### Orthogonal and symmetric maps: Unitary maps

### Diagonal form for unitary maps

*Previously* we saw that, if an orthogonal map leaves invariant a subspace, it also leaves invariant a complementary subspace. The same is true of unitary maps.

Invariant orthogonal complements Let #W# be a linear subspace of a complex finite-dimensional inner product space #V#, and #L:V\rightarrow V# a unitary map that leaves #W# invariant. Then also #W^\perp# is invariant under #L#.

*Earlier* we also saw that orthogonal maps are complex diagonalizable. This also holds for unitary maps.

Unitary maps are diagonalizable Let #L:V\to V# be a unitary map on a complex finite-dimensional inner product space #V#. Then there is an orthonormal basis #\alpha# for #V# such that \(L_\alpha \) is a diagonal matrix with only complex numbers of *absolute value* #1# on the diagonal.

The statements about classifications of orthogonal maps have their analogue unitary versions.

Conjugate unitary maps Let #V# be a complex inner product space of finite dimension and suppose that #L# and #M# are unitary maps #V\to V#. Then the following statements about #L# and #M# are equivalent:

- There is a unitary map #X:V\to V# such that #M = X\, L\, X^{-1}#.
- There is an invertible linear map #Y: V\to V# such that #M = Y\, L\, Y^{-1}#.
- The characteristic polynomials of #L# and #M# are equal.
- The eigenvalues with multiplicities of #L# are the same as those of #M#.

The characteristic polynomial of #A# is \[ p_A(x) = \det(A-x\, I_2) = \left(x-1\right)\cdot \left(x+1\right)\] Therefore, the eigenvalues of #A# are #-1# and #1#. Because #A# is unitary, it is diagonalizable, so #A# is conjugate to \[ D =\matrix{-1&0\\ 0& 1}\] In order to find the requested unitary matrix #T#, we first determine a basis of #\mathbb{C}^2# consisting of eigenvectors of #A#.

The eigenspace of #A# corresponding to #-1# is found by the calculating the nullspace of #A+1\cdot I_2#. A spanning vector is #\rv{ \complexi , 1 } #. Likewise the eigenspace of #A# corresponding to #1# is found by calculating the nullspace of #A- 1\cdot I_2#. A spanning vector is #\rv{ \complexi , -1 } #.

The two spanning vectors of the eigenspaces found are perpendicular to each other. We find an orthonormal basis #\alpha# of #\mathbb{C}^2# by dividing these two vectors by their lengths:

\[\begin{array}{rclcl}\vec{a}_1 &=& \dfrac{1}{\sqrt{2}}\cdot \rv{ \complexi , 1 } &=&\displaystyle \rv{ {{\complexi}\over{\sqrt{2}}} , {{1}\over{\sqrt{2}}} } \\ \vec{a}_2 &=& \dfrac{1}{\sqrt{2}} \cdot \rv{ \complexi , -1 } &=& \displaystyle \rv{ {{\complexi}\over{\sqrt{2}}} , -{{1}\over{\sqrt{2}}} }

\end{array}\] Thus, the basis #\alpha=\basis{\vec{a}_1, \vec{a}_2}# is given by the following matrix #T# whose columns are the two vectors #\vec{a}_1# and # \vec{a}_2#: \[T = \matrix{{{\complexi}\over{\sqrt{2}}} & {{\complexi}\over{\sqrt{2}}} \\ {{1}\over{\sqrt{2}}} & -{{1}\over{\sqrt{2}}} \\ } \]The answer is not unique: the columns of #T# may be swapped because the order of the diagonal elements in #D# is irrelevant; furthermore, the columns of #T# may be multiplied by #-1# independently of each other because the signs of the basis vectors in #\alpha# are irrelevant.

\[\begin{array}{rcl}T^{-1}\, A\, T &=& {\matrix{{{\complexi}\over{\sqrt{2}}} & {{\complexi}\over{\sqrt{2}}} \\ {{1}\over{\sqrt{2}}} & -{{1}\over{\sqrt{2}}} \\ }}^{-1}\, \matrix{0 & -\complexi \\ \complexi & 0 \\ } \, \matrix{{{\complexi}\over{\sqrt{2}}} & {{\complexi}\over{\sqrt{2}}} \\ {{1}\over{\sqrt{2}}} & -{{1}\over{\sqrt{2}}} \\ }\\

&&\phantom{xxx}\color{blue}{\text{matrices substituted}}\\

&=& \matrix{-{{\complexi}\over{\sqrt{2}}} & {{1}\over{\sqrt{2}}} \\ -{{\complexi}\over{\sqrt{2}}} & -{{1}\over{\sqrt{2}}} \\ }\, \matrix{-{{\complexi}\over{\sqrt{2}}} & {{\complexi}\over{\sqrt{2}}} \\ -{{1}\over{\sqrt{2}}} & -{{1}\over{\sqrt{2}}} \\ } \\ &&\phantom{xxx}\color{blue}{\text{inverse determined and }A\, T\text{ computed}}\\

&=& \matrix{-1 & 0 \\ 0 & 1 \\ }

\end{array}\]

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