Linear Independence and the Wronskian

To build the general solution of an $n$ th order linear equation you need $n$ solutions that are genuinely different, not disguised copies of one another. Linear independence makes that idea precise, and the Wronskian, a determinant built from the functions and their derivatives, provides a computational test for it.

Quick Reference

Item	Statement
Linearly independent on $I$	$c_{1} ϕ_{1} + \dots + c_{m} ϕ_{m} = 0$ on $I$ forces $c_{1} = \dots = c_{m} = 0$
Linearly dependent on $I$	The same identity holds for some constants not all zero
Wronskian	$W (ϕ_{1}, \dots, ϕ_{m}) (x) = det [ϕ_{j}^{(i - 1)} (x)]$ , an $m \times m$ determinant
Two-function case	W\left(y_1,y_2\right) = y_1 y_2' - y_2 y_1'
Key implication	$W (x_{0}) \neq 0$ at some point $⟹$ the functions are linearly independent
Converse	Holds only for solutions of a common linear equation, not for arbitrary functions
Two-function shortcut	Two functions are dependent exactly when one is a constant multiple of the other

Definition

We say $m$ functions $ϕ_{1} (x), ϕ_{2} (x), \dots, ϕ_{m} (x)$ are linearly independent on an interval $I$ if

c_{1} ϕ_{1} (x) + c_{2} ϕ_{2} (x) + \dots + c_{m} ϕ_{m} (x) = 0 everywhere on I

implies

c_{1} = c_{2} = \dots = c_{m} = 0 .

A set ${ϕ_{1}, \dots, ϕ_{m}}$ is called linearly dependent if the same identity also holds for some constants $c_{1}, \dots, c_{m}$ not all zero.

Two points about the definition are worth emphasizing. First, the identity must hold at every point of $I$ , not merely at a few. Second, the constants $c_{i}$ are genuine constants, not functions of $x$ ; allowing them to vary would make the definition vacuous.

Informally, a dependent set contains redundancy: at least one of its members can be written in terms of the others. If $c_{1} \neq 0$ , then

ϕ_{1} = - \frac{c_{2}}{c_{1}} ϕ_{2} - \dots - \frac{c_{m}}{c_{1}} ϕ_{m},

so $ϕ_{1}$ contributes nothing new. An independent set has no such redundancy.

Examples

The set ${e^{a x}, e^{b x}}$ with $a \neq b$ is linearly independent on every interval $I$ .

Solution

Suppose

c_{1} e^{a x} + c_{2} e^{b x} = 0

for every $x$ in $I$ . Dividing both sides by $e^{b x}$ , which is never zero,

c_{1} e^{(a - b) x} + c_{2} = 0 .

If we differentiate this identity with respect to $x$ , the constant $c_{2}$ drops out:

c_{1} (a - b) e^{(a - b) x} = 0 .

Because $a \neq b$ , the factor $(a - b)$ is nonzero, and an exponential function is never zero. Therefore $c_{1} = 0$ . Substituting back into

\underset{= 0}{\underset{⏟}{c_{1} e^{(a - b) x}}} + c_{2} = 0

gives $c_{2} = 0$ .

Since both constants must vanish, the set is linearly independent.

The set ${e^{x}, e^{- x}, \cosh x}$ is linearly dependent on every interval.

Solution

To show this, we exhibit constants, not all zero, for which

c_{1} e^{x} + c_{2} e^{- x} + c_{3} \cosh x = 0 .

Simply choose

c_{1} = \frac{1}{2}, c_{2} = \frac{1}{2}, c_{3} = - 1,

and the equation

\frac{1}{2} e^{x} + \frac{1}{2} e^{- x} - \cosh x = 0

always holds, recalling that $\cosh x = \frac{e^{x} + e^{- x}}{2}$ .

Since a nontrivial choice of constants works, the set is dependent. The redundancy is plain: $\cosh x$ carries no information beyond what $e^{x}$ and $e^{- x}$ already provide.

Show that ${1, x, x^{2}}$ is linearly independent on any interval.

Solution

Suppose $c_{1} + c_{2} x + c_{3} x^{2} = 0$ for every $x$ in the interval. A polynomial of degree at most $2$ that is not identically zero has at most $2$ roots, so it cannot vanish on a whole interval unless all its coefficients are zero. Therefore $c_{1} = c_{2} = c_{3} = 0$ .

Alternatively, differentiate twice: $c_{2} + 2 c_{3} x = 0$ , then $2 c_{3} = 0$ , giving $c_{3} = 0$ , then $c_{2} = 0$ , then $c_{1} = 0$ . The same argument shows ${1, x, x^{2}, \dots, x^{n}}$ is independent for every $n$ .

Deriving the Wronskian Test

Suppose ${ϕ_{1}, \dots, ϕ_{m}}$ is linearly dependent. How can we determine constants $c_{1}, \dots, c_{m}$ , not all zero, such that

c_{1}\phi_{1}(x) + c_{2}\phi_{2}(x) + \cdots + c_{m}\phi_{m}(x) = 0 \tag{1}

for every $x$ in $I$ ?

There are $m$ unknowns, $c_{1}$ through $c_{m}$ , and only one equation. The way forward is to generate additional equations, and differentiation is what supplies them: since (1) holds identically on an interval, so does every derivative of it.

Differentiating both sides of (1),

c_{1}\phi_{1}'(x) + c_{2}\phi_{2}'(x) + \cdots + c_{m}\phi_{m}'(x) = 0 \tag{2}

Notice that because $c_{1}, \dots, c_{m}$ are constants,

\frac{d}{dx}\left(c_{i}\phi_{i}(x)\right) = c_{i}\frac{d\phi_{i}(x)}{dx} = c_{i}\phi_{i}'(x) ,

so the unknowns pass through the differentiation untouched. This is exactly why the trick works.

Differentiating (2) again,

c_{1}\phi_{1}''(x) + c_{2}\phi_{2}''(x) + \cdots + c_{m}\phi_{m}''(x) = 0 \tag{3}

and so on:

\begin{aligned} c_{1}\phi_{1}'''(x) + c_{2}\phi_{2}'''(x) + \cdots + c_{m}\phi_{m}'''(x) &= 0 \\ &\ \ \vdots \\ c_{1}\phi_{1}^{(m-1)}(x) + c_{2}\phi_{2}^{(m-1)}(x) + \cdots + c_{m}\phi_{m}^{(m-1)}(x) &= 0 \end{aligned}

Stopping at the $(m - 1)$ st derivative gives exactly $m$ equations in the $m$ unknowns, a square system. We can write them all at once as

\begin{bmatrix} \phi_{1}(x) & \phi_{2}(x) & \cdots & \phi_{m}(x) \\ \phi_{1}'(x) & \phi_{2}'(x) & \cdots & \phi_{m}'(x) \\ \vdots & \vdots & \ddots & \vdots \\ \phi_{1}^{(m-1)}(x) & \phi_{2}^{(m-1)}(x) & \cdots & \phi_{m}^{(m-1)}(x) \end{bmatrix} \begin{bmatrix} c_{1} \\ c_{2} \\ \vdots \\ c_{m} \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ \vdots \\ 0 \end{bmatrix}

The determinant of this $m \times m$ matrix has a specific name.

The Wronskian of $ϕ_{1}, \dots, ϕ_{m}$ is

W\left(\phi_{1}, \cdots, \phi_{m}\right)(x) = \begin{vmatrix} \phi_{1}(x) & \phi_{2}(x) & \cdots & \phi_{m}(x) \\ \phi_{1}'(x) & \phi_{2}'(x) & \cdots & \phi_{m}'(x) \\ \vdots & \vdots & \ddots & \vdots \\ \phi_{1}^{(m-1)}(x) & \phi_{2}^{(m-1)}(x) & \cdots & \phi_{m}^{(m-1)}(x) \end{vmatrix}

From linear algebra, a homogeneous square system $A 𝐜 = 0$ has a nonzero solution $𝐜 \neq 0$ if and only if $det A = 0$ . Applied here:

If $W (ϕ_{1}, \dots, ϕ_{m}) (x_{0}) \neq 0$ at even a single point $x_{0}$ of $I$ , then $ϕ_{1}, \dots, ϕ_{m}$ are linearly independent on $I$ .

Proof

Suppose the functions were dependent. Then constants $c_{1}, \dots, c_{m}$ , not all zero, satisfy the system above at every $x$ , in particular at $x_{0}$ . But a square homogeneous system with a nonzero solution must have zero determinant, contradicting $W (x_{0}) \neq 0$ . Therefore the functions are independent.

Read the direction of this theorem carefully. A nonzero Wronskian proves independence. A zero Wronskian proves nothing in general: there exist independent functions whose Wronskian vanishes identically. The classic example is $ϕ_{1} = x^{2}$ and $ϕ_{2} = x | x |$ on $(- 1, 1)$ , which are independent but have $W \equiv 0$ . The converse is rescued in Section 3.3 by restricting to solutions of a common linear equation.

The Case of Two Functions

For two functions the Wronskian is small enough to use constantly:

W\left(y_{1}, y_{2}\right) = \begin{vmatrix} y_{1}(x) & y_{2}(x) \\ y_{1}'(x) & y_{2}'(x) \end{vmatrix} = y_{1}(x)y_{2}'(x) - y_{2}(x)y_{1}'(x) .

There is also a shortcut that needs no determinant at all.

Two functions are linearly dependent precisely when one is a constant multiple of the other. So ${e^{x}, 3 e^{x}}$ is dependent while ${e^{x}, x e^{x}}$ is independent, since $x$ is not a constant. This test applies only to pairs; for three or more functions the relation can involve all of them at once, as the $\cosh x$ example showed.

Compute $W (\sin x, \cos x)$ and conclude.

Solution

W = | \begin{matrix} \sin x & \cos x \\ \cos x & - \sin x \end{matrix} | = - \sin^{2} x - \cos^{2} x = - 1 .

The Wronskian is $- 1$ , never zero, so $\sin x$ and $\cos x$ are linearly independent on every interval. This is why $y = c_{1} \sin x + c_{2} \cos x$ is the general solution of y'' + y = 0 rather than a redundant description of a smaller family.

Show that $ϕ_{1} = e^{x}$ , $ϕ_{2} = x e^{x}$ , and $ϕ_{3} = e^{2 x}$ , which are solutions of

y''' - 4y'' + 5y' - 2y = 0 ,

are linearly independent.

Solution

First confirm the claim about the equation. Its characteristic polynomial is

r^{3} - 4 r^{2} + 5 r - 2 = (r - 1)^{2} (r - 2),

with the double root $r = 1$ and the simple root $r = 2$ , which is exactly what produces $e^{x}$ , $x e^{x}$ , and $e^{2 x}$ .

Now compute the Wronskian. The derivatives are

\begin{aligned} \phi_1 &= e^{x}, & \phi_1' &= e^{x}, & \phi_1'' &= e^{x} \\ \phi_2 &= xe^{x}, & \phi_2' &= (1+x)e^{x}, & \phi_2'' &= (2+x)e^{x} \\ \phi_3 &= e^{2x}, & \phi_3' &= 2e^{2x}, & \phi_3'' &= 4e^{2x} \end{aligned}

W = | \begin{matrix} e^{x} & x e^{x} & e^{2 x} \\ e^{x} & (1 + x) e^{x} & 2 e^{2 x} \\ e^{x} & (2 + x) e^{x} & 4 e^{2 x} \end{matrix} | .

Factor $e^{x}$ from the first column, $e^{x}$ from the second, and $e^{2 x}$ from the third, giving a common factor $e^{4 x}$ :

W = e^{4 x} | \begin{matrix} 1 & x & 1 \\ 1 & 1 + x & 2 \\ 1 & 2 + x & 4 \end{matrix} | .

Subtract row $1$ from rows $2$ and $3$ :

| \begin{matrix} 1 & x & 1 \\ 0 & 1 & 1 \\ 0 & 2 & 3 \end{matrix} | = 1 \cdot (3 - 2) = 1 .

Therefore $W = e^{4 x}$ , which is never zero. The three functions are linearly independent, so they form a fundamental set for this third-order equation.

Exercises

Determine whether ${x, | x |}$ is linearly independent on $(- 1, 1)$ and on $(0, 1)$ .

Solution

On $(0, 1)$ we have $| x | = x$ , so the two functions are identical and therefore dependent: take $c_{1} = 1$ , $c_{2} = - 1$ .

On $(- 1, 1)$ suppose $c_{1} x + c_{2} | x | = 0$ for all $x$ . At $x = \frac{1}{2}$ this gives $\frac{1}{2} c_{1} + \frac{1}{2} c_{2} = 0$ , so $c_{1} = - c_{2}$ . At $x = - \frac{1}{2}$ it gives $- \frac{1}{2} c_{1} + \frac{1}{2} c_{2} = 0$ , so $c_{1} = c_{2}$ . Together these force $c_{1} = c_{2} = 0$ , so the set is independent.

Independence depends on the interval, which is why the definition names one.

Compute $W (e^{2 x}, e^{- 3 x})$ and conclude.

Solution

W = | \begin{matrix} e^{2 x} & e^{- 3 x} \\ 2 e^{2 x} & - 3 e^{- 3 x} \end{matrix} | = - 3 e^{- x} - 2 e^{- x} = - 5 e^{- x},

which is never zero. The functions are linearly independent, consistent with the general result about $e^{a x}$ and $e^{b x}$ with $a \neq b$ .

Show that ${\sin^{2} x, \cos^{2} x, 1}$ is linearly dependent, and verify that its Wronskian is identically zero.

Solution

The Pythagorean identity $\sin^{2} x + \cos^{2} x - 1 = 0$ gives the nontrivial relation with $c_{1} = 1$ , $c_{2} = 1$ , $c_{3} = - 1$ , so the set is dependent.

For the Wronskian, the third column is ${(1, 0, 0)}^{T}$ , and the first two columns have derivative rows

\left(\sin^{2}x\right)' = \sin 2x, \quad \left(\cos^{2}x\right)' = -\sin 2x, \quad \left(\sin^{2}x\right)'' = 2\cos 2x, \quad \left(\cos^{2}x\right)'' = -2\cos 2x .

Columns one and two of the lower two rows are exact negatives of each other, so adding column one to column two produces a column of the form ${(1, 0, 0)}^{T}$ , identical to column three. A determinant with two equal columns is zero, so $W \equiv 0$ .

Frequently Asked Questions

What does linear independence mean for functions?

That no nontrivial constant-coefficient combination of them vanishes identically on the interval. Equivalently, none of them can be built from the others by scaling and adding. It is the same notion as independence of vectors, with functions playing the role of vectors.

What is the Wronskian and why is it a determinant?

It is the determinant of the matrix whose rows are the functions and their successive derivatives up to order $m - 1$ . It arises because differentiating the dependence relation repeatedly produces a square homogeneous linear system in the unknown constants, and a determinant is exactly the tool that decides whether such a system has a nonzero solution.

Does a zero Wronskian mean the functions are dependent?

Not for arbitrary functions. The standard counterexample is $x^{2}$ and $x | x |$ on $(- 1, 1)$ : their Wronskian vanishes everywhere, yet no constants make $c_{1} x^{2} + c_{2} x | x | = 0$ throughout. The converse becomes true only when the functions are known to be solutions of the same linear differential equation, which is proved in the next section.

At how many points do I need to check the Wronskian?

For proving independence, one point suffices: if $W (x_{0}) \neq 0$ for any single $x_{0}$ in $I$ , the functions are independent on all of $I$ . This is why the test is practical: pick whatever value of $x$ makes the determinant easiest, often $x = 0$ .

Is there a faster test for two functions?

Yes. Two functions are dependent exactly when one is a constant multiple of the other, so you can often decide by inspection. The determinant is worth computing anyway when the answer is not obvious, and it becomes essential for three or more functions, where dependence can involve all of them simultaneously.

Why does the Wronskian stop at the

(m - 1)

st derivative?

Because $m$ unknowns require $m$ equations, and the original identity plus $m - 1$ derivatives gives exactly $m$ . Going further would produce a non-square system with no determinant to compute, and the extra equations carry no new information about the constants anyway.

Quick Reference

Definition

Examples

Deriving the Wronskian Test

The Case of Two Functions

Exercises

Frequently Asked Questions

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