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

Let vectors x and y be two column vectors (or an n 1 matrix) defined by

The inner product of x and y is a scalar quantity written as x · y defined by

If x, y and z are vectors in R^{n} and
k_{1} and k_{2}
are scalars, then

- x · y = y · x
- (x + y)z = x · z + y · z
- (k
_{1}x) · (k_{2}y) = k_{1}k_{2}(x · y)

Vectors x and y are called orthogonal if

x · y = 0

\( \) \( \) \( \)
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.

Vectors \( \textbf x \) and \( \textbf y \) are orthogonal if and only if

\[ ||x+y||^2 = ||x||^2 + ||y||^2 \]

The distance between vectors \( \textbf x \) and \( \textbf y \) is defined as

\[ dist(\textbf x,\textbf y) = || \textbf x - \textbf y || \]

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.

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