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An inner product is a map

$\langle\cdot,\cdot\rangle:V\times V\to F$

of two vectors producing a scalar in the field $F$ (either the field of real numbers $\R$ , or complex numbers $\C$) that satisfies the following conditions for all vectors in V and all scalars in F:

$\langle u,v\rangle=\overline{\langle v,u\rangle}$
$\langle au,v\rangle=a\langle u,v\rangle$
$\langle u+v,w\rangle=\langle u,w\rangle+\langle v,w\rangle$
• Positive-definiteness:
$\langle u,u\rangle\ge0$
$\langle u,u\rangle=0$ iff u = 0.

For the vectors

$u=(u_1,\ldots,u_i)\ ,\ v=(v_1,\ldots,v_i)$

the inner product is computed as

$\langle u,v\rangle=\sum_{n=0}^i u_i\overline{v_i}$

Since the conjugate of $z$ is equal to $z$ for real numbers, if all elements of both vectors have no imaginary components this merely reduces to the dot product. In this sense, the inner product can be thought of an extension of the dot product to the complex plane. They are also similar to the outer product in that whereas an inner product is given by

$\langle u,v\rangle=u^\mathrm{H}v$

where H is the conjugate transpose, the outer product is given by

$u\otimes v=uv^\mathrm{H}$

As such, the inner product of two vectors is the trace of the outer product.

Functions can be treated as vectors with continuous, rather than discrete components; as such, they have an inner product. The inner product of $f$ and $g$ with the domain between $a$ and $b$ is

$\langle f,g\rangle=\int\limits_a^b f\cdot\overline{g}\,dx$

If this is equal to 0, the functions are said to be orthogonal on the interval (unlike with vectors, this has no geometric significance). This definition is useful in Fourier analysis.

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