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 #(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:
\[
A^{-1}=\dfrac{1}{25} \matrix{24 & -7 \\ 7 & 24 \\ }
\]
Orthogonal matrices
