sum
\[ (1,1) \]lies on neither axis. Thus the union is not closed under addition.
Vector Spaces as Models of Multidimensional Systems
A multidimensional system can often be represented by a state vector
\[ \mathbf{x}=(x_1,x_2,\ldots,x_n), \]where each coordinate records one quantity in the system.
Examples include:
- positions and velocities in mechanics;
- inventories across many facilities;
- signals sampled at many times;
- economic quantities across sectors;
- features used in statistical or machine-learning models;
- flows across a network.
Vector spaces provide the ambient structure in which these states can be combined, transformed, constrained, and analyzed.
What Makes the Structure Linear?
The word linear reflects preservation of addition and scaling.
If a system allows us to combine states using expressions such as
\[ a\mathbf{u}+b\mathbf{v}, \]and those combinations remain inside the space under study, linear methods become available.
This does not mean every real-world system is perfectly linear. Rather, linear structure often provides a fundamental model, approximation, or local description.
Linear versus Affine Sets
A vector subspace must contain the origin.
An affine set is often a translated copy of a subspace.
For example,
\[ y=2x \]is a line subspace of \(\mathbb{R}^2\), while
\[ y=2x+3 \]is a parallel affine line obtained by shifting the first line upward.
The geometric shape is similar, but the algebraic structure is different.
Common Error: Forgetting the Origin Requirement
Students often see a straight line or plane and assume it must be a subspace.
A straight geometric shape is not enough.
The line
\[ y=4x-7 \]is linear-looking geometrically, but it does not contain \((0,0)\).
Therefore, it is not a vector subspace of \(\mathbb{R}^2\).
Common Error: Checking Only One Example
Showing that one particular pair of vectors adds correctly does not prove closure.
A proof must begin with arbitrary vectors in the set and show that their sum remains in the set.
Counterexamples work differently: one failed case is enough to disprove closure.
Common Error: Changing the Operations
To test whether \(W\) is a subspace of \(V\), use the same vector addition and scalar multiplication that \(V\) uses.
A subset cannot be declared a subspace by inventing a different addition or scaling rule just for that subset.
Subspaces inherit their linear operations from the ambient vector space.
Common Error: Confusing Algebraic Closure with Geometric Closedness
In linear algebra, saying that a set is closed under addition means that adding allowed vectors produces another allowed vector.
This is different from the topological meaning of a closed set.
The two uses of the word closed refer to different mathematical ideas.
A Reliable Procedure for Testing a Subspace
- Identify the ambient vector space and its operations.
- Check whether the zero vector belongs to the subset.
- Take arbitrary vectors \(\mathbf{u}\) and \(\mathbf{v}\) in the subset.
- Show that \(\mathbf{u}+\mathbf{v}\) still satisfies the defining conditions.
- Take an arbitrary scalar \(c\).
- Show that \(c\mathbf{u}\) still satisfies the defining conditions.
- Conclude that the subset is a subspace.
If any condition fails, provide the failed condition or a counterexample and conclude that the set is not a subspace.
Example: Full Subspace Test
Determine whether
\[ W=\{(x,y,z)\in\mathbb{R}^3:x-2y+z=0\} \]is a subspace of \(\mathbb{R}^3\).
Step 1: Zero Vector
\[ 0-2(0)+0=0, \]so \(\mathbf{0}\in W\).
Step 2: Addition
Let
\[ \mathbf{u}=(x_1,y_1,z_1), \qquad \mathbf{v}=(x_2,y_2,z_2) \]be in \(W\). Then
\[ x_1-2y_1+z_1=0 \]and
\[ x_2-2y_2+z_2=0. \]Add:
\[ (x_1+x_2)-2(y_1+y_2)+(z_1+z_2)=0. \]Therefore, \(\mathbf{u}+\mathbf{v}\in W\).
Step 3: Scalar Multiplication
For any \(c\in\mathbb{R}\),
\[ c(x_1-2y_1+z_1)=0. \]Thus,
\[ cx_1-2cy_1+cz_1=0, \]so \(c\mathbf{u}\in W\).
All three conditions hold. Therefore,
\[ \boxed{W\text{ is a subspace of }\mathbb{R}^3.} \]Example: A Set That Fails the Subspace Test
Determine whether
\[ S=\{(x,y)\in\mathbb{R}^2:x+y=1\} \]is a subspace of \(\mathbb{R}^2\).
Test the zero vector:
\[ 0+0=0\ne1. \]Therefore,
\[ \mathbf{0}\notin S. \]No further work is required.
\[ \boxed{S\text{ is not a subspace of }\mathbb{R}^2.} \]Recognition Patterns
Certain forms should immediately suggest a possible subspace.
- homogeneous linear equations;
- systems of the form \(A\mathbf{x}=\mathbf{0}\);
- coordinate restrictions equal to zero;
- polynomials missing selected coefficients;
- functions satisfying homogeneous linear conditions;
- matrices satisfying homogeneous linear entry conditions.
Conditions involving a fixed nonzero constant often signal an affine set rather than a subspace.
Why the Abstract Definition Is Worth Learning
It may seem easier to define vectors only as coordinate lists. That would work for many elementary calculations, but it would hide the full power of linear algebra.
The abstract definition lets one theorem apply to many spaces at once.
A statement proved for vector spaces can often be used immediately for coordinate vectors, matrices, polynomials, functions, and other linear objects without proving the same idea separately in each setting.
Coordinates Are a Representation, Not the Whole Idea
Coordinates describe a vector relative to some chosen reference structure.
In \(\mathbb{R}^3\), we naturally write
\[ (x,y,z). \]In a polynomial space, the polynomial
\[ a+bx+cx^2 \]can also be associated with the coefficient vector
\[ (a,b,c). \]Later, bases will make this relationship precise and show how one abstract vector can receive different coordinate descriptions.
Connection to Systems of Equations
Linear systems can be written in matrix form as
\[ A\mathbf{x}=\mathbf{b}. \]When
\[ \mathbf{b}=\mathbf{0}, \]the system is homogeneous and its solution set forms a subspace.
When \(\mathbf{b}\ne\mathbf{0}\), a consistent solution set is generally a translated version of a subspace.
This distinction will become central when systems and elimination are revisited from the linear-algebra perspective.
Practice 1: Vector Addition
Let
\[ \mathbf{u}=(2,-1,4), \qquad \mathbf{v}=(-3,5,2). \]Find \(\mathbf{u}+\mathbf{v}\).
Practice 2: Scalar Multiplication
Let
\[ \mathbf{v}=(3,-2,1). \]Find \(-4\mathbf{v}\).
Practice 3: Zero Vector Test
Determine whether the set
\[ S=\{(x,y)\in\mathbb{R}^2:x-y=5\} \]can be a subspace of \(\mathbb{R}^2\).
Practice 4: Line Through the Origin
Determine whether
\[ W=\{(x,y)\in\mathbb{R}^2:y=-3x\} \]is a subspace of \(\mathbb{R}^2\).
Practice 5: Plane Constraint
Determine whether
\[ W=\{(x,y,z)\in\mathbb{R}^3:2x-y+4z=0\} \]is a subspace of \(\mathbb{R}^3\).
Practice 6: Nonhomogeneous Plane
Determine whether
\[ S=\{(x,y,z)\in\mathbb{R}^3:2x-y+4z=7\} \]is a subspace of \(\mathbb{R}^3\).
Practice 7: Polynomial Subspace
In \(P_3\), determine whether
\[ W=\{ax+bx^2+cx^3:a,b,c\in\mathbb{R}\} \]is a subspace.
Practice 8: Polynomial Nonexample
In \(P_2\), determine whether
\[ S=\{1+ax+bx^2:a,b\in\mathbb{R}\} \]is a subspace.
Practice 9: Closure Counterexample
Consider
\[ S=\{(x,y)\in\mathbb{R}^2:xy=0\}. \]Give two vectors in \(S\) whose sum is not in \(S\).
Practice 10: Parametric Description
Rewrite every vector satisfying
\[ x-y+z=0 \]in the form
\[ s\mathbf{u}+t\mathbf{v}. \]Practice 11: Function Subspace
Let \(V\) be the vector space of all real-valued functions. Determine whether
\[ W=\{f:f(2)=0\} \]is a subspace of \(V\).
Practice 12: Matrix Subspace
Let \(M_{2\times2}\) be the vector space of all real \(2\times2\) matrices. Determine whether the set
\[ W=\left\{ \begin{bmatrix} a&b\\ c&d \end{bmatrix} :a+d=0 \right\} \]is a subspace.
Practice Answers
Practice 1
\[ \mathbf{u}+\mathbf{v} =(2-3,-1+5,4+2) =\boxed{(-1,4,6)}. \]Practice 2
\[ -4(3,-2,1)=\boxed{(-12,8,-4)}. \]Practice 3
The zero vector does not satisfy the equation:
\[ 0-0\ne5. \]Therefore, \(S\) is not a subspace.
Practice 4
Every vector in the set has the form
\[ (t,-3t)=t(1,-3). \]The set contains the zero vector and is closed under addition and scalar multiplication. Therefore,
\[ \boxed{W\text{ is a subspace of }\mathbb{R}^2.} \]Practice 5
The equation is homogeneous. If two vectors satisfy \(2x-y+4z=0\), their sum also satisfies it, and multiplying the equation by any scalar preserves the equality to zero. Therefore,
\[ \boxed{W\text{ is a subspace of }\mathbb{R}^3.} \]Practice 6
The zero vector gives
\[ 2(0)-0+4(0)=0\ne7. \]Therefore, \(S\) is not a subspace.
Practice 7
The zero polynomial is included. Adding two polynomials with zero constant term preserves the zero constant term, and scalar multiplication does the same. Therefore, \(W\) is a subspace of \(P_3\).
Practice 8
The zero polynomial does not have constant term \(1\). Therefore, \(S\) is not a subspace.
Practice 9
One choice is
\[ (1,0),\qquad(0,1). \]Both satisfy \(xy=0\), but
\[ (1,0)+(0,1)=(1,1), \]and \(1\cdot1\ne0\). Therefore, the set is not closed under addition.
Practice 10
From
\[ x-y+z=0, \]solve for \(x\):
\[ x=y-z. \]Let \(y=s\) and \(z=t\). Then
\[ (x,y,z)=(s-t,s,t). \]Therefore,
\[ (s-t,s,t) =s(1,1,0)+t(-1,0,1). \]So one valid representation is
\[ \boxed{s(1,1,0)+t(-1,0,1)}. \]Practice 11
The zero function satisfies \(f(2)=0\). If \(f(2)=g(2)=0\), then
\[ (f+g)(2)=0 \]and for any scalar \(c\),
\[ (cf)(2)=0. \]Therefore, \(W\) is a subspace.
Practice 12
The condition \(a+d=0\) is homogeneous. The zero matrix satisfies it. Adding two matrices whose diagonal entries sum to zero preserves the condition, and scalar multiplication preserves it as well. Therefore,
\[ \boxed{W\text{ is a subspace of }M_{2\times2}.} \]What to Remember
A vector space is a collection of objects that supports consistent vector addition and scalar multiplication.
- Vectors are not limited to geometric arrows.
- \(\mathbb{R}^n\), polynomial spaces, matrix spaces, and function spaces can all be vector spaces.
- Every vector space contains a zero vector and additive inverses.
- A subspace is a subset that remains a vector space under the inherited operations.
- A subspace must contain the zero vector.
- A subspace must be closed under vector addition.
- A subspace must be closed under scalar multiplication.
- Homogeneous linear equations naturally produce subspaces.
- Translated lines and planes generally produce affine sets rather than subspaces.
The core habit is to ask whether linear combinations remain inside the set. That single question connects vector spaces, subspaces, systems, and the rest of linear algebra.
Where We Go Next
Vector spaces tell us which objects can participate in linear structure. The next question is how vectors can be built from other vectors.
The next lesson, Linear Combinations, Span, and Independence, develops the language for combining vectors, describing the sets they generate, and determining whether some vectors contribute genuinely new directions or merely repeat information already present.