# Linear Independence and Wronskian

## Related calculator: Wronskian Calculator

A set of functions ${\left\{{y}_{{1}}{\left({x}\right)},{y}_{{2}}{\left({x}\right)},\ldots,{y}_{{n}}{\left({x}\right)}\right\}}$ is linearly dependent on ${a}\le{x}\le{b}$, if there exist constants ${c}_{{1}}$, ${c}_{{2}}$, ... , ${c}_{{n}}$, not all zero, such that ${c}_{{1}}{y}_{{1}}{\left({x}\right)}+{c}_{{2}}{y}_{{2}}{\left({x}\right)}+\ldots+{c}_{{n}}{y}_{{n}}{\left({x}\right)}\equiv{0}$ on ${a}\le{x}\le{b}$.

Note that ${c}_{{1}}={c}_{{2}}=\ldots={c}_{{n}}={0}$ is a set of constants that always satisfies the given equation. A set of functions is linearly dependent, if there exists another set of constants, not all zero, that also satisfies the given equation. If the only solution to the equation is ${c}_{{1}}={c}_{{2}}=\ldots={c}_{{n}}={0}$, the set of functions ${\left\{{y}_{{1}}{\left({x}\right)},{y}_{{2}}{\left({x}\right)},\ldots,{y}_{{n}}{\left({x}\right)}\right\}}$ is linearly independent on ${a}\le{x}\le{b}$.

Example 1. The set ${\left\{{2}{x},{x},{\sin{{\left({x}\right)}}}\right\}}$ is linearly dependent on ${\left[-{1},{1}\right]}$, since there exist constants ${c}_{{1}}={1}$, ${c}_{{2}}=-{2}$, ${c}_{{3}}={0}$, such that ${c}_{{1}}\cdot{2}{x}+{c}_{{2}}{x}+{c}_{{3}}{\sin{{\left({x}\right)}}}={2}{x}-{2}{x}+{0}\cdot{\sin{{\left({x}\right)}}}\equiv{0}$.

In general, it is not always easy to test whether the given set of functions is linearly independent by definition. Linear independence can be tested with the Wronskian.

Definition. The Wronskian of a set of functions $\left\{z_1\left(x\right),z_2\left(x\right),\ldots,z_n\left(x\right)\right\}$ on the interval ${a}\le{x}\le{b}$ having the property that each function possesses ${n}-{1}$ derivatives on this interval is the determinant ${W}{\left({z}_{{1}},{z}_{{2}},\ldots,{z}_{{n}}\right)}={\left|\begin{array}{cccc}{z}_{{1}}&{z}_{{2}}&\ldots&{z}_{{n}}\\{z}_{{1}}'&{z}_{{2}}'&\ldots&{z}_{{n}}'\\{z}_{{1}}''&{z}_{{2}}''&\ldots&{z}_{{n}}''\\\ldots&\ldots&\ldots&\ldots\\{{z}_{{1}}^{{{\left({n}-{1}\right)}}}}&{{z}_{{2}}^{{{\left({n}-{1}\right)}}}}&\ldots&{{z}_{{n}}^{{{\left({n}-{1}\right)}}}}\\ \end{array}\right|}$.

Example 2. The Wronskian of the set ${\left\{{1},{\sin{{\left({x}\right)}}}\right\}}$ is ${W}={\left|\begin{array}{cc}{1}&{\sin{{\left({x}\right)}}}\\\frac{{{d}{\left({1}\right)}}}{{{d}{x}}}&\frac{{{d}{\left({\sin{{\left({x}\right)}}}\right)}}}{{{d}{x}}}\\ \end{array}\right|}={\left|\begin{array}{cc}{1}&{\sin{{\left({x}\right)}}}\\{0}&{\cos{{\left({x}\right)}}}\\ \end{array}\right|}={1}\cdot{\cos{{\left({x}\right)}}}-{0}\cdot{\sin{{\left({x}\right)}}}={\cos{{\left({x}\right)}}}$.

Theorem. If the Wronskian of a set of ${n}$ functions defined on the interval ${a}\le{x}\le{b}$ is non-zero for at least one point in this interval, the set of functions is linearly independent there. If the Wronskian is zero on this interval and if each of the functions is a solution to the same linear differential equation, the set of functions is linearly dependent.

CAUTION: The theorem is silent when the Wronskian is zero and the functions are not known to be solutions to the same linear differential equation. In this case, one should test independence directly by definition.

Example 3. Determine whether the set ${\left\{{{e}}^{{x}},{\cos{{\left({x}\right)}}}\right\}}$ is linearly independent on ${\left(-\infty,\infty\right)}$.

The Wronskian of this set is ${W}={\left|\begin{array}{cc}{{e}}^{{{x}}}&{\cos{{\left({x}\right)}}}\\\frac{{{d}{\left({{e}}^{{x}}\right)}}}{{{d}{x}}}&\frac{{{d}{\left({\cos{{\left({x}\right)}}}\right)}}}{{{d}{x}}}\\ \end{array}\right|}={\left|\begin{array}{cc}{{e}}^{{x}}&{\cos{{\left({x}\right)}}}\\{{e}}^{{x}}&-{\sin{{\left({x}\right)}}}\\ \end{array}\right|}=-{{e}}^{{x}}{\sin{{\left({x}\right)}}}-{{e}}^{{x}}{\cos{{\left({x}\right)}}}$.

Since the Wronskian is non-zero for at least one point in the interval of interest, it follows from the theorem that the set is linearly independent.

Theorem. The nth-order linear homogeneous differential equation ${L}{\left({y}\right)}={0}$ always has ${n}$ linearly independent solutions. If ${y}_{{1}}{\left({x}\right)}$, ${y}_{{2}}{\left({x}\right)}$, ..., ${y}_{{n}}{\left({x}\right)}$ represent these solutions, the general solution of ${L}{\left({y}\right)}={0}$ is ${y}{\left({x}\right)}={c}_{{1}}{y}_{{1}}{\left({x}\right)}+{c}_{{2}}{y}_{{2}}{\left({x}\right)}+\ldots+{c}_{{n}}{y}_{{n}}{\left({x}\right)}$, where ${c}_{{1}}$, ${c}_{{2}}$, ..., ${c}_{{n}}$ are arbitrary constants.

Example 4. Since ${y}_{{1}}{\left({x}\right)}={\sin{{\left({x}\right)}}}$ and ${y}_{{2}}{\left({x}\right)}={\cos{{\left({x}\right)}}}$ are solutions to the differential equation ${y}''+{y}={0}$ and these solutions are linearly independent (because their Wronskian ${W}={\left|\begin{array}{cc}{\sin{{\left({x}\right)}}}&{\cos{{\left({x}\right)}}}\\{\cos{{\left({x}\right)}}}&-{\sin{{\left({x}\right)}}}\\ \end{array}\right|}=-{{\sin}}^{{2}}{\left({x}\right)}-{{\cos}}^{{2}}{\left({x}\right)}=-{1}\ne{0}$), the general solution to the differential equation is ${y}{\left({x}\right)}={c}_{{1}}{\sin{{\left({x}\right)}}}+{c}_{{2}}{\cos{{\left({x}\right)}}}$, where ${c}_{{1}}$ and ${c}_{{2}}$ are arbitrary constants.