Span, Basis, and Column Space
This note follows Gilbert Strang's Section 3.4, Independence, Basis and Dimension, focusing on the connection between span, basis, linear independence, \(Ax=b\), and the column space.
1. What does span practically mean?
Suppose we have vectors
\[ v_1,v_2,\ldots,v_n. \]
Their linear combinations have the form
\[ c_1v_1+c_2v_2+\cdots+c_nv_n. \]
The numbers
\[ c_1,c_2,\ldots,c_n \]
are coefficients. They tell us how much of each vector to use.
The span of the vectors is the set of everything that can be produced by choosing all possible coefficient values.
So
\[ \boxed{\operatorname{span}\{v_1,\ldots,v_n\} =\text{all vectors that can be built from }v_1,\ldots,v_n.} \]
The practical reason span matters is that we often want a set of building blocks capable of representing every state or vector in the space we care about.
For example, in \(\mathbb R^2\), if we use only
\[ v_1= \begin{bmatrix} 1\\ 0 \end{bmatrix}, \]
then its combinations are
\[ c_1v_1= \begin{bmatrix} c_1\\ 0 \end{bmatrix}. \]
This reaches only the horizontal axis. It cannot produce
\[ \begin{bmatrix} 0\\ 1 \end{bmatrix} \]
or
\[ \begin{bmatrix} 3\\ 5 \end{bmatrix}. \]
Therefore one vector does not span all of \(\mathbb R^2\).
But with
\[ v_1= \begin{bmatrix} 1\\ 0 \end{bmatrix}, \qquad v_2= \begin{bmatrix} 0\\ 1 \end{bmatrix}, \]
we obtain
\[ c_1v_1+c_2v_2 = \begin{bmatrix} c_1\\ c_2 \end{bmatrix}. \]
Since \(c_1\) and \(c_2\) can be any real numbers, every vector in \(\mathbb R^2\) can be produced.
Thus
\[ \boxed{\operatorname{span}\{v_1,v_2\}=\mathbb R^2.} \]
2. Span and independence answer different questions
Span asks:
\[ \boxed{\text{What entire space can these vectors generate?}} \]
Independence asks:
\[ \boxed{\text{Is there redundancy among these vectors?}} \]
For example,
\[ v_1= \begin{bmatrix} 1\\ 0 \end{bmatrix}, \quad v_2= \begin{bmatrix} 0\\ 1 \end{bmatrix}, \quad v_3= \begin{bmatrix} 1\\ 1 \end{bmatrix}. \]
These three vectors span \(\mathbb R^2\), because the first two already span the whole plane.
But
\[ v_3=v_1+v_2. \]
Therefore
\[ v_1+v_2-v_3=0. \]
The coefficients
\[ 1,1,-1 \]
are not all zero, so this is a nontrivial zero combination.
Therefore the vectors are dependent.
So vectors can span a space while still containing unnecessary redundancy.
3. Trivial and nontrivial zero combinations
The combination
\[ 0v_1+0v_2+\cdots+0v_n=0 \]
always works. This is called the trivial zero combination.
It tells us nothing about dependence.
A nontrivial zero combination is
\[ c_1v_1+\cdots+c_nv_n=0 \]
where the coefficients are not all zero.
Finding such a combination proves that the vectors are dependent.
So
\[ \boxed{\text{nontrivial zero combination}=\text{evidence of redundancy}.} \]
4. What is a basis?
A basis for a vector space must satisfy two conditions:
\[ \boxed{1.\ \text{The vectors are linearly independent.}} \]
and
\[ \boxed{2.\ \text{The vectors span the space.}} \]
The intuitive meaning is
\[ \boxed{\text{basis}=\text{enough building blocks to represent everything, with no redundancy}.} \]
Independence ensures that there are no unnecessary vectors.
Span ensures that no vector in the target space is missing from what can be represented.
5. Example: basis of a plane inside \(\mathbb R^3\)
Consider
\[ v_1= \begin{bmatrix} 1\\ 1\\ 2 \end{bmatrix}, \qquad v_2= \begin{bmatrix} 2\\ 2\\ 5 \end{bmatrix}. \]
Test independence:
\[ c_1v_1+c_2v_2=0. \]
This gives
\[ c_1+2c_2=0, \]
\[ c_1+2c_2=0, \]
\[ 2c_1+5c_2=0. \]
From the first equation,
\[ c_1=-2c_2. \]
Substitute into the third:
\[ 2(-2c_2)+5c_2=0, \]
so
\[ c_2=0, \]
and therefore
\[ c_1=0. \]
Thus the only solution is the trivial one, so the vectors are independent.
Because there are two independent vectors in \(\mathbb R^3\), their span is a two-dimensional plane through the origin.
Therefore
\[ \boxed{\{v_1,v_2\}\text{ is a basis for that plane}.} \]
They are not a basis for all of \(\mathbb R^3\), because two vectors cannot span all three independent directions of \(\mathbb R^3\).
6. Basis and unique representation
A basis does more than generate every vector. It gives every vector a unique coordinate representation.
Suppose basis vectors \(v_1,\ldots,v_n\) produced the same vector \(v\) in two different ways:
\[ v=a_1v_1+\cdots+a_nv_n \]
and
\[ v=b_1v_1+\cdots+b_nv_n. \]
Subtracting gives
\[ (a_1-b_1)v_1+\cdots+(a_n-b_n)v_n=0. \]
Because the basis vectors are independent, every coefficient must be zero:
\[ a_1-b_1=0,\ldots,a_n-b_n=0. \]
Therefore
\[ a_i=b_i \]
for every \(i\).
So:
\[ \boxed{\text{Every vector has exactly one coordinate description in a given basis}.} \]
Span guarantees that a representation exists.
Independence guarantees that the representation is unique.
7. A basis is not unique
The standard basis of \(\mathbb R^2\) is
\[ e_1= \begin{bmatrix} 1\\ 0 \end{bmatrix}, \qquad e_2= \begin{bmatrix} 0\\ 1 \end{bmatrix}. \]
But another valid basis is
\[ v_1= \begin{bmatrix} 1\\ 1 \end{bmatrix}, \qquad v_2= \begin{bmatrix} 1\\ -1 \end{bmatrix}. \]
These are independent and also span all of \(\mathbb R^2\).
Therefore a vector space can have infinitely many different bases.
The basis vectors can change, while the space itself remains the same.
8. Invertible matrices and bases
If \(A\) is an invertible \(n\times n\) matrix, then its columns form a basis for \(\mathbb R^n\).
First, the columns are independent because
\[ Ax=0 \]
implies
\[ x=A^{-1}0=0. \]
Second, the columns span \(\mathbb R^n\) because for every \(b\),
\[ Ax=b \]
has the solution
\[ x=A^{-1}b. \]
Thus
\[ \boxed{\text{invertible}\Rightarrow\text{independent columns + spanning columns}\Rightarrow\text{basis}.} \]
9. Pivot columns give a basis for the column space
If the columns of \(A\) are dependent, not all of them are needed to describe the column space.
The pivot columns identify the independent columns that contain the genuinely new directions.
For example,
\[ A= \begin{bmatrix} 1&2\\ 2&4 \end{bmatrix}. \]
Column 2 satisfies
\[ A_2=2A_1. \]
Reduction gives
\[ A\longrightarrow \begin{bmatrix} 1&2\\ 0&0 \end{bmatrix}. \]
Only column 1 is a pivot column.
Therefore
\[ \boxed{ \left\{ \begin{bmatrix} 1\\ 2 \end{bmatrix} \right\} } \]
is a basis for the column space of \(A\).
The second column adds no new direction.
A crucial rule is:
\[ \boxed{\text{For }C(A),\text{ take the pivot columns from the original }A,\text{ not from }R.} \]
Row operations change the column vectors, so \(C(A)\) and \(C(R)\) are generally different spaces.
10. Connecting span directly to \(Ax=b\)
Let
\[ A=[v_1\ v_2\ \cdots\ v_n] \]
and
\[ x= \begin{bmatrix} x_1\\x_2\\\vdots\\x_n \end{bmatrix}. \]
Then
\[ Ax=x_1v_1+x_2v_2+\cdots+x_nv_n. \]
So the components of \(x\) are exactly the coefficients used to combine the columns of \(A\).
As \(x\) ranges over all of \(\mathbb R^n\), the outputs \(Ax\) run through every possible linear combination of the columns.
Therefore
\[ \boxed{ C(A) = \{Ax:x\in\mathbb R^n\} = \operatorname{span}\{v_1,\ldots,v_n\}. } \]
This gives a very useful interpretation of
\[ Ax=b. \]
The vector \(b\) is one particular output.
The column space is the set of all possible outputs:
\[ \boxed{C(A)=\{b:Ax=b\text{ is solvable}\}.} \]
So \(b\) itself is not the column space.
Instead,
\[ \boxed{b\text{ is one vector that may or may not belong to }C(A).} \]
If
\[ b\in C(A), \]
then \(Ax=b\) has at least one solution.
If
\[ b\notin C(A), \]
then \(Ax=b\) has no solution.
11. Quantum computing connection
The mathematical meaning of basis here is the same idea used in quantum computing.
For a single qubit, the computational basis is
\[ |0\rangle= \begin{bmatrix} 1\\ 0 \end{bmatrix}, \qquad |1\rangle= \begin{bmatrix} 0\\ 1 \end{bmatrix}. \]
Any qubit state can be written as
\[ |\psi\rangle=\alpha|0\rangle+\beta|1\rangle. \]
Thus \(|0\rangle\) and \(|1\rangle\) span the single-qubit state space and are independent.
Another basis is
\[ |+\rangle=\frac{1}{\sqrt2}(|0\rangle+|1\rangle), \qquad |-\rangle=\frac{1}{\sqrt2}(|0\rangle-|1\rangle). \]
The basis can change while the underlying vector space remains the same.
Core mental model
The main ideas can be compressed to:
\[ \boxed{\text{span}=\text{everything the vectors can generate}} \]
\[ \boxed{\text{independence}=\text{no redundancy}} \]
\[ \boxed{\text{basis}=\text{spans the space + independent}} \]
and
\[ \boxed{C(A)=\text{all possible }b=Ax.} \]
A basis is therefore a complete, nonredundant coordinate system for the vector space being studied.
Note metadata
- Note type: learning-note
- Subject: learning-notes
- Source: Introduction to Linear Algebra, 5th Edition — Section 3.4: Independence, Basis and Dimension
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