Linear independence

Lecture 14

Author

Affiliation

Minjae Park

Auburn University
MATH 2660 - Spring 2026

Published

February 11, 2026

Spanning sets

Every vector in a vector space can be written as a linear combination of vectors in a spanning set.
A spanning set plays the role of “coordinate directions” for the space.
In this sense, a spanning set generates a “grid” that allows us to visualize the entire vector space.

The set \(S_1=\left\{\langle 1,0 \rangle,\langle 0,1 \rangle\right\}\) spans \(\mathbb{R}^2\).
Add another vector: \(S_2=\left\{\langle 1,0 \rangle,\langle 0,1 \rangle,\langle 1,1 \rangle\right\}\). This still spans \(\mathbb{R}^2\).
But \(\langle 1,1 \rangle\) is redundant, since \[ \langle 1,1 \rangle=\langle 1,0 \rangle+\langle 0,1 \rangle. \]
Any linear combination \[ \vec{u}=c_1\langle 1,0 \rangle+c_2\langle 0,1 \rangle+c_3\langle 1,1 \rangle \] can be rewritten using only \(\langle 1,0 \rangle\) and \(\langle 0,1 \rangle\): \[ \vec{u}=(c_1+c_3)\langle 1,0 \rangle+(c_2+c_3)\langle 0,1 \rangle. \]

Ideally, a spanning set should act like genuine coordinate directions forming a clean grid.
However, a spanning set may contain redundant vectors.
A vector is redundant if it is already inside the space generated by the others.
In other words, one vector can be written as a linear combination of the remaining vectors.
When such redundancy exists, the set is called linearly dependent.

A set \(\{\vec{v}_1,\ldots,\vec{v}_k\}\) is linearly dependent if there exist scalars \(c_1,\ldots,c_k\), not all zero, such that \[ c_1\vec{v}_1+\cdots+c_k\vec{v}_k=\vec{0}. \]
Equivalently, at least one vector can be written as a linear combination of the others.
If the only solution is \(c_1=\cdots=c_k=0\), the vectors are linearly independent.

Linear dependence means the equation \[ c_1\vec{v}_1+\cdots+c_k\vec{v}_k=\vec{0} \] has more than one solution, including the trivial (all-zero) solution.
This is equivalent to solving \[ A\vec{c}=\vec{0}, \] where \(A=[\vec{v}_1\ \cdots\ \vec{v}_k]\).
The vectors are linearly independent if and only if the reduced row-echelon form of \(A\) has a pivot in every column.
Therefore, compute the RREF of \(A\) using Gauss–Jordan elimination.

The standard coordinate vectors \(\vec{e}_1,\ldots,\vec{e}_n\) are linearly independent because \[ [\vec{e}_1\ \cdots\ \vec{e}_n]=I_n \] already has a pivot in every column.
Let \(\{\vec{v}_1,\ldots,\vec{v}_m\}\subset\mathbb{R}^n\).
- If \(m>n\), the set is always dependent (more vectors than dimensions).
- If \(m\le n\), it depends:
  - If the RREF has a pivot in every column, the vectors are independent.
  - If some column has no pivot, the vectors are dependent.

Determine whether the vectors \[ \{\langle 1,2,0 \rangle,\ \langle -1,-2,1 \rangle,\ \langle 2,4,1 \rangle\} \] are linearly independent.

Form the matrix \[ A= \begin{bmatrix} 1 & -1 & 2\\ 2 & -2 & 4\\ 0 & 1 & 1 \end{bmatrix}. \]

Row-reducing gives \[ \text{RREF}(A)= \begin{bmatrix} 1 & 0 & 3\\ 0 & 1 & 1\\ 0 & 0 & 0 \end{bmatrix}. \]

Since there are fewer pivots than columns, the system has infinitely many solutions. Therefore, the vectors are linearly dependent.

Consider two vectors \(\vec{v}_1,\vec{v}_2\in \mathbb{R}^3\).
Let \(A=[\vec{v}_1\ \vec{v}_2]\) be the \(3\times 2\) matrix whose columns are \(\vec{v}_1\) and \(\vec{v}_2\).
Suppose \[ \text{RREF}(A)= \begin{bmatrix} 1 & 0 \\ 0 & 1 \\ 0 & 0 \end{bmatrix}. \] Is \(\{\vec{v}_1,\vec{v}_2\}\) linearly independent?
Answer: Yes. There is a pivot in every column, so the system \(A\vec{c}=\vec{0}\) has only the trivial solution. (The RREF does not need to be a square identity matrix; it only needs a pivot in each column.)

A set \(S=\{\vec v_1,\ldots,\vec v_k\}\) is a basis of \(V\) if
- it spans \(V\), and
- it is linearly independent.
Equivalently, a basis is a spanning set with no redundancy.
The number of vectors in a basis is called the dimension of \(V\), denoted by \(\dim(V)\).
Fact: Although a vector space may have many different bases, all bases have the same number of vectors. Therefore, the dimension is well-defined.
Recall: The empty set \(\emptyset\) is a spanning set of the zero vector space, so \(\dim(\{ \vec 0 \})=0\).

\(\mathbb{R}^n\) has the standard basis \(\{\vec{e}_1,\ldots,\vec{e}_n\}\), so \(\dim(\mathbb{R}^n)=n\).
The set \[ V=\{t\langle a,b,c \rangle:t\in\mathbb{R}\}\subset\mathbb{R}^3 \] is a line through the origin spanned by \(\{\langle a,b,c \rangle\}\), so \(\dim(V)=1\).
The set \[ V=\{ax^2+bx: a,b\in\mathbb{R}\}\subset \mathcal P_2 \] is spanned by \(\{x,x^2\}\), so \(\dim(V)=2\).

Let \(V\) be the set of \(3\times 3\) upper triangular matrices. What is the \(\dim(V)\)?

Scan the QR code or go to join.iclicker.com/MBNJ.