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\begin{center}
  {\bf University of Kentucky, Physics 306 \\
    Homework \#5, Rev. B, Bonus }
\end{center}

\vspace{.3 in}

{\bf 1. The good, \ldots}

{\bf \quad a)} The \href{https://en.wikipedia.org/wiki/Commutator#Ring_theory}{\it commutator} $[A,B]$ of two matrices $A$ and $B$ is defined as the matrix $[A,B]\equiv AB-BA$.  Calculate $[\sigma_{j},\sigma_{k}]$ for the \href{https://en.wikipedia.org/wiki/pauli_matrices}{\it Pauli matrices} $\sigma_{i}$, $i=x,y,z$.  Do they commute? Compare with H02\#2:
$\sigma_{j}\sigma_{k}=I\sigma_{j}\cdot\sigma_{k} + i\sigma_{j}\times\sigma_{k}$, treating the Pauli matrices $\sigma_{k}\sim \bh e_{k}$ as unit vectors.

{\bf \quad b)} Show that the commutator of $A$ and $B$ is antisymmetric: $[A,B]=-[B,A]$, and bilinear: $[\alpha_{i}A_{i},B_{j}\beta_{j}]=\alpha_{i}[A_{i},B_{j}]\beta_{j}$. Thus it is a `matrix version' of the vector cross product: $[A,A]=0$ and two matrices commute if they are in the `same direction' (both diagonal in the same basis).

{\bf \quad c)} Verify that the two matrices %
$A=\scriptsize{\pmatrix{ %
    \phantom{-}9 & -2 & -6 \cr %
    -2 & \phantom{-}9 & -6 \cr %
    -6 & -6 & -7 }}$ %
and %
$B=\scriptsize{\pmatrix{ %
    54 & \phantom{-}10 & -3 \cr %
    10 & -45 & 30 \cr %
    -3 & \phantom{-}30 & 46 }}$.\\ %
commute.  Simultaneously diagonalize both matrices.  Are the eigenvectors
orthogonal?\\ {\it Hint:} \href{https://www.gnu.org/software/octave}{Octave}
or \href{http://www.wolfram.com/solutions/education/students}{Mathematica} is
your friend!

{\bf \quad d)} By adding and subtracting $M=H+iK$ and its adjoint $M\ad=H-iK$, where $H\ad=H$ and $K\ad=K$ are Hermitian (note that $A=iK$ is antisymmetric: $A\ad=-A$), show that any matrix $M$ can be decomposed into its symmetric and antisymmetric parts, similar to how $z=x+iy$ projects into its real and imaginary parts.  

{\bf \quad e)} A \href{https://en.wikipedia.org/wiki/Normal_matrix}{\it normal} matrix is one that commutes with its adjoint: $[N\ad,N]=0$.  Setting $N=H+iK$ as in d), show that $[N\ad,N]=2i[H,K]$.  This shows that a normal matrix is one whose symmetric and antisymmetric parts commute.  Thus normal matrices are the most general matrices that can be diagonalized with orthogonal eigenvectors.  The eigenvalues of $N$ are $n_{j}=h_{j}+ik_{j}$ where $h_{j}$ and $k_{j}$ are the eigenvalues of $H$ and $K$.  Thus $N$ acts like independent complex numbers!

{\bf \quad f)} What are the properties of the eigenvalues of the following types of matrices: i) [anti]Hermitian, $H\ad=\pm H$; ii) Unitary, $U\ad=U^{-1}$; iii) Zero, $Z\vec v=\vec 0$; iv) Identity, $I\vec v=\vec v$; v) Traceless, $\mathrm{tr}\ A=0$; vi) Special, $\mathrm{det}\ A=1$; vii) Projection, $P^{2}=P$; viii) Involution, $A=A^{-1}$.

{\bf \quad g)} In analogy with polar complex numbers $z=e^{\sigma+i\phi}=\rho u$, where $\rho=e^{\sigma}>0$ and $u=e^{i\phi}$ is a $unit$: $|u|^{2}=u^{*}u=1$, any matrix $A$ has a \href{https://en.wikipedia.org/wiki/Polar_decomposition#Matrix_polar_decomposition}{\it polar decomposition} $A=RS$ into a rotation $R$ and a stretch $S$, which are the building blocks of all linear operators.  By taking the exponential $M=e^{N}$ of $N=H+iK$, show that any normal matrix can be decomposed into $M=UP$, where $U$ is unitary, $U\ad U=I$, and $P$ is positive definite, $P\ad =P$ with non-negative eigenvalues.
%relation between A=UP and A=P'U'?

{\bf \quad h)} Using the polar decomposition $A=RS$, where $R\ad R=I$ and $S\ad=S$, and diagonalizing $S=VWV\ad$, show that any matrix may be decomposed into $A=UWV\ad$, where $U=RV$ and $V$ are unitary, and $W$ is diagonal.  This \href{https://en.wikipedia.org/wiki/Singular_value_decomposition}{\it singular value decomposition} of matrices is the generalization the spectral theorem for all matrices.  Because it involves both stretches and rotations, every {\it singular value} (eigenvalue) $w_{i}$ has two eigenvectors: a {\it left singular vector} $\vec u_{i}$ and a {\it right singular vector} $\vec v_{i}$.  This is useful for transformations from one vector space into another, not just operators.


\newpage
{\bf 2.  the bad, \ldots} 

{\bf \quad a)} The Pauli \href{https://en.wikipedia.org/wiki/Ladder_operator}{\it ladder operators} $\sigma_{\pm}=(\sigma_{x}\pm i\sigma_{y})/2$ transform the unit vectors $(1,0)^{T}$ and $(0,1)^{T}$ from one to the next along a chain, {\it annihilating} the last one.  Calculate $\sigma_{\pm}$, $\sigma_{\pm}^{2}$, $\sigma_{\pm}^{3}, \ldots, \sigma^{n}$.  This is a {\it nilpotent} matrix: $Z^n=0$ for some $n$.  By definition, $Z^{0}=I$ is the identity.

{\bf \quad b)} Calculate the doubly {\it degenerate} eigenvalue $\lambda$ of $\sigma_{\pm}$ and show that they have only one eigenvector $\vec v_{1}$.  A matrix with less eigenvectors than eigenvalues is called {\it defective} and is the unique sum of two commuting matrices: one {\it diagonalizable} and the other {\it nilpotent}, by the \href{https://en.wikipedia.org/wiki/Jordan\%E2\%80\%93Chevalley_decomposition}{\it Jordan-Chevalley decomposition} .
The other vectors associated with $\lambda$ are called {\it generalized eigenvectors}, defined by $M\vec v_{k}=\vec v_{k}\lambda+\vec v_{k-1}$. Identify each term with the corresponding matrix part above.  Note that $M \vec v_{1}$ follows the pattern if we define $\vec v_{0}=\vec 0$.  Find the matrix $J$ of $M$ in the basis $V=\{\vec v_{k}\}$ (called a \href{https://en.wikipedia.org/wiki/Jordan_matrix}{\it Jordan block}), which gives rise to the {\it Jordan decomposition} (similarity transform) $M=VJV^{-1}$.

{\bf \quad c)} By expanding the Taylor series, show that [wiki:Jordan Matrix]
$$ f(J_{\lambda,n})= \sum_{k=0}^{n-1} \frac{f^{(k)}(\lambda) Z^k}{k!} = \\
\pmatrix{
f(\lambda) & f^\prime (\lambda) & \frac{f^{\prime\prime}(\lambda)}{2} & \cdots & \frac{f^{(n-2)}(\lambda)}{(n-2)!} & \frac{f^{(n-1)}(\lambda)}{(n-1)!} \cr
0      & f(\lambda) & f^\prime (\lambda) & \cdots & \frac{f^{(n-3)}(\lambda)}{(n-3)!} & \frac{f^{(n-2)}(\lambda)}{(n-2)!} \cr
0      & 0      & f(\lambda) & \cdots & \frac{f^{(n-4)}(\lambda)}{(n-4)!} & \frac{f^{(n-3)}(\lambda)}{(n-3)!} \cr
\vdots & \vdots & \vdots & \ddots & \vdots     & \vdots \cr
0      & 0      & 0      & \cdots & f(\lambda) & f^\prime (\lambda) \cr
0      & 0      & 0      & \cdots & 0          & f(\lambda) \cr}$$
so $f(A) = Vf(J)V^{-1} = V\left(\bigoplus_{k=1}^N f\left(J_{\lambda_k ,m_k}\right)\right)V^{-1}$, where 
$A = VJV^{-1} = V \left(\bigoplus_{k=1}^N J_{\lambda_k,m_k}\right) V^{-1}$.

{\bf \quad d)} Calculate the adjoint $\sigma_{\pm}^{\ad}$, the \href{https://en.wikipedia.org/wiki/Projection_(linear_algebra)}{\it projections} $P_{+}=\sigma_{+}\sigma_{-}$ and $P_{-}=\sigma_{-}\sigma_{+}$, and the commutators $[\sigma_{+},\sigma_{-}]$, $[I,\sigma_{\pm}]$, and $[\sigma_{z},\sigma_{\pm}]$.  Note that $\sigma_{\pm}$ commutes with $I$ because its eigenvalues are all the same, but not with $\sigma_{z}$, which has distinct eigenvalues.  This is true in general.  Use the last commutation relation to show that if $\vec v$ is an eigenvector of $\sigma_{z}$, then $\sigma_{\pm}\vec v$ is also an eigenvector.  Thus $\sigma_{\pm}$ raise an lower one one eigenvector of $\sigma_{z}$ into another, like climbing up or down the rungs of a ladder, which is where their name comes from.

%NOT TRUE! {\bf \quad b)} Let the operators $Z_{1}$ and $Z_{2}$ be {\it nilpotent}: $Z_{i}^{n_{i}}$ for some $n_{i}\in \blackboard{N}$.  Use the binomial expansion to show that the sum $Z=Z_{1}+Z_{2}$, is also nilpotent: $Z^{n}=0$ where $n=n_{1}+n_{2}$.  Thus the set of nilpotent operators form a subspace.

{\bf \quad e)} Show that $\sigma_{x}\sigma_{\pm}\sigma_{x}^{-1}=\sigma_{\mp}$  Find the most general $2\times2$ nilpotent operator via the similarity transform $V\sigma_{+}V^{-1}$ of $\sigma_{+}$ into a generic basis $V=\pmatrix{a & b \cr c & d}$.  Represent it as an outer product.  What is its geometrical interpretation?  Do the same for a generic {\it diagonalizable} operator to represent it as the sum of two outer products (its {\it spectral decomposition}).

{\bf \quad f)} The derivative operator $D=d/dx$ is a nilpotent operator in the linear space of polynomials $A(x)=a_{0}+a_{1}x+a_{2}x^{2}+\ldots +a_{n}x^{n}$.  i) Find the matrix of $D$, in the basis of monomials $\bh e_{k} = x^{k}$. Show that $D^{n}=0$.  ii) What is the eigenvalue of $D$ in this space?  iii) In what basis is its matrix in the normal Jordan form described in b)?  Show how this relates to the Taylor series $A(x)=\sum_{k}\frac{1}{k!}A^{(k)}(0) x^{k}$.  iv) $D$ acts like the identity on $e^{x}$ and thus is not nilpotent in this space.  What is different about $e^{x}$?  v) What are the eigenvalues and eigenfunctions of $D$?


\newpage
{\bf 3. \ldots and the gnarly.}

We have seen that diagonal matrices behave nicely, adding {\it and}
multiplying like $n$ independent complex numbers.  By extension, a function
of any diagonalizable matrix acts independently on each eigenvalue in its
eigenbasis.  We saw in \#1 that commuting matrices can be {\it simultaneously
  diagonalized} to perform operations in their common diagonal
basis.  This leads to th {\it normal matrix analogy}, where normal matrices act like $n$ independent complex numbers in the
common orthogonal eigenbasis of their {\it Hermitian} plus {\it anti-Hermitian}, or {\it positive semidefinite} times {\it unitary} parts.

On the other hand we saw in \#2 that matrices with a {\it nilpotent} component were particularly nasty, and could not be diagonalized at all, but only reduced to the {\it Jordan normal form}. These {\it defective matrices} are worse than {\it singular matrices}, which are still potentially diagonalizable, but cannot be inverted because of one or more null eigenvalues $\lambda_{k}=0$.

Only a small slice of matrices are `bad' (defective), while an even smaller sliver are `good' (normal). But what about the vast expanse of `gnarly' (abnormal but diagonalizable) matrices, which are neither `good' nor `bad'?  In this exercise, we will show there exists a {\it generalized adjoint}, in which any such matrix is normal, with an orthonormal eigenbasis in the corresponding metric (the {\it generalized eigenvalue problem}).  Thus, {\it ``beauty is in the eye of the beholder''}, and any non-defective matrix is normal in its natural setting of appropriately defined distance and angle.

And what about pairs of matrices that do not commute?  For these we consider a second form of {\it simultaneous diagonalization}, which is possible for any two Hermitian matrices.  We diagonalize one matrix, treating is as a {\it metric}, with a {\it congruency transform} $I=V\ad g V$, resulting in the {\it identity} matrix $I$, signifying an orthonormal basis.  This is always preserved during diagonalization of the second {\it operator} matrix using the more common {\it similarity transform} $D=V^{-1}MV$.  Note that eigenvalues of $g$ can be $1$ by scaling the eigenvectors $V$, while eigenvalues of $M$ are invariant.

{\bf \quad a)} Given an eigenbasis $V$ of a matrix $M=VDV\ad$, describe a procedure for finding the class of metrics $g$ in which $V$ is orthonormal:  $V\cdot V=V\ad g V=I$.  This can be thought of as the inverse of the {\it Gram-Schmidt} method.  Show that $M$ is normal using the {\it adjoint} $M\ad \equiv g^{-1}M^{T}g$.  For what other metrics is $M$ normal?  Show that if $M\bv v_{i}=g\bv v_{i}\lambda_{i}$ or $MV=gVD$ (the {\it generalized eigenvalue problem}), then $V\ad g V=I$, that is $V$ can be chosen to be orthonormal.

{\bf \quad b)} Working backwards, show that any two Hermitian matrices $A\ad=A$ and $B\ad=B$ can be simultaneously diagonalized in the following form: $V\ad AV=D$ and $V\ad B V=I$.  This is useful to solve a system of particles at coordinates $x^{k}$, connected with masses and springs, having kinetic plus potential energy $H=\frac12 \dot{\bv x}\ad M\dot{\bv x} + \frac12 \bv x\ad K \bv x$.  Here, $M$ acts as the metric $v^{2}=\dot{\bs x}\cdot\dot{\bs x}$.

%  "any two symmetric matrices can be simultaneously diagonalized"
%by invoking the metric and replacing Hermitian conjugate with the general metric and then make the connection to normal matrices above.

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