numpy.vdot() in Python (original) (raw)

Last Updated : 08 Mar, 2024

Prerequisite – numpy.dot() in Python
numpy.vdot(vector_a, vector_b) returns the dot product of vectors a and b. If first argument is complex the complex conjugate of the first argument(this is where vdot() differs working of dot() method) is used for the calculation of the dot product. It can handle multi-dimensional arrays but working on it as a flattened array.

Parameters –

  1. vector_a : [array_like] if a is complex its complex conjugate is used for the calculation of the dot product.
  2. vector_b : [array_like] if b is complex its complex conjugate is used for the calculation of the dot product.

Return – dot Product of vectors a and b.

Code 1 :

import numpy as geek

vector_a = 2 + 3j

vector_b = 4 + 5j

product = geek.vdot(vector_a, vector_b)

print ( "Dot Product : " , product)

Output :

Dot Product : (23-2j)

How Code1 works ?
vector_a = 2 + 3j
vector_b = 4 + 5j

As per method, take conjugate of vector_a i.e. 2 – 3j

now dot product = 2(4 – 5j) + 3j(4 – 5j)
= 8 – 10j + 12j + 15
= 23 – 2j

Code 2 :

import numpy as geek

vector_a = geek.array([[ 1 , 4 ], [ 5 , 6 ]])

vector_b = geek.array([[ 2 , 4 ], [ 5 , 2 ]])

product = geek.vdot(vector_a, vector_b)

print ( "Dot Product : " , product)

product = geek.vdot(vector_b, vector_a)

print ( "\nDot Product : " , product)

Output :

Dot Product : 55

Dot Product : 55

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