Simultaneous diagonalization of hermitian and anti-hermitian matrices

linear algebramatrices

I have an unitary matrix U, that i can decompose on the hermitian and anti-hermitian parts.

$$U = A+B$$

How to prove that two matrices are diagonalizable simultaneously without knowledge, that unitary matrices are diagonalizable.

I tried to solve this problem like this:

1)$S^HAS = D_1$, where $D_1$ – diagonal matrix

2)Then i have a matrix: $V = S^HBS $

3)I know, that all symmetric matrices are diagonazible, but V is not symmetric, so i don`t know what to do.

Best Answer

Note that the Hermitian and anti-Hermitian parts of $U$ can be written as "polynomials" (more accurately, rational functions) of $U$. In particular, we have $$ A = \frac 12 (U + U^*) = \frac 12 (U + U^{-1}), \quad B = \frac 12 (U - U^*) = \frac 12 (U - U^{-1}). $$ It follows that $A$ and $B$ commute. In particular, we can expand the product of the $\frac{U \pm U^{-1}}{2}$ matrices to get $$ AB = BA = \frac 14 (U^2 - U^{-2}). $$ Because $A,B$ are diagonalizable with $AB = BA$, it follows that $A$ and $B$ are simultaneously diagonalizable.

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[Math] Geometric interpretation of normal and anti-hermitian matrices

Let $V$ be a finite-dimensional complex inner product space.

By the finite-dimensional spectral theorem, one has that $T \in L(V) := L(V,V)$ is normal if and only if there exists an orthonormal basis for $V$ consisting of eigenvectors of $T$. Geometrically, this means that $T$ acts on individual each coordinate axis (with respect to this orthonormal basis) by rescaling by a complex scaling factor (i.e., the eigenvalue for the eigenvector spanning that axis). In other words, $T$ is normal if and only if there exists some orthonormal coordinate system for $V$ such that $T$ fixes each coordinate axis, i.e., for each coordinate axis $\mathbb{C} v_k$, $T(\mathbb{C} v_k) \subseteq \mathbb{C} v_k$.
A $1$-parameter groups of unitaries is a map $U : \mathbb{R} \to U(V)$, the group of unitaries in $L(V)$, such that $U(0) = I$ and $U(t_1+t_2) = U(t_1)U(t_2)$ for any $t_1$, $t_2$; geometrically, given a fixed orthonormal basis defining your initial orthonormal coordinate system, you can view $U$ as defining a a time-dependent orthonormal coordinate system on $V$.

So, let $U : \mathbb{R} \to U(V)$ be a $1$-parameter group of unitaries. Observe that for $t \in \mathbb{R}$ and $h \neq 0$, $$ \frac{1}{h}(U(t+h)-U(t)) = \left(\frac{1}{h}\left(U(h)-I\right)\right)U(t), $$ so that $U$ is differentiable at $t$ if and only if it is differentiable at $0$. So, suppose that $U$ is differentiable at $0$. Then $$ (U^\prime(0))^\ast = \left(\lim_{h \to 0}\frac{1}{h}\left(U(h)-I)\right)\right)^\ast = \lim_{h\to 0}\frac{1}{h}\left(U(-h)-I\right) = -\lim_{k\to 0} \frac{1}{k} \left(U(k)-I\right) = -U^\prime(0), $$ so that $U(t)$ satisfies the ODE $$ U^\prime(t) = A U(t), $$ where $A := U^\prime(0)$ is skew-adjoint. Thus, from a geometric standpoint, skew-adjoint operators are precisely the infinitesimal generators of differentiable time-dependent orthonormal coordinate systems.

In fact, bearing in mind that $T$ is skew-adjoint if and only if $iT$ is self-adjoint (Hermitian), and that in mathematical physics and theoretical physics one generally uses $e^{iA}$ instead of $e^A$, this is essentially why self-adjoint operators (on infinite-dimensional Hilbert space!) become essential in quantum mechanics. This is because in quantum mechanics, the time evolution (in the so-called Heisenberg picture) is effected by a $1$-parameter group of unitaries whose infinitesimal generator is the all-important Hamiltonian of your quantum mechanical system.

Now, suppose that $V$ is a finite-dimensional real inner product space.

By the finite-dimensional spectral theorem one has that $T \in L(V)$ is symmetric (i.e., self-adjoint) if and only if there exists an orthonormal basis for $V$ consisting of eigenvectors. In other words, $T$ is symmetric if and only if there exists some orthonormal coordinate system for $V$ such that $T$ fixes each coordinate axis, i.e., for each coordinate axis $\mathbb{R}v_k$, $T(\mathbb{R}v_k) \subseteq \mathbb{R} v_k$. Observe that the notion of normal operator is no longer nearly as useful as in the complex case, since the finite-dimensional spectral theorem now characterizes precisely the symmetric operators. In particular, even though orthogonal operators are normal, examples of orthogonal operators with non-real complex eigenvalues are plentiful.
When it comes to skew-symmetric (i.e., skew-adjoint) operators, everything carries over exactly as before, except that we now find that they are precisely the infinitesimal generators of $1$-parameter groups of orthogonal operators, viz, time-dependent orthonormal coordinate systems. Note, however, that skew-symmetric and symmetric operators are now fundamentally different now, since we can no longer multiply by $i$ to go between the skew-adjoints and self-adjoint, as we could in the complex case. In particular, symmetric operators necessarily have real eigenvalues, whilst examples of skew-symmetric operators with non-real complex eigenvalues are absolutely plentiful.

[Math] How to a matrix be Hermitian, unitary, and diagonal all at once

Hint : Here I have done for $2 \times 2$ matrix.

Let $A = \left( \begin{array}{cc} a & 0 \\ 0 & b \\ \end{array} \right)$

be a diagonal matrix with complex entries. Its eigenvalues are precisely $a$, $b$. Because $A$ is Hermitian, they must be real. Also $A$ is unitary, they must each be of absolute value $1$. There are exactly four matrices satisfying these conditions:

Let $A_1 = \left( \begin{array}{cc} 1 & 0 \\ 0 & 1 \\ \end{array} \right)$, $A_2 = \left( \begin{array}{cc} 1 & 0 \\ 0 & -1 \\ \end{array} \right)$, $A_3 = \left( \begin{array}{cc} -1 & 0 \\ 0 & 1 \\ \end{array} \right)$, $A_4 = \left( \begin{array}{cc} -1 & 0 \\ 0 & -1 \\ \end{array} \right)$

I hope this may help you.

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[Math] Geometric interpretation of normal and anti-hermitian matrices

[Math] How to a matrix be Hermitian, unitary, and diagonal all at once

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