The Central Idea
The previous lesson separated two important properties:
- spanning means a collection can generate the whole space;
- independence means the collection contains no redundant vector.
A basis has both properties at once.
\[ \boxed{\text{basis}=\text{spanning}+\text{linear independence}} \]A basis therefore gives an efficient set of building directions for the entire space.
Definition of a Basis
Let \(V\) be a vector space. A collection
\[ B=\{v_1,v_2,\ldots,v_n\} \]is a basis for \(V\) if:
- \(B\) spans \(V\); and
- \(B\) is linearly independent.
Both conditions are necessary.
Why Both Conditions Matter
A spanning set may contain unnecessary vectors.
An independent set may fail to generate the whole space.
A basis sits exactly between these two failures:
\[ \boxed{ \text{enough vectors to span} \quad+\quad \text{no redundant vectors} }. \]The Standard Basis of \(\mathbb{R}^2\)
The familiar coordinate vectors
\[ e_1= \begin{bmatrix} 1\\ 0 \end{bmatrix}, \qquad e_2= \begin{bmatrix} 0\\ 1 \end{bmatrix} \]form a basis for \(\mathbb{R}^2\).
They span the plane because every vector
\[ \begin{bmatrix} a\\ b \end{bmatrix} \]can be written
\[ a e_1+b e_2. \]They are also linearly independent.
The Standard Basis of \(\mathbb{R}^3\)
The standard basis is
\[ e_1= \begin{bmatrix} 1\\ 0\\ 0 \end{bmatrix}, \qquad e_2= \begin{bmatrix} 0\\ 1\\ 0 \end{bmatrix}, \qquad e_3= \begin{bmatrix} 0\\ 0\\ 1 \end{bmatrix}. \]Every vector in \(\mathbb{R}^3\) can be written uniquely as
\[ x e_1+y e_2+z e_3. \]The Standard Basis of \(\mathbb{R}^n\)
More generally, the standard basis of \(\mathbb{R}^n\) consists of \(n\) vectors:
\[ e_1,e_2,\ldots,e_n, \]where \(e_i\) has a \(1\) in position \(i\) and \(0\) in every other position.
Thus
\[ \boxed{\dim(\mathbb{R}^n)=n}. \]A Basis Does Not Have to Be Standard
Consider
\[ v_1= \begin{bmatrix} 1\\ 1 \end{bmatrix}, \qquad v_2= \begin{bmatrix} 1\\ -1 \end{bmatrix}. \]These vectors are not the standard coordinate vectors, but they are not scalar multiples of one another, so they are independent.
Two independent vectors in \(\mathbb{R}^2\) span all of \(\mathbb{R}^2\). Therefore
\[ \boxed{ \left\{ \begin{bmatrix}1\\1\end{bmatrix}, \begin{bmatrix}1\\-1\end{bmatrix} \right\} \text{ is a basis for }\mathbb{R}^2}. \]Vector Spaces Usually Have Many Bases
A vector space generally does not have one unique basis.
For \(\mathbb{R}^2\), all of the following may serve as bases:
\[ \left\{ \begin{bmatrix}1\\0\end{bmatrix}, \begin{bmatrix}0\\1\end{bmatrix} \right\}, \] \[ \left\{ \begin{bmatrix}1\\1\end{bmatrix}, \begin{bmatrix}1\\-1\end{bmatrix} \right\}, \] \[ \left\{ \begin{bmatrix}2\\1\end{bmatrix}, \begin{bmatrix}-1\\3\end{bmatrix} \right\}. \]What remains fixed is not the particular basis, but the number of vectors in every basis.
What Is Dimension?
If a vector space \(V\) has a finite basis containing \(n\) vectors, then the dimension of \(V\) is
\[ \boxed{\dim(V)=n}. \]Dimension measures the number of independent directions needed to describe the space.
Dimension Is Well Defined
A finite-dimensional vector space can have many different bases, but every basis has the same number of vectors.
This fact makes dimension meaningful.
We do not say that a space has dimension \(2\) because one convenient basis happens to contain two vectors. We say it has dimension \(2\) because every basis contains exactly two vectors.
Dimension of the Zero Space
The vector space containing only the zero vector,
\[ \{0\}, \]has no nonzero directions.
Its basis is the empty set, so
\[ \boxed{\dim(\{0\})=0}. \]Dimension of a Line Through the Origin
Let
\[ L=\operatorname{span}\{v\}, \qquad v\neq0. \]The single vector \(v\) is independent and spans \(L\), so
\[ \boxed{\dim(L)=1}. \]Dimension of a Plane Through the Origin
If
\[ P=\operatorname{span}\{v_1,v_2\} \]and \(v_1,v_2\) are linearly independent, then they form a basis for \(P\).
Therefore,
\[ \boxed{\dim(P)=2}. \]Dimension as Degrees of Freedom
Dimension can also be understood as the number of independent parameters needed to describe an arbitrary vector in the space.
A line through the origin can be written
\[ tv, \]using one free scalar \(t\).
A plane through the origin can be written
\[ sv_1+tv_2, \]using two independent scalar parameters.
Unique Representation Is the Payoff
If \(B=\{v_1,\ldots,v_n\}\) is a basis for \(V\), then every vector \(v\in V\) can be written in exactly one way as
\[ \boxed{v=c_1v_1+\cdots+c_nv_n}. \]Spanning guarantees that at least one representation exists.
Independence guarantees that two different coefficient lists cannot represent the same vector.
Why the Coordinates Are Unique
Suppose
\[ v=c_1v_1+\cdots+c_nv_n \]and also
\[ v=d_1v_1+\cdots+d_nv_n. \]Subtracting gives
\[ (c_1-d_1)v_1+\cdots+(c_n-d_n)v_n=0. \]Since the basis vectors are independent,
\[ c_1-d_1=\cdots=c_n-d_n=0. \]Therefore \(c_i=d_i\) for every \(i\).
Coordinates Relative to a Basis
If
\[ B=\{v_1,\ldots,v_n\} \]is a basis and
\[ v=c_1v_1+\cdots+c_nv_n, \]then the coordinate vector of \(v\) relative to \(B\) is
\[ \boxed{ [v]_B= \begin{bmatrix} c_1\\ \vdots\\ c_n \end{bmatrix}}. \]Example: Coordinates in a Nonstandard Basis
Let
\[ B= \left\{ \begin{bmatrix} 1\\ 1 \end{bmatrix}, \begin{bmatrix} 1\\ -1 \end{bmatrix} \right\} \]and let
\[ v= \begin{bmatrix} 5\\ 1 \end{bmatrix}. \]We seek \(c_1,c_2\) satisfying
\[ c_1 \begin{bmatrix} 1\\ 1 \end{bmatrix} + c_2 \begin{bmatrix} 1\\ -1 \end{bmatrix} = \begin{bmatrix} 5\\ 1 \end{bmatrix}. \]Therefore,
\[ c_1+c_2=5, \] \[ c_1-c_2=1. \]Solving gives
\[ c_1=3,\qquad c_2=2. \]Hence
\[ \boxed{ [v]_B= \begin{bmatrix} 3\\ 2 \end{bmatrix}}. \]Coordinates Depend on the Basis
The vector
\[ v= \begin{bmatrix} 5\\ 1 \end{bmatrix} \]has standard coordinates
\[ \begin{bmatrix} 5\\ 1 \end{bmatrix}, \]but relative to the basis
\[ B= \left\{ \begin{bmatrix}1\\1\end{bmatrix}, \begin{bmatrix}1\\-1\end{bmatrix} \right\} \]its coordinates are
\[ \begin{bmatrix} 3\\ 2 \end{bmatrix}. \]The vector itself has not changed. Only its coordinate description has.
A Basis Is a Coordinate System
Choosing a basis is equivalent to choosing a coordinate system for the vector space.
In ordinary geometry, we often use horizontal and vertical axes by habit. Linear algebra allows us to replace those directions with any independent spanning directions that are useful for the problem.
Basis for a Subspace
A subspace can have its own basis even when it sits inside a larger vector space.
For example, let
\[ W= \left\{ \begin{bmatrix} x\\ y\\ z \end{bmatrix} : z=0 \right\}. \]Then
\[ \left\{ \begin{bmatrix} 1\\ 0\\ 0 \end{bmatrix}, \begin{bmatrix} 0\\ 1\\ 0 \end{bmatrix} \right\} \]is a basis for \(W\), so
\[ \boxed{\dim(W)=2}. \]Finding a Basis from a Parametric Description
Suppose
\[ W= \left\{ \begin{bmatrix} s+2t\\ 3s-t\\ t \end{bmatrix} :s,t\in\mathbb{R} \right\}. \]Separate the parameters:
\[ \begin{bmatrix} s+2t\\ 3s-t\\ t \end{bmatrix} = s \begin{bmatrix} 1\\ 3\\ 0 \end{bmatrix} + t \begin{bmatrix} 2\\ -1\\ 1 \end{bmatrix}. \]The two generating vectors are independent, so they form a basis for \(W\).
\[ \boxed{\dim(W)=2}. \]Finding a Basis from an Equation
Consider the plane
\[ x+y+z=0. \]Solve for \(x\):
\[ x=-y-z. \]Let \(y=s\) and \(z=t\). Then
\[ \begin{bmatrix} x\\ y\\ z \end{bmatrix} = \begin{bmatrix} -s-t\\ s\\ t \end{bmatrix} = s \begin{bmatrix} -1\\ 1\\ 0 \end{bmatrix} + t \begin{bmatrix} -1\\ 0\\ 1 \end{bmatrix}. \]Therefore a basis is
\[ \boxed{ \left\{ \begin{bmatrix}-1\\1\\0\end{bmatrix}, \begin{bmatrix}-1\\0\\1\end{bmatrix} \right\}}. \]Reducing a Spanning Set to a Basis
Suppose a spanning set contains dependent vectors.
We may remove redundant vectors one at a time without changing the span until the remaining set becomes independent.
The result is a basis for the same subspace.
Example: Remove a Redundant Vector
Consider
\[ v_1= \begin{bmatrix} 1\\ 0 \end{bmatrix}, \qquad v_2= \begin{bmatrix} 0\\ 1 \end{bmatrix}, \qquad v_3= \begin{bmatrix} 1\\ 1 \end{bmatrix}. \]Since
\[ v_3=v_1+v_2, \]the third vector is redundant.
Removing it leaves
\[ \boxed{\{v_1,v_2\}}, \]a basis for the same span, namely \(\mathbb{R}^2\).
Extending an Independent Set to a Basis
Suppose we begin with an independent set that does not yet span the whole vector space.
We may add vectors that lie outside the current span. Each such vector contributes a new independent direction.
In a finite-dimensional space, this process can continue until a basis is obtained.
Example: Extend to a Basis of \(\mathbb{R}^3\)
Start with
\[ v_1= \begin{bmatrix} 1\\ 1\\ 0 \end{bmatrix}. \]One possible extension is
\[ v_2= \begin{bmatrix} 0\\ 1\\ 0 \end{bmatrix}, \qquad v_3= \begin{bmatrix} 0\\ 0\\ 1 \end{bmatrix}. \]The collection
\[ \boxed{\{v_1,v_2,v_3\}} \]is independent and contains three vectors in \(\mathbb{R}^3\), so it forms a basis.
The Basis Size Test in \(\mathbb{R}^n\)
Since \(\dim(\mathbb{R}^n)=n\), a collection of exactly \(n\) vectors in \(\mathbb{R}^n\) has a useful shortcut:
If the \(n\) vectors are linearly independent, then they automatically span \(\mathbb{R}^n\).
Equivalently, if they span \(\mathbb{R}^n\), then they are automatically independent.
Why the Basis Size Shortcut Works
In an \(n\)-dimensional space:
- no independent set can contain more than \(n\) vectors;
- no spanning set can contain fewer than \(n\) vectors.
Therefore, once a set has exactly \(n\) vectors, proving either independence or spanning is enough to establish that it is a basis.
Too Few Vectors
In an \(n\)-dimensional space, fewer than \(n\) vectors cannot span the entire space.
For example, two vectors cannot form a basis for \(\mathbb{R}^3\).
Even if those two vectors are independent, their span has dimension at most \(2\).
Too Many Vectors
In an \(n\)-dimensional space, more than \(n\) vectors cannot be linearly independent.
For example, four vectors cannot form a basis for \(\mathbb{R}^3\).
At least one vector must be redundant.
Dimension of a Subspace Cannot Exceed the Ambient Dimension
If
\[ W\subseteq V \]and \(V\) is finite-dimensional, then
\[ \boxed{\dim(W)\leq\dim(V)}. \]A subspace cannot contain more independent directions than the space in which it lives.
Same Dimension Can Force Equality
Suppose
\[ W\subseteq V \]and both spaces are finite-dimensional.
If
\[ \dim(W)=\dim(V), \]then
\[ \boxed{W=V}. \]There is no room for a proper subspace with the full dimension of the ambient finite-dimensional space.
Basis for Polynomial Spaces
Let \(P_2\) be the vector space of real polynomials of degree at most \(2\).
The set
\[ \boxed{\{1,x,x^2\}} \]is a basis for \(P_2\).
Every polynomial in \(P_2\) has the unique form
\[ a+bx+cx^2. \]Therefore,
\[ \boxed{\dim(P_2)=3}. \]Dimension of \(P_n\)
The polynomial space
\[ P_n \]consists of polynomials of degree at most \(n\).
A standard basis is
\[ \{1,x,x^2,\ldots,x^n\}. \]This basis contains \(n+1\) vectors, so
\[ \boxed{\dim(P_n)=n+1}. \]Why \(P_n\) Has Dimension \(n+1\)
The notation \(P_n\) can tempt us to guess dimension \(n\), but the constant term counts as an independent coefficient.
A polynomial
\[ a_0+a_1x+\cdots+a_nx^n \]has the \(n+1\) independent coefficients
\[ a_0,a_1,\ldots,a_n. \]A Nonstandard Polynomial Basis
Consider
\[ B=\{1+x,\;x+x^2,\;1-x^2\}. \]If these three polynomials are linearly independent in \(P_2\), then they automatically form a basis because \(\dim(P_2)=3\).
This illustrates the basis-size shortcut outside coordinate-vector spaces.
Basis for Matrix Spaces
Let \(M_{2\times2}\) be the vector space of all real \(2\times2\) matrices.
A standard basis is
\[ E_{11}= \begin{bmatrix} 1&0\\ 0&0 \end{bmatrix}, \quad E_{12}= \begin{bmatrix} 0&1\\ 0&0 \end{bmatrix}, \] \[ E_{21}= \begin{bmatrix} 0&0\\ 1&0 \end{bmatrix}, \quad E_{22}= \begin{bmatrix} 0&0\\ 0&1 \end{bmatrix}. \]Therefore,
\[ \boxed{\dim(M_{2\times2})=4}. \]Dimension of \(M_{m\times n}\)
An \(m\times n\) matrix has \(mn\) independently adjustable entries.
A standard basis contains one matrix for each entry position, with a single \(1\) in that position and zeros elsewhere.
Therefore,
\[ \boxed{\dim(M_{m\times n})=mn}. \]Basis for Diagonal Matrices
Consider the subspace of \(2\times2\) diagonal matrices:
\[ D= \left\{ \begin{bmatrix} a&0\\ 0&b \end{bmatrix} :a,b\in\mathbb{R} \right\}. \]A basis is
\[ \left\{ \begin{bmatrix} 1&0\\ 0&0 \end{bmatrix}, \begin{bmatrix} 0&0\\ 0&1 \end{bmatrix} \right\}. \]Hence
\[ \boxed{\dim(D)=2}. \]Infinite-Dimensional Vector Spaces
Not every vector space has a finite basis.
For example, the space of all real polynomials
\[ P \]requires arbitrarily high powers
\[ 1,x,x^2,x^3,\ldots \]to represent all possible polynomial degrees.
Such a space is infinite-dimensional.
Finite Versus Infinite Dimension
A vector space is finite-dimensional if it has a finite basis.
It is infinite-dimensional if no finite basis can span the entire space.
Many elementary matrix and coordinate spaces are finite-dimensional, while many spaces of functions are infinite-dimensional.
Dimension and Information
Dimension can be interpreted as the number of independent pieces of information needed to identify an arbitrary vector in the space.
A vector in \(\mathbb{R}^3\) needs three coordinates.
A polynomial in \(P_2\) needs three coefficients.
A \(2\times2\) matrix needs four entries.
Dimension and Constraints
Linear constraints often reduce dimension.
For example, vectors in \(\mathbb{R}^3\) satisfy three free coordinate choices before constraints are imposed.
The single independent homogeneous equation
\[ x+y+z=0 \]leaves two free parameters, producing a two-dimensional subspace.
Independent Constraints and Dimension
Each genuinely independent linear constraint can reduce the number of free directions.
Redundant constraints do not reduce dimension further.
This distinction becomes important when studying systems, row reduction, rank, and solution spaces.
Dimension Is Not Physical Size
A one-dimensional subspace can contain infinitely many vectors.
A two-dimensional subspace can also contain infinitely many vectors.
Dimension does not count how many vectors are in the space. It counts the number of independent directions needed to generate it.
Dimension Is Not the Number of Coordinates Written Down
A subspace of \(\mathbb{R}^5\) may be described using vectors with five entries while still having dimension \(2\).
The ambient coordinates tell us where the vectors live.
The dimension tells us how many independent directions the subspace itself possesses.
Example: A Two-Dimensional Subspace of \(\mathbb{R}^4\)
Let
\[ W= \operatorname{span} \left\{ \begin{bmatrix} 1\\ 0\\ 1\\ 0 \end{bmatrix}, \begin{bmatrix} 0\\ 1\\ 0\\ 1 \end{bmatrix} \right\}. \]The two generators are independent.
Therefore,
\[ \boxed{\dim(W)=2} \]even though every vector in \(W\) is written with four coordinates.
Basis Selection Can Simplify a Problem
Different bases describe the same vector space in different ways.
A strategically chosen basis can make a pattern more visible, reduce the complexity of formulas, or align coordinates with the natural directions of a system.
Later topics such as diagonalization and singular-value methods rely heavily on choosing useful bases.
Common Error: Calling Any Spanning Set a Basis
A spanning set is not automatically a basis.
For example,
\[ \left\{ \begin{bmatrix}1\\0\end{bmatrix}, \begin{bmatrix}0\\1\end{bmatrix}, \begin{bmatrix}1\\1\end{bmatrix} \right\} \]spans \(\mathbb{R}^2\), but it is dependent.
Therefore it is not a basis.
Common Error: Calling Any Independent Set a Basis
Independence alone is not enough.
The single vector
\[ \begin{bmatrix} 1\\ 0 \end{bmatrix} \]is independent in \(\mathbb{R}^2\), but it does not span the entire plane.
Therefore it is not a basis for \(\mathbb{R}^2\).
Common Error: Confusing a Basis with the Space
A basis is a finite or infinite collection of selected vectors.
The vector space contains every allowed linear combination of those basis vectors.
The basis is a generating coordinate framework, not the full set of vectors.
Common Error: Assuming the Standard Basis Is the Only Basis
Standard bases are convenient, but they are not privileged by the abstract definition.
Any independent spanning set is a basis.
Choosing a nonstandard basis is often one of the most useful moves in linear algebra.
Common Error: Counting Generators Before Removing Redundancy
Suppose a subspace is written as
\[ W=\operatorname{span}\{v_1,v_2,v_3\}. \]We cannot immediately conclude that \(\dim(W)=3\).
First we must determine whether the generators are independent.
Common Error: Saying \(\dim(P_n)=n\)
The standard basis of \(P_n\) is
\[ 1,x,x^2,\ldots,x^n. \]There are \(n+1\) vectors because the powers begin at \(0\).
Therefore,
\[ \boxed{\dim(P_n)=n+1}. \]Common Error: Treating Coordinates as the Vector Itself
The coordinate column
\[ [v]_B \]depends on the chosen basis \(B\).
The underlying vector \(v\) is the same object even when a different basis gives it a different coordinate column.
A Reliable Procedure for Verifying a Basis
- Identify the target vector space or subspace.
- Check that every proposed basis vector belongs to that space.
- Verify linear independence.
- Verify that the collection spans the target space.
- Use dimension shortcuts when the dimension is already known.
- Conclude that the set is a basis only when both requirements are satisfied.
A Reliable Procedure for Finding Coordinates
- Write the ordered basis \(B=(v_1,\ldots,v_n)\).
- Write \(v=c_1v_1+\cdots+c_nv_n\).
- Translate the vector equation into scalar equations.
- Solve for \(c_1,\ldots,c_n\).
- Place the coefficients in basis order.
- Write \([v]_B=(c_1,\ldots,c_n)^T\).
The order of the basis vectors matters when recording coordinates.
A Reliable Procedure for Finding Dimension
- Find a spanning set for the space.
- Remove redundant vectors until the set is independent.
- The remaining vectors form a basis.
- Count the basis vectors.
- That count is the dimension.
Example: Full Basis Workflow
Determine whether
\[ B= \left\{ \begin{bmatrix} 1\\ 1\\ 0 \end{bmatrix}, \begin{bmatrix} 0\\ 1\\ 1 \end{bmatrix}, \begin{bmatrix} 1\\ 0\\ 1 \end{bmatrix} \right\} \]is a basis for \(\mathbb{R}^3\).
Test
\[ c_1 \begin{bmatrix} 1\\ 1\\ 0 \end{bmatrix} + c_2 \begin{bmatrix} 0\\ 1\\ 1 \end{bmatrix} + c_3 \begin{bmatrix} 1\\ 0\\ 1 \end{bmatrix} = 0. \]Coordinate comparison gives
\[ c_1+c_3=0, \] \[ c_1+c_2=0, \] \[ c_2+c_3=0. \]The only solution is
\[ c_1=c_2=c_3=0. \]So the vectors are independent. Since there are three independent vectors in \(\mathbb{R}^3\), they automatically form a basis.
Example: Full Dimension Workflow
Let
\[ W= \operatorname{span} \left\{ \begin{bmatrix} 1\\ 0\\ 1 \end{bmatrix}, \begin{bmatrix} 0\\ 1\\ 1 \end{bmatrix}, \begin{bmatrix} 1\\ 1\\ 2 \end{bmatrix} \right\}. \]The third vector satisfies
\[ \begin{bmatrix} 1\\ 1\\ 2 \end{bmatrix} = \begin{bmatrix} 1\\ 0\\ 1 \end{bmatrix} + \begin{bmatrix} 0\\ 1\\ 1 \end{bmatrix}. \]Therefore the third generator is redundant.
The first two vectors are independent, so they form a basis for \(W\).
\[ \boxed{\dim(W)=2}. \]Why Bases Matter in Multidimensional Systems
A multidimensional system may contain many variables, measurements, or states, but the effective number of independent directions may be smaller than the raw number of quantities recorded.
A basis isolates those independent directions. Dimension records how many there are. Coordinates relative to a basis then describe every state using a minimal set of coefficients.
This framework is central to matrix models, state spaces, data reduction, linear transformations, eigenvector methods, least squares, and singular values.
Practice 1: Standard Dimension
What is the dimension of \(\mathbb{R}^5\)?
Practice 2: Basis or Not?
Determine whether
\[ \left\{ \begin{bmatrix} 1\\ 0 \end{bmatrix}, \begin{bmatrix} 2\\ 0 \end{bmatrix} \right\} \]is a basis for \(\mathbb{R}^2\).
Practice 3: Nonstandard Basis
Determine whether
\[ \left\{ \begin{bmatrix} 1\\ 2 \end{bmatrix}, \begin{bmatrix} 2\\ 1 \end{bmatrix} \right\} \]is a basis for \(\mathbb{R}^2\).
Practice 4: Coordinates Relative to a Basis
Let
\[ B= \left\{ \begin{bmatrix} 1\\ 1 \end{bmatrix}, \begin{bmatrix} 1\\ -1 \end{bmatrix} \right\}. \]Find the coordinates of
\[ v= \begin{bmatrix} 7\\ 3 \end{bmatrix} \]relative to \(B\).
Practice 5: Dimension of a Span
Find the dimension of
\[ W= \operatorname{span} \left\{ \begin{bmatrix} 1\\ 2 \end{bmatrix}, \begin{bmatrix} 2\\ 4 \end{bmatrix} \right\}. \]Practice 6: Polynomial Dimension
What is the dimension of \(P_4\)?
Practice 7: Polynomial Basis
Give a standard basis for \(P_3\).
Practice 8: Matrix Dimension
What is the dimension of \(M_{3\times2}\)?
Practice 9: Basis for a Plane
Find a basis for the subspace
\[ x+y+z=0 \]in \(\mathbb{R}^3\).
Practice 10: Too Many Vectors
Can a set of six vectors be a basis for \(\mathbb{R}^5\)? Explain.
Practice 11: Too Few Vectors
Can a set of three vectors be a basis for \(\mathbb{R}^4\)? Explain.
Practice 12: Ambient Space versus Dimension
Let
\[ W= \operatorname{span} \left\{ \begin{bmatrix} 1\\ 0\\ 0\\ 1 \end{bmatrix}, \begin{bmatrix} 0\\ 1\\ 1\\ 0 \end{bmatrix} \right\}. \]If the two generators are independent, what is \(\dim(W)\)?
Practice Answers
Practice 1
The standard basis of \(\mathbb{R}^5\) contains five vectors, so
\[ \boxed{\dim(\mathbb{R}^5)=5}. \]Practice 2
No. The second vector is twice the first, so the collection is linearly dependent. It spans only the \(x\)-axis.
Practice 3
Yes. The two vectors are not scalar multiples of one another, so they are independent. Two independent vectors in \(\mathbb{R}^2\) form a basis.
Practice 4
Solve
\[ c_1 \begin{bmatrix} 1\\ 1 \end{bmatrix} + c_2 \begin{bmatrix} 1\\ -1 \end{bmatrix} = \begin{bmatrix} 7\\ 3 \end{bmatrix}. \]This gives
\[ c_1+c_2=7, \qquad c_1-c_2=3. \]Therefore,
\[ c_1=5,\qquad c_2=2. \]Hence
\[ \boxed{ [v]_B= \begin{bmatrix} 5\\ 2 \end{bmatrix}}. \]Practice 5
Since
\[ \begin{bmatrix} 2\\ 4 \end{bmatrix} = 2 \begin{bmatrix} 1\\ 2 \end{bmatrix}, \]only one independent direction remains. Thus
\[ \boxed{\dim(W)=1}. \]Practice 6
A standard basis is
\[ \{1,x,x^2,x^3,x^4\}. \]Therefore,
\[ \boxed{\dim(P_4)=5}. \]Practice 7
\[ \boxed{\{1,x,x^2,x^3\}}. \]Practice 8
A \(3\times2\) matrix has six independently adjustable entries, so
\[ \boxed{\dim(M_{3\times2})=6}. \]Practice 9
Write
\[ x=-y-z. \]Then
\[ \begin{bmatrix} x\\ y\\ z \end{bmatrix} = y \begin{bmatrix} -1\\ 1\\ 0 \end{bmatrix} + z \begin{bmatrix} -1\\ 0\\ 1 \end{bmatrix}. \]One basis is
\[ \boxed{ \left\{ \begin{bmatrix}-1\\1\\0\end{bmatrix}, \begin{bmatrix}-1\\0\\1\end{bmatrix} \right\}}. \]Practice 10
No. A basis for \(\mathbb{R}^5\) must contain exactly five vectors. Any six vectors in \(\mathbb{R}^5\) are linearly dependent.
Practice 11
No. Three vectors cannot span the four-dimensional space \(\mathbb{R}^4\).
Practice 12
Two independent vectors form a basis for their span, so
\[ \boxed{\dim(W)=2}. \]The fact that the vectors have four coordinates only tells us that \(W\subseteq\mathbb{R}^4\).
What to Remember
A basis is a linearly independent spanning set:
\[ \boxed{\text{basis}=\text{independent}+\text{spanning}}. \]- A basis contains no redundant vectors.
- A basis generates the entire target space.
- Every vector has a unique representation relative to a basis.
- The coefficients in that representation are the vector's basis coordinates.
- Every basis of a finite-dimensional space has the same number of vectors.
- That number is the dimension of the space.
- \(\dim(\mathbb{R}^n)=n\).
- \(\dim(P_n)=n+1\).
- \(\dim(M_{m\times n})=mn\).
The most useful interpretation is:
\[ \boxed{\text{dimension counts independent directions}}. \]Where We Go Next
A basis gives us a compact coordinate description of every vector in a space. The next question is what happens when one vector space is mapped into another while preserving the linear structure.
The next lesson, Linear Transformations, develops these structure-preserving maps and shows how their behavior can be understood through vectors, bases, and eventually matrices.