Diagonal form for unitary maps

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.

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 $L$ .

The proof of this statement is very similar to the real case.

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

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.

Suppose that $\lambda$ is a complex eigenvalue of $L$ . We show that the generalized eigenspace for $\lambda$ coincides with the eigenspace $\ker{L-\lambda\,I_V}$ . For this purpose, it is sufficient to show that $\ker{(L-\lambda\,I_V)^2}$ is equal to $\ker{L-\lambda\,I_V}$ . Let $\vec{w}$ be a vector in $\ker{(L-\lambda\,I_V)^2}$ and write $\vec{v} = (L-\lambda\, I_V)(\vec{w})$ . Then $\vec{v}$ belongs to $\ker{L-\lambda\,I_V}$ , so $\vec{v}$ is an eigenvector of $L$ with eigenvalue $\lambda$ .

We examine what it means for $L$ to preserve the inner product of $\vec{v}$ and $\vec{w}$ :

$\begin{array}{rcl}\dotprod{\vec{v}}{\vec{w}}& =& \dotprod{L(\vec{v})}{L(\vec{w})} \\&&\phantom{xxx}\color{blue}{L\text{ is unitary}}\\&=& \dotprod{(\lambda\cdot\vec{v})}{(\lambda\,\vec{w}+\vec{v})} \\&&\phantom{xxx}\color{blue}{\vec{v}\text{ lies in }\ker{L-\lambda\,I_V}\text{ and }\vec{v} = L(\vec{w})-\lambda\, \vec{w}}\\&=& {\lambda\cdot\overline\lambda}\cdot\dotprod{\vec{v}}{\vec{w}}+\lambda\cdot\dotprod{\vec{v}}{\vec{v}} \\&&\phantom{xxx}\color{blue}{\text{hermiticity of the inner product }}\\&=& \dotprod{\vec{v}}{\vec{w}}+\lambda\cdot\dotprod{\vec{v}}{\vec{v}} \\&&\phantom{xxx}\color{blue}{|\lambda| = 1}\end{array}$

By subtracting $\dotprod{\vec{v}}{\vec{w}}$ from both sides, we see that $\lambda\cdot\dotprod{\vec{v}}{\vec{v}}=0$ . Because $|\lambda| = 1$ , we have $\lambda\ne0$ , so $\dotprod{\vec{v}}{\vec{v}}=0$ . Because of the positive definiteness of the inner product, this implies $\vec{v} = \vec{0}$ . We conclude that $\vec{w}$ belongs to $\ker{L-\lambda\,I_V}$ . This shows that $\ker{(L-\lambda\,I_V)^2}=\ker{L-\lambda\,I_V}$ . By induction with respect to the exponent $j$ we derive immediately from this that $\ker{(L-\lambda\,I_V)^j}=\ker{L-\lambda\,I_V}$ for all natural numbers $j$ , so the generalized eigenspace of $L$ for $\lambda$ is equal to its eigenspace.

According to theorems Direct sum decomposition and Generalized eigenspaces, $V$ is the direct sum of the generalized eigenspaces of $L$ . Because of the above, $V$ is the direct sum of the eigenspaces of $L$ . The eigenspaces are perpendicular to each other: If $\lambda$ and $\mu$ are eigenvalues with eigenvectors $\vec{v}$ and $\vec{w}$ , respecively, then we have $\dotprod{\vec{v}}{\vec{w}} = \dotprod{L(\vec{v})}{L(\vec{w})} = \dotprod{(\lambda\cdot \vec{v})}{(\mu\cdot \vec{w})}=\lambda\cdot\overline{\mu}\cdot \dotprod{\vec{v}}{\vec{w}}$ After multiplication by $\mu$ (which has absolute value $1$ because it is the eigenvalue of a unitary map), this leads to $(\lambda-\mu)\cdot (\dotprod{\vec{v}}{\vec{w}} )= 0$ Therefore, if $\lambda\ne \mu$ , then $\dotprod{\vec{v}}{\vec{w}} = 0$ , that is, $\vec{v}$ is perpendicular to $\vec{w}$ .

So, if we choose orthonormal bases for each of the eigenspaces, then $\alpha$ , the composition of these orthonormal bases, is an orthonormal basis of $V$ and the matrix $L_\alpha$ is diagonal.

According to the theorem, eigenvectors of $L$ with different eigenvalues are perpendicular to each other. This makes it easy to find an orthonormal basis of eigenvectors.

Whereas the orthogonal Jordan normal form still may have $(2\times2)$ -submatrices corresponding to rotations, the theorem tells us that unitary Jordan (normal) form of a unitary map is always diagonal. On the diagonal are the eigenvalues, which all have absolute value $1$ .

The converse is also true: Suppose that $L:V\to V$ is a linear map on a complex finite-dimensional inner product space $V$ . If there is an orthonormal basis $\alpha$ for $V$ such that $L_\alpha$ is a diagonal matrix with only complex numbers with absolute value $1$ on the diagonal, then $L$ is unitary.

The absolute value of the determinant of a unitary matrix is equal to $1$ . Indeed, the determinant of a matrix does not depend on the basis chosen for the vectorspace $V$ of the map $L:V\to V$ determined by that matrix. Thanks to the above theorem, we may choose the basis $\alpha$ such that the matrix $L_\alpha$ of a unitary map $L$ with respect to $\alpha$ is diagonal with only numbers of absolute value $1$ on the diagonal. The matrix $L_\alpha$ is conjugate with the original matrix and therefore has the same determinant. The determinant of a diagonal matrix is equal to the product of the diagonal elements, so that the absolute value of the determinant of $L_\alpha$ equals $1$ .

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

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 $M$ .

We prove the implications according to the scheme $1\Rightarrow 2\Rightarrow 3\Rightarrow 4\Rightarrow 1$ . This suffices for the proof of all equivalences.

$1\Rightarrow 2$ is trivial.

$2\Rightarrow 3$ holds because it is known from The characteristic polynomial of a linear mapping that the polynomial does not depend on the choice of a basis.

$3\Rightarrow 4$ Statements 3 and 4 are equivalent because the characteristic polynomial is the product of the linear factors $x-\lambda$ , where $\lambda$ runs over all eigenvalues (with the appropriate multiplicities).

$4\Rightarrow 1$ Suppose that statement 4 holds. From the above theorem Unitary maps are diagonalizable, it follows that there are orthonormal bases $\alpha$ and $\beta$ such that $L_\alpha$ and $M_\beta$ are diagonal matrices. These diagonal matrices for $L$ and $M$ have the same diagonal entries, namely the eigenvalues. We may assume that $L_\alpha = M_\beta$ so $\alpha \,L\,\alpha^{-1}=\beta \,M\,\beta^{-1}$ . Put $X =\beta^{-1} \alpha$ . Because $\alpha$ and $\beta$ are orthonormal bases, the corresponding linear transformations are unitary. Because of Properties of isometries, also $X$ is unitary. Finally, we have

$X\, L\, X^{-1} = \beta^{-1} \alpha \, L\, (\beta^{-1} \alpha)^{-1}= \beta^{-1} \alpha \, L\, \alpha^{-1} \beta = \beta^{-1}\beta \,M\,\beta^{-1} \beta = M$ Thus statement 1 is derived from statement 4.

We say that two square matrices $A$ and $B$ are unitary conjugate if there is a unitary matrix $T$ with $A = T\, B \, T^{-1}$ . This defines an equivalence relation. The proof of transitivity is similar to the proof of the implication $4\Rightarrow 1$ of the theorem.

The matrix $A = \matrix{0 & -\complexi \\ \complexi & 0 \\ }$ is unitary. Give a unitary matrix $T$ such that $T^{-1} \, A \, T$ is diagonal.

$T =$ $\matrix{{{\complexi}\over{\sqrt{2}}} & {{\complexi}\over{\sqrt{2}}} \\ {{1}\over{\sqrt{2}}} & -{{1}\over{\sqrt{2}}} \\ }$

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.

The answer can be verified by conjugation of $A$ by the inverse of the answer and verifying whether this is the diagonal matrix $D$ . We carry this check out for the answer $T = \matrix{{{\complexi}\over{\sqrt{2}}} & {{\complexi}\over{\sqrt{2}}} \\ {{1}\over{\sqrt{2}}} & -{{1}\over{\sqrt{2}}} \\ }$ :
$\begin{array}{rcl}T^{-1}\, A\, T &=& {\matrix{{{\complexi}\over{\sqrt{2}}} & {{\complexi}\over{\sqrt{2}}} \\ {{1}\over{\sqrt{2}}} & -{{1}\over{\sqrt{2}}} \\ }}^{-1}\, \matrix{0 & -\complexi \\ \complexi & 0 \\ } \, \matrix{{{\complexi}\over{\sqrt{2}}} & {{\complexi}\over{\sqrt{2}}} \\ {{1}\over{\sqrt{2}}} & -{{1}\over{\sqrt{2}}} \\ }\\ &&\phantom{xxx}\color{blue}{\text{matrices substituted}}\\ &=& \matrix{-{{\complexi}\over{\sqrt{2}}} & {{1}\over{\sqrt{2}}} \\ -{{\complexi}\over{\sqrt{2}}} & -{{1}\over{\sqrt{2}}} \\ }\, \matrix{-{{\complexi}\over{\sqrt{2}}} & {{\complexi}\over{\sqrt{2}}} \\ -{{1}\over{\sqrt{2}}} & -{{1}\over{\sqrt{2}}} \\ } \\ &&\phantom{xxx}\color{blue}{\text{inverse determined and }A\, T\text{ computed}}\\ &=& \matrix{-1 & 0 \\ 0 & 1 \\ } \end{array}$

New example