Orthogonal Matrices and their examples

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Orthogonal Matrices and their examples

A square matrix A over R fot which [katex]A^T=A^{-1}[/katex], or equivalently [katex]AA^T=A^TA=I[/katex], is called an orthogonal matrix . Clearly, an orthogonal matrix is non singular.

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Therorem: The following conditions for a square matrix A are equivalent:

    1. A is orthogonal.
    1. The rows of A form an orthogonal set.
    1. The columns of A form an orthogonal set.

Example of Orthogonal Matrices

Show that the rows (columns) of the matrix

A=\left[\begin{array}{ccc}\cos \theta & -\sin \theta & 0 \\ \sin \theta & \cos \theta & 0 \\ 0 & 0 & 1\end{array}\right]

form an orthonormal set.

Solution. Since the rows (columns) of A form an orthonormal set if and only if A is orthogonal, we have only to show that [katex]A A^{T}=I[/katex]. Now

A A^2=\left[\begin{array}{ccc}\cos \theta & -\sin \theta & 0 \\ \sin \theta & \cos \theta & 0 \\ 0 & 0 & 1\end{array}\right]\left[\begin{array}{ccc}\cos \theta & \sin \theta & 0 \\ -\sin \theta & \cos \theta & 0 \\ 0 & 0 & 1\end{array}\right]
=\left[\begin{array}{ccc}\cos ^2 \theta+\sin ^2 \theta & \cos \theta \sin \theta-\cos \theta \sin \theta & 0 \\ \sin \theta \cos \theta-\cos \theta \sin \theta & \sin ^2 \theta+\cos ^2 \theta & 0 \\ 0 & 0 & 0\end{array}\right]
=\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]=I.

Similarly, [katex]\quad A^{T}A=I[/katex]. Thus A is orthogonal and so rows (columns) of A form an orthonormal set.

Example of an orthogonal matrix

Example 14. Find an orthogonal matrix A whose first row is [katex]\left(\frac{1}{3}, \frac{2}{3}, \frac{2}{3}\right)[/katex]. Solution. Let [katex]v_1=\left(\frac{1}{3}, \frac{2}{3}, \frac{2}{3}\right)[/katex] and [katex]w_2=(x, y, z)[/katex] be orthogonal to [katex]v_1[/katex].

Then [katex]\left\langle v_1, w_2\right\rangle=0[/katex] 

\Rightarrow \quad \frac{1}{3} x+\frac{2}{3} y+\frac{2}{3} z=0 \text { of } x+2 y+2 z=0 \text {. }

If we take [katex]\quad x=0,2 y+2 z=0[/katex] or z=-y, then

\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right]=\left[\begin{array}{c}
0 \\
y \\
-y
\end{array}\right]=y\left[\begin{array}{r}
0 \\
1 \\
-1
\end{array}\right]

So we may take [katex]w_2=(0,1,-1)[/katex]. Normalize [katex]w_2[/katex] to obtain [katex]v_2=\left(0, \frac{1}{\sqrt{2}},-\frac{1}{\sqrt{2}}\right)[/katex] as second row of A. Take [katex]w_3=(x, y, z)[/katex] so that [katex]w_3[/katex] is orthogonal to both [katex]v_1[/katex] and [katex]v_2[/katex]. Now [katex]\left\langle v_1, w_3\right\rangle=0[/katex] 

\Rightarrow \frac{1}{3} x+\frac{2}{3} y+\frac{2}{3} z=0\\
\left\langle w_2, w_3\right\rangle=0 \\
\Rightarrow \frac{1}{\sqrt{2}} y-\frac{1}{\sqrt{2}} z=0
\Rightarrow x+2 y+2 z=0
\Rightarrow \quad y=z=0
\Rightarrow \quad y=z
\Rightarrow x+4 z=0 \Rightarrow x=-4z

Thus

\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\left[\begin{array}{r}-4 z \\ z \\ z\end{array}\right]=-z\left[\begin{array}{r}4 \\ -1 \\ -1\end{array}\right]

Thus we can take [katex]w_3=(4,-1,-1)[/katex]. Normalizing [katex]w_3[/katex], we get [katex]y=\left(\frac{4}{\sqrt{18}}, \frac{-1}{\sqrt{18}}, \frac{-1}{\sqrt{18}}\right)[/katex] as third row of A.

\text {Hence } A=\left[\begin{array}{ccc}\frac{1}{3} & \frac{2}{3} & \frac{2}{3} \\ 0 & \frac{1}{\sqrt{2}} & \frac{-1}{\sqrt{2}} \\ \frac{4}{3 \sqrt{2}} & \frac{-1}{3 \sqrt{2}} & \frac{-1}{3 \sqrt{2}}\end{array}\right]

Note that [katex]v_3[/katex] could also be obtained as:

v_3 =\left[\begin{array}{ccc}
e_1 & e_2 & e_3 \\
\frac{1}{3} & \frac{2}{3} & \frac{2}{3} \\
0 & \frac{1}{\sqrt{2}} & \frac{-1}{\sqrt{2}}
\end{array}\right]\\
=\frac{-4}{3 \sqrt{2}} c_1+\frac{1}{3 \sqrt{2}} e_2+\frac{1}{3 \sqrt{2}} e_3 \\

=\left(\frac{-4}{3 \sqrt{2}}, \frac{1}{3 \sqrt{2}}, \frac{1}{3 \sqrt{2}}\right)

instead of solving the linear equations. Note: The above matrix A is not unique.

Alternative Method

Let [katex]R_1=\left(\frac{1}{3}, \frac{2}{3}, \frac{2}{3}\right)[/katex]. A vector (row) orthogonal to [katex]R_1[/katex] is

R_1^* =\left|\begin{array}{cc}
e_1 & e_2 \\
\frac{1}{3} & \frac{2}{3}
\end{array}\right|+\left|\begin{array}{cc}
e_2 & e_3 \\
\frac{2}{3} & \frac{2}{3}
\end{array}\right|+\left|\begin{array}{cc}
e_3 & e_1 \\
\frac{2}{3} & \frac{1}{3}
\end{array}\right|
=\frac{1}{3} e_2-\frac{1}{3} e_3=\left(0, \frac{1}{3}, \frac{-1}{3}\right)

and

R_3^{\prime} =\left|\begin{array}{ccc}
e_1 & e_2 & e_3 \\
\frac{1}{3} & \frac{2}{3} & \frac{2}{3} \\
0 & \frac{1}{3} & -\frac{1}{3}
\end{array}\right|
=-\frac{4}{9} e_1+\frac{1}{9} e_2+\frac{1}{9} e_3=\left(\frac{-4}{9}, \frac{1}{9}, \frac{1}{9}\right) .

Now

R_1=\left(\frac{1}{3}, \frac{2}{3}, \frac{2}{3}\right),
R_2=\frac{R_1^*}{\left||R_1^*\right||}=\left(0, \frac{1}{\sqrt{2}}, \frac{-1}{\sqrt{2}}\right)
R_3=\frac{R_3^{\prime}}{\left|R_3^{\prime}\right|}=\left(\frac{-4}{3 \sqrt{2}}, \frac{1}{3 \sqrt{2}}, \frac{1}{3 \sqrt{2}}\right)

give the required matrix as

A=\left[\begin{array}{ccc}
\frac{1}{3} & \frac{2}{3} & \frac{2}{3} \\
\\
0 & \frac{1}{\sqrt{2}} & \frac{-1}{\sqrt{2}} \\
\\
\frac{-4}{3 \sqrt{2}} & \frac{1}{3 \sqrt{2}} & \frac{1}{3 \sqrt{2}}
\end{array}\right]

EXERCISE

    1. If A is an orthogonal matrix, show that det A=1 or -1.
    1. Find an orthogonal matrix whose first row is [katex]\left(\frac{1}{\sqrt{5}}, \frac{2}{\sqrt{5}}\right)[/katex]
    1. Find an orthogonal matrix whose first row is a multiple of (1,1,1).
    1. Find an orthogonal matrix whose first row is [katex]\left(0, \frac{1}{\sqrt{5}}, \frac{2}{\sqrt{5}}\right)[/katex]
    1. Show that the products and inverses of orthogonal matrices are orthogonal. Hence show that orthogonal matrices form a group under multiplication.

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