Mathematical principles in image processing 18--inner product and outer product

mathematical Principles in image processing 18--inner product and outer productTags: image processing mathematical internal product 2015-12-09 15:42 5172 people read reviews (1) Favorite Report Category: Mathematics (44)

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inner product and outer product of 1.3.2

Because cos (Π/2) = 0. Of course, this is also a number of textbooks on the introduction of vector product at the beginning of the most often used to define the way. But it must be made clear that this representation is only a very narrow definition. If we introduce the inner product of vector from this definition, it is actually putting the cart before the horse. Because for high-dimensional vectors, the meaning of the angle is ambiguous. For example, in three-dimensional coordinate space, and then introduce one-dimensional time coordinate to form a four-dimension space, then how to explain the angle between the time vector and the space vector. So the reader must be clear, first of all should be given as this section at the beginning of the definition of the inner product, and then the two-dimensional or three-dimensional space under the definition of the angle. On this basis, we will prove the cosine law.

If according to a B = |a| | b|cosθ this definition, because 0<=cosθ<=1, apparently Cauchy-Schwartz inequality is established. But such a way of proving the same also made the mistake of putting the cart before the horse. The Cauchy-Schwartz inequality does not define the dimensions of the vector, in other words it is a vector of any dimension, and the definition of the angle is ambiguous. The right idea should also prove the Cauchy-Schwartz inequality from the very beginning of this section, because there is an inequality relationship, and then we think that there is a coefficient between the product and the vector modulus between 0 and 1, and then we use cosθ to express this coefficient, so we get a B = |a| | b|cosθ the expression. Here's a proof of the Cauchy-Schwartz inequality.

Prove:

Similar to the inner product, the outer product of the vector A, B, can also be narrowly defined as

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