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#.

Some 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 #(3\times 3)#-matrix:

\[

A= \dfrac{1}{7} \matrix{2 & 3 & 6 \\ -3 & 6 & -2 \\ -6 & -2 & 3 \\ }

\] 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}{7} \matrix{2 & -3 & -6 \\ 3 & 6 & -2 \\ 6 & -2 & 3 \\ }

\]