3.3 Linear Independence · Linear Algebra Refresher

Independence is the property of 'no redundancy' — no vector in the set is expressible from the others. It is the condition that makes coordinates unique, which is why bases require it.

Definition · Linear dependence / independence

The vectors

v_{1}, \dots, v_{n}

are linearly dependent if there exist scalars

c_{1}, \dots, c_{n}

, not all zero, with

c_{1} v_{1} + \dots + c_{n} v_{n} = 0

. If the only such combination is the trivial one, they are linearly independent.

Geometry in

R^{2}

R^{3}

Two vectors are dependent iff they lie on the same line through the origin. Three vectors in

R^{3}

are dependent iff they lie in a common plane through the origin.

Instant dependence

Any set containing

0

is dependent (take

c = 1

on the zero vector). Any set of more than

dim V

vectors is dependent.

Theorem 3.3.1 · The square case

Let

x_{1}, \dots, x_{n} \in R^{n}

and let

X = (x_{1}, \dots, x_{n})

. The vectors are linearly dependent if and only if

X

is singular, i.e.

det (X) = 0

Because $c_{1} x_{1} + \dots + c_{n} x_{n} = 0$ is exactly $X c = 0$ , and a nontrivial solution exists iff $X$ is singular.

!Exam trap

The determinant test needs the matrix to be square — exactly

n

vectors in

R^{n}

. For 3 vectors in

R^{4}

or 5 vectors in

R^{3}

there is no determinant to take: row reduce and count pivots instead.

Testing a set for linear independence

0
Count first
$k$ vectors in a space of dimension $d$ . If $k > d$ , they are automatically dependent — stop.
1
Build the matrix $X$ with the vectors as columns
The question is always whether $X c = 0$ has a nontrivial solution.
2
Is $X$ square?
yesCompute $det (X)$ . Nonzero ⟹ independent; zero ⟹ dependent.
noRow reduce $X$ to echelon form.
3
Count pivots $r$ against the number of vectors $k$
r = kIndependent — no free variables, so only $c = 0$ .
r < kDependent — free variables give a nontrivial relation.
4
If dependent and you need the actual relation, finish to reduced form
Set a free variable to $1$ and read the other coefficients off — that gives an explicit dependency.
fn
Functions instead of vectors?
In $P_{n}$ , equate coefficients of like powers. In $C^{(n - 1)} [a, b]$ , use the Wronskian: if $W (x_{0}) \neq = 0$ for some $x_{0}$ , they're independent.

Worked example

Dependent — and finding the relation

0/4 steps

Are

v_{1} = (1, - 1, 2)^{T}

v_{2} = (2, 1, 1)^{T}

v_{3} = (4, - 1, 5)^{T}

linearly independent? If not, exhibit a dependency.

Theorem 3.3.3 · Uniqueness of representation

v_{1}, \dots, v_{n}

span

V

and are linearly independent iff every

v \in V

has a unique representation

v = c_{1} v_{1} + \dots + c_{n} v_{n}

This is the real reason bases require independence: without it, 'coordinates' would be ambiguous, and §3.5 would collapse.

Functions: the Wronskian

Definition · Wronskian

For

f_{1}, \dots, f_{n} \in C^{(n - 1)} [a, b]

W [f_{1}, \dots, f_{n}] (x) = f_{1} (x) f_{1}^{'} (x) ⋮ f_{1}^{(n - 1)} (x) f_{2} (x) f_{2}^{'} (x) f_{2}^{(n - 1)} (x) \dots \dots \dots f_{n} (x) f_{n}^{'} (x) ⋮ f_{n}^{(n - 1)} (x)

Theorem 3.3.4

W [f_{1}, \dots, f_{n}] (x_{0}) \neq = 0

for some

x_{0} \in [a, b]

, then

f_{1}, \dots, f_{n}

are linearly independent on

[a, b]

One direction only. $W \equiv 0$ proves nothing — it does not imply dependence.

Check yourself

You want to test whether four vectors in

R^{4}

are linearly independent. The fastest correct test is:

Five vectors in

R^{3}

are:

T / F

If the Wronskian

W [f_{1}, \dots, f_{n}] (x)

is identically zero on

[a, b]

, the functions are linearly dependent.

3.3Linear Independence

What you must be able to do

Testing a set for linear independence

Dependent — and finding the relation

Functions: the Wronskian

Check yourself