### Orthogonal and symmetric maps: Unitary maps

### The notion of unitary map

The complex analog of an orthogonal map is a unitary map. As discussed *previously*, the vectors of a complex inner product space have length. Unitary maps are linear maps that retain this length. We will show that results about orthogonal maps have analogs for unitary maps.

Unitary map

Let #V# and #W# be complex inner product spaces.

A map #L :V\rightarrow W# is called an **isometry** if #\norm{L(\vec{x}-\vec{y})}=\norm{\vec{x}-\vec{y}}# for all #\vec{x}#, #\vec{y}# in #V#.

As a consequence, a linear map #L :V\rightarrow W# is an isometry if and only if #\norm{L(\vec{x})}=\norm{\vec{x}}# for all #\vec{x}\in V#.

A linear isometry #L:V\to V# is called a **unitary map**.

The following properties of isometries of complex inner product spaces correspond to the real case.

Properties of isometries Let #U#, #V#, #W# be complex inner product spaces.

- If #L:V\rightarrow W# and #M :U\rightarrow V# are linear isometries, then the composition is #L\,M:U\rightarrow W# a linear isometry.
- If #L:V\rightarrow W# is a linear isometry, then #L# is injective.
- If #L:V\rightarrow V# is unitary and #V# finite dimensional, then #L# is invertible and

also #L^{-1}# is unitary. - The map #L:V\to V# is unitary if and only if, for every orthonormal system #\vec{a}_1,\ldots ,\vec{a}_n# in #V#, the system #L(\vec{a}_1),\ldots , L(\vec{a}_n)# in #V# also is orthonormal.
- If #\vec{a}_1,\ldots ,\vec{a}_n# is an orthonormal basis for #V#, then the map #L:V\to V# is unitary if and only if #L(\vec{a}_1),\ldots , L(\vec{a}_n)# is an orthonormal basis in #V# as well.

Also, the correspondence with matrices runs parallel to the real case.

Unitary matrix A complex #(n\times n)#-matrix #A# is called** unitary** if the columns of #A# form an orthonormal system in #\mathbb{C}^n#.

If #A# is a complex #(n\times n)#-matrix, then we denote by #A^\star# the matrix #{\overline A}^\top#, where #{\overline A}# is the #(n\times n)#-matrix each entry of which is the *complex conjugate* of the corresponding entry of #A#.

By use of the matrix with respect to an orthonormal basis, it can be verified whether a linear map on a finite-dimensional complex inner product space is unitary or not:

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

- The linear map #L# is unitary.
- The matrix #A# is unitary.
- #A^{\star}\, A=I_n#.
- The columns of #A# form an orthonormal system.
- The rows of #A# form an orthonormal system.

Now consider the case where #V = \mathbb{C}^3# and \[\vec{a} = \left[ 1 , \complexi , \complexi+1 \right] \phantom{xxx}\text{ and }\phantom{xxx} \lambda = -1\] Determine the matrix of the reflection #S_{\vec{a},\lambda}#.

We use the mapping rule

\[S_{\vec{a},\lambda}(\vec{x}) = \vec{x} - (1-\lambda)\frac{\dotprod{\vec{x}}{\vec{a}}}{\dotprod{\vec{a}}{\vec{a}}}\cdot \vec{a}\] The proof of this formula follows from the fact that the map defined by the requirement fixes each vector of the hyperplane #\vec{a}^{\perp}# and that #\vec{a}# is an eigenvector with eigenvalue #\lambda#.

Now we substitute the given eigenvector #\vec{a}=\left[ 1 , \complexi , \complexi+1 \right] # and corresponding eigenvalue #\lambda=-1# in the above mapping rule:

\[\begin{array}{rcl}S_{\vec{a},\lambda}(\vec{x})& =&\displaystyle \rv{x_1,x_2,x_3} - 2 \frac{x_{1}-\complexi\cdot x_{2}+\left(1-\complexi\right)\cdot x_{3}}{4}\cdot \left[ 1 , \complexi , \complexi+1 \right] \\ &=& \left[ {{x_{1}+\complexi\cdot x_{2}+\left(\complexi-1\right)\cdot x_{3}}\over{2}} , {{-\complexi\cdot x_{1}+x_{2}-\left(\complexi+1\right)\cdot x_{3}}\over{2}} , {{\left(-\complexi-1\right)\cdot x_{1}+\left(\complexi-1\right)\cdot x_{2}}\over{2}} \right] \end{array}\] Therefore,

\[\begin{array}{rcl}S_{\vec{a},\lambda}(\rv{1,0,0})& =&\displaystyle \left[ {{1}\over{2}} , -{{\complexi}\over{2}} , {{-\complexi-1}\over{2}} \right] \\

S_{\vec{a},\lambda}(\rv{0,1,0})& =&\displaystyle \left[ {{\complexi}\over{2}} , {{1}\over{2}} , {{\complexi-1}\over{2}} \right] \\

S_{\vec{a},\lambda}(\rv{0,0,1})& =&\displaystyle \left[ {{\complexi-1}\over{2}} , {{-\complexi-1}\over{2}} , 0 \right]

\end{array}\]We conclude that the matrix of \(S_{\vec{a},\lambda}(\vec{x})\) equals \[ \matrix{{{1}\over{2}} & {{\complexi}\over{2}} & {{\complexi-1}\over{2}} \\ -{{\complexi}\over{2}} & {{1}\over{2}} & {{-\complexi-1}\over{2}} \\ {{-\complexi-1}\over{2}} & {{\complexi-1}\over{2}} & 0 \\ } \]

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