Complexification

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In mathematics, the complexification of a vector space V over the real number field is the 'corresponding' vector space V^C over the complex number field. That is, it shares the same dimension as V; and a basis for V over the real numbers can serve as a basis for V^C over the complex numbers.

For example, if V consists of the m×n matrices with real coefficients, V^C would consist of m×n complex matrices.

For the sake of having a basis-free definition, one can take

[V^C=V\otimes_} \mathbb],

the tensor product over the real field of V and the complex numbers.

[V^C] is a complex vector space with additional structure: a canonical complex conjugation [\phi]. Indeed, since [V] is included in [V^C] by [v\mapsto v\otimes 1], the complex conjugation is defined by [\phi(v\otimes z) = v\otimes z^*]. This operation is usually denoted by [w^*] or [\overline.]

Conversely, given a complex vector space [W] with a complex conjugation [\phi], [W] is isomorphic as a complex vector space to the complexification [S^C] of the real subspace of [W]: [ S = \ ]. In other words, all complex vector spaces with complex conjugation are the complexification of a real vector space.

For example, when [W=\mathbb] with the standard complex conjugation [\phi(z) = z^*], [S=\mathbb].

Example

Here is how complexification can be used to solve the inhomogeneous differential equation:

[y''(t) + 25y(t) = 3\sin 5t]

First, the solution to the homogeneous equation [y''(t) + 25y(t) = 0] is found. The notation [y_h(t)] denotes the solution to the homogeneous equation.

[s^2 + 25 = 0]

[s = \pm 5i]

[y_h(t) = ke^]

[y_h(t) = c_1\sin 5t + c_2\cos 5t] (by Euler's formula, [c_1] and [c_2] are arbitrary constants)

Now, the particular solution to [3\sin 5t] is solved by moving it to the complex plane. It can be verified using Euler's formula that [\sin 5t] is the imaginary part of [e^]. Note that [e^] is the solution to the homogeneous equation, so we multiply our particular solution by [t].

[y_c(t) = Ate^] (where [A] is a constant)

In order to obtain the constant [A], we plug [y_c(t)], [y_c'(t)], and [y_c''(t)] into the original differential equation.

[y_c'(t) = Ae^ + 5Aite^]

[y_c''(t) = 5Aie^ + 5Aie^ - 25Ate^]

We then obtain:

[(5Ai + 5Ai - 25At + 25At)e^ = 3e^]

[10Ai = 3]

[A = ]

So our complex solution [y_c(t)] becomes:

[y_c(t) = te^]

Because the forcing function is a sine function, we use the imaginary part of the complex solution.

First, we break [e^] into [\cos 5t + i\sin 5t] by Euler's formula.

[y_c(t) = t(\cos 5t + i\sin 5t)]

[y_c(t) = t\cos 5t]

The final solution is the sum of the homogeneous solution and the complex (particular) solution.

[y(t) = c_1\sin 5t + c_2\cos 5t + t\cos 5t]

 

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