Orthogonal matrices

Orthogonal and symmetric maps: Orthogonal maps

Orthogonal matrices

We now focus on orthogonal maps in inner product spaces $\mathbb{R}^n$ (with standard inner product) and their matrices. Because the standard basis is an orthonormal basis of $\mathbb{R}^n$ , it follows from theorem Orthogonal maps and orthonormal systems that a linear map $L:\mathbb{R}^n\rightarrow\mathbb{R}^n$ is orthogonal if and only if $L(\vec{e}_1),\ldots ,L(\vec{e}_n)$ is an orthonormal system. This explains the following definition of orthogonality for a matrix.

A real $(n\times n)$ -matrix $A$ is called orthogonal if the columns form an orthonormal system in $\mathbb{R}^n$ .

ExamplesSome examples of orthogonal matrices are

$\matrix{\frac{1}{\sqrt{2}}&\frac{1}{\sqrt{2}}\\\frac{1}{\sqrt{2}}&-\frac{1}{\sqrt{2}}}\quad\quad \matrix{\frac{1}{3}&\frac{2}{3}&-\frac{2}{3}\\-\frac{2}{3}&\frac{2}{3}&\frac{1}{3}\\\frac{2}{3}&\frac{1}{3}&\frac{2}{3}}$

This can be verified directly via the definition. The following theorem gives a number of other methods.

Permutation matrices are orthogonal. After a suitable re-ordering, their columns form the standard basis of $\mathbb{R}^n$ .

If the columns of an $(n\times n)$ -matrix $A$ are independent, the matrix $Q$ of a QR decomposition $A=Q\,R$ is orthogonal.

We formulate the relationship between orthogonality of the matrix $A$ and map $L_A$ determined by $A$ and some other characteristics of orthogonality.

Let $L: \mathbb{R}^n \rightarrow \mathbb{R}^n$ be a linear map with matrix $A$ . Then the following statements are equivalent:

The linear map $L$ is orthogonal.
The matrix $A$ is orthogonal.
$A^{\top}\, A=I_n$ .
The matrix $A$ is invertible with inverse $A^{-1} = A^\top$ .
The matrix $A$ is invertible and $A^{-1}$ is orthogonal.
The rows of $A$ form an orthonormal system.

$1.\Leftrightarrow 2.$ As discussed, it follows from theorem Orthogonal maps and orthonormal systems that $L_A:\mathbb{R}^n\rightarrow\mathbb{R}^n$ is orthogonal if and only if $A\,\vec{e}_1,\ldots ,A\vec{e}_n$ is an orthonormal system, so if and only if $A$ is orthogonal.

$2.\Leftrightarrow 3.$ The $(i,j)$ -entry of the matrix product $A^\top\, A$ is the inner product of the $i$ -th and the $j$ -th column of $A$ . Thus, verifying that the columns of $A$ form an orthonormal system amounts to checking that $A^{\top}\, A=I_n$ .

$3.\Leftrightarrow 4.$ This follows at once from the theorem Inverse of a matrix.

$4.\Rightarrow 5.$ Suppose that $A$ is invertible with inverse $A^{-1} = A^\top$ . Then, by the Rules for the inverse of a matrix,

$\begin{array}{rcl} (A^{-1})^\top\, A^{-1}&=& (A^\top)^{-1}\, A^{-1} = (A\,A^\top)^{-1} = I_n^{-1} = I_n\end{array}$ This shows that $A^{-1}$ is invertible with inverse $(A^{-1})^{\top}$ . The matrix $A^{-1}$ thus satisfies statement 3 and so, because of the equivalence of statements 2 and 3, is orthogonal.

$5.\Rightarrow 4.$ Suppose that $A$ is invertible and that $A^{-1}$ is orthogonal. According to the Rules for the inverse of a matrix, $A$ is invertible with inverse $((A^{-1})^\top)^{-1} =A^\top$ . This shows that $A$ satisfies statement 4.

$5.\Leftrightarrow 6.$ The foregoing items of the proof show that statement 5 is equivalent to the orthogonality of $A$ and the orthogonality of $A^\top$ . By the definition of orthogonality of a matrix, statement 5 is therefore equivalent to the statement that the columns of $A^\top$ form an orthonormal system, and so also to the statement that the rows of $A$ form an orthonormal system. This is the content of statement 6.

By statement 4, $A^\top$ is the inverse of $A$ . This shows that the inverse of an orthogonal matrix is easy to calculate.

Also, the inverse of $A$ is orthogonal if $A$ is orthogonal. After all, because ${(A^\top)}^\top = A$ , we have

$I_n=A^\top\, A= A^\top {(A^\top)}^\top$ This shows that the transpose of $A^\top$ is the inverse of $A^\top$ , so $A^\top$ is orthogonal. But (again because of statement 3) $A^\top$ is the inverse of $A$ , so $A^{-1}$ is orthogonal.

Let $A$ be the following $(2\times 2)$ -matrix:

$A= \dfrac{1}{25} \matrix{24 & 7 \\ -7 & 24 \\ }$ Is $A$ orthogonal?

Yes

Each column of the matrix $A$ has length $1$ and the inner product of each pair of distinct columns equals $0$ . Therefore, the matrix $A$ is orthogonal.

Also, the rows of $A$ have length $1$ and their mutual inner products are equal to $0$ . In order to determine the inverse of $A$ , we only need to transpose the matrix:

$A^{-1}=\dfrac{1}{25} \matrix{24 & -7 \\ 7 & 24 \\ }$

New example