Inner Product, Orthogonality and Length of Vectors

The inner product , the condition of orthogonality and the length of vectors are presented through examples including their detailed solutions.

Definition of the Inner Product of two Vectors

Let vectors \( \textbf {x} \) and \( \textbf y \) be two column vectors (or an \( n \) by \( 1 \) matrix) defined by
\( \textbf x = \begin{bmatrix} x_1 \\ x_2 \\ . \\ . \\ . \\ x_n \end{bmatrix} \) and \( \textbf y = \begin{bmatrix} y_1 \\ y_2 \\ . \\ . \\ . \\ y_n \end{bmatrix} \)
The inner product of \( \textbf x \) and \(\textbf y \) is a scalar quantity written as \( \textbf x \cdot \textbf y \) defined by
\[ \textbf x. \textbf y = \textbf {x}^T \textbf y = \begin{bmatrix} & x_1 & x_2 & . . . & x_n\\ \end{bmatrix} \begin{bmatrix} y_1 \\ y_2 \\ . \\ . \\ . \\ y_n \end{bmatrix} = x_1 y_1 + x_2 y_2 + ... + x_n y_n\] where \( \textbf {x}^T \) is the transpose of vector \( \textbf {x} \)


Properties of the Inner Product

If \( \textbf x \) , \( \textbf y \) and \( \textbf z \) are vectors in \( R^n \) and \( k_1 \) and \( k_2 \) are scalars, then

  1. \( \textbf x \cdot \textbf y = \textbf y \cdot \textbf x \)
  2. \( (\textbf x + \textbf y) \cdot \textbf z = \textbf x \cdot \textbf z + \textbf y \cdot \textbf z \)
  3. \( (k_1 \textbf x) \cdot (k_2 \textbf y) = k_1 k_2 (\textbf x \cdot \textbf y) \)


Orthogonal Vectors

Vectors \( \textbf x \) and \( \textbf y \) are called orthogonal if
\[ \textbf x \cdot \textbf y = 0 \]


Definition of the Length (or Norm) of a Vector and Unit Vector

The length (or norm ) of vector \( \textbf x = \begin{bmatrix} x_1 \\ x_2 \\ . \\ . \\ . \\ x_n \end{bmatrix} \) written as \( || \textbf x || \) is given by
\[ || \textbf x || = \sqrt {x_1^2 + x_2^2 + .... + x_n^2} = \sqrt {\textbf x \cdot \textbf x} \]
From the above definition, we can easily conclude that
\( || \textbf x || \ge 0 \) and \( || \textbf x ||^2 = \textbf x \cdot \textbf x \)
A unit vector is a vector whose length (or norm) is equal to 1.


Pythagorean Theorem

Vectors \( \textbf x \) and \( \textbf y \) are orthogonal if and only if
\[ ||x+y||^2 = ||x||^2 + ||y||^2 \]


Distance Between two Vectors

The distance between vectors \( \textbf x \) and \( \textbf y \) is defined as
\[ dist(\textbf x,\textbf y) = || \textbf x - \textbf y || \]


Examples with Solutions

Example 1
Given the vectors \( \textbf x = \begin{bmatrix} -2 \\ 3 \\ 0 \\ -1 \end{bmatrix} \) , \( \textbf y = \begin{bmatrix} 3 \\ -1 \\ 4 \\ 0 \end{bmatrix} \) , \( \textbf z = \begin{bmatrix} 2 \\ -1 \\ 4 \\ -7 \end{bmatrix} \), find
a) \( \textbf x \cdot \textbf y \) and \( \textbf y \cdot \textbf x \)
b) \( \textbf x \cdot \textbf y + \textbf x \cdot \textbf z \) and \( \textbf x \cdot (\textbf y + \textbf z) \)
c) \( (3 \textbf x ) \cdot (-2\textbf z) \)

Solution to Example 1
Use the definition given above
a)
\( \textbf x \cdot \textbf y = (-2)(3) + 3(-1) + 0(4) + (-1)0 = - 9 \)
Using property 1 of the inner product above
\( \textbf y \cdot \textbf x = \textbf x \cdot \textbf y = - 9 \)

b)
\( \textbf x \cdot \textbf y + \textbf x \cdot \textbf z = - 9 + 23 = 14 \)
According to property 2 of the inner product
\( \textbf x \cdot (\textbf y + \textbf z) = \textbf x \cdot \textbf y + \textbf x \cdot \textbf z = 14 \)

c)
According to property 3 of the inner product
\( (3 \textbf x ) \cdot (-2\textbf z) = (3)(-2) \textbf x \cdot \textbf y = -6 (-9) = 54 \)



Example 2
a) Show that vectors \( \textbf x = \begin{bmatrix} -2 \\ 3 \\ 0 \end{bmatrix} \) and \( \textbf y = \begin{bmatrix} 3 \\ 2 \\ 4 \end{bmatrix} \) are orthogonal.
b) Find the constant \( a \) and \( b \) so that the vector \( \textbf z = \begin{bmatrix} a \\ b \\ 4 \end{bmatrix} \) is orthogonal to both vectors \( \textbf x \) and \( \textbf y \)

Solution to Example 2
Calculate the inner product of vectors \( \textbf x \) and \( \textbf y \)
a)
\( \textbf x \cdot \textbf y = (-2)(3) + 3(2) + 0(4) = 0 \)
Since the inner product of vectors \( \textbf x \) and \( \textbf y \) is equal to zero, the two vectors are orthogonal.

b)
We first calculate the following inner product
\( \textbf x \cdot \textbf z = -2(a) + 3(b) + 0(4) = -2a + 3b \)
\( \textbf y \cdot \textbf z = 3(a) + 2(b) + 4(4) = 3a + 2b + 16 \)
For vector \( \textbf z \) to be orthogonal to both \( \textbf x \) and \( \textbf y \), both inner product calculated above must be equal to zero. Hence the system of equations to solve
\( -2a + 3b =0 \\ 3a + 2b + 16 = 0 \)
Solve the above system to to obtain
\( a = -\dfrac{48}{13} , b = - \dfrac{32}{13} \)



Example 3
Let \( \textbf x = \begin{bmatrix} \sqrt 2 \\ 0 \\ 1 \end{bmatrix} \) and \( \textbf y = \begin{bmatrix} 0 \\ \sqrt 5 \\ 0 \end{bmatrix} \).
Find \( || \textbf x || \), \( || \textbf y || \) and \( || \textbf x + \textbf y|| \) and compare \( || \textbf x ||^2 + || \textbf y ||^2 \) and \( || \textbf x + \textbf y||^2 \)

Solution to Example 3
Use formula for the definition of the length of a vector
\( || \textbf x || = \sqrt { (\sqrt 2)^2 + 0^2 + 1^2 } = \sqrt 3 \)
\( || \textbf y || = \sqrt { 0^2 + (\sqrt 5)^2 + 0^2 } = \sqrt 5 \)
\( || \textbf x + \textbf y|| = \sqrt { (\sqrt 2)^2 + (\sqrt 5)^2 + 1} = \sqrt 8\)
\( || \textbf x ||^2 + || \textbf y ||^2 = 3 + 5 = 8 \)
\( || \textbf x + \textbf y||^2 = 8 \)
We notice that \( || \textbf x + \textbf y||^2 = || \textbf x ||^2 + || \textbf y ||^2 \) and that is becaues vectors \( \textbf x \) and \( \textbf y \) are orthogonal (the inner product of two vectors is equal to 0) which verify the Pythagorean theorem above.


More References and links

  1. Vector Spaces - Questions with Solutions
  2. Linear Algebra and its Applications - 5 th Edition - David C. Lay , Steven R. Lay , Judi J. McDonald
  3. Elementary Linear Algebra - 7 th Edition - Howard Anton and Chris Rorres