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\begin{center}
  {\bf University of Kentucky, Physics 306 \\
    Homework \#5, Rev. A, due Monday, 2025-02-18 }
\end{center}

\vspace{.3 in}

{\bf 1.  Stretches}.  In analogy with the \href{https://en.wikipedia.org/wiki/Polar_decomposition#Matrix_polar_decomposition}{\it polar decomposition} $w=x+iy=\rho e^{i\phi}$ of complex numbers, any matrix $A$ can be decomposed $A=RS$ into a stretch $S$ and a rotation $R$, which are the building blocks of all linear operators.  This problem set explores the structure of stretches.

{\bf \quad a)} For any eigenvalue problem $M\bv v_{i}=\bv v_{i}\lambda_{i}$, augment these $n$ equations to obtain $M V=VW$, where $W=\rm{diag}(\lambda_{1},\lambda_{2},\ldots)$, and thus show $M=VWV^{-1}$ and $W=V^{-1} MV$.  This is the {\it similarity transform} to the eigenbasis of $M$, in which the operator's matrix is diagonal, and its inverse.

{\bf \quad b)} Diagonalize $\sigma_{x}$ and $\sigma_{y}$ by calculating their eigenbasis $U_{x,y}=(\vec v_{1}|\vec v_{2})$ of H04\#1c to justify the similarity transforms $\sigma_{x}U_{x}=U_{x}\sigma_{z}$ of H04\#1d.  Thus, $\sigma_{x,y}$ are matrices of the same operator $\sigma_{z}$ in different bases. Calculate the eigenvalues and eigenvectors of $\sigma_{z}$, which is already diagonal.

{\bf \quad c)} Calculate all eigenvalues and eigenvectors of the {\it ladder operators} $\sigma_{\pm}=(\sigma_{x}\pm i \sigma_{y})/2$ of H04\#1b to show that they are not diagonalizable.  Such matrices with less eigevectors than eigenvalues are called {\it defective} (see H08\#2).  Show that they are also {\it nilpotent}: $\sigma_{\pm}^{n}=0$ for some value $n$.

{\bf \quad d)} Calculate the eigenvalues and eigenvectors of the {\it projections} $P_{+}=$\scalebox{.8}{$\pmatrix{1&0\cr0&0}$} and $P_{-}=$\scalebox{.8}{$\pmatrix{0&0\cr0&1}$} of H04\#1b. [{\it bonus}: show that projections, $P^{2}=P$, can only have eigenvalues of 0 or 1.]

{\bf \quad e)} A {\it symmetric} matrix $S$ is guaranteed to have a complete set of eigenvectors $V=(\bv{v}_{1}, \bv{v}_{2}, \ldots )$ and corresponding eigenvalues $\lambda_{1}, \lambda_{2}, \ldots$, such that $S \bv{v}_{i}= \bv{v}_{i}\lambda_{i}$ with the following special properties: show that i) the eigenvalues of $S$ must be real: $\lambda^{*}=\lambda$, and that ii) two eigenvectors $\bv{v}_{i},\bv{v}_{j}$ of $S$ with distinct eigenvalues $\lambda_{i}\neq\lambda_{j}$ must be orthogonal: $\bv{v}_{i}\cdot\bv{v}_{j}=0$.  Thus the matrix of eigenvectors is unitary: $V\ad V=I$, so that $S=VWV\ad$ and $W=V\ad S V$.  Interpret these similarity transforms geometrically.

{\bf \quad f)} In addition to symmetric matrices, complex {\it normal matrics}, $N\ad N=NN\ad$ (see H08\#1), also have an othogonal eigenbasis but only have real eigenvalues if they are {\it Hermitian}, $H\ad=H$.  Calculate the eigenvalues and eigenvectors of $M_z=$\scalebox{.8}{$\pmatrix{0&-1\cr 1&0}$}. How were these used previously?  Show that $M_z = VWV\ad = \frac12$\scalebox{.8}{$\pmatrix{1&1 \cr i&-i} \pmatrix{-i&0\cr0&i} \pmatrix{1&-i\cr1&i}$} and $e^{M_z \phi} = V e^{W\phi}V\ad = \frac12$\scalebox{.8}{$\pmatrix{1&1 \cr i&-i} \pmatrix{e^{-i\phi}&0\cr0&e^{i\phi}} \pmatrix{1&-i\cr1&i}$}.
Multiply this out to verify H04\#2b.  This is an example of the \href{https://en.wikipedia.org/wiki/Normal_matrix#Normal_matrix_analogy}{\it normal matrix analogy}, which relates matrices and complex numbers.  Hermitian matrices have real eigenvalues while anti-Hermitian matrices have zeros on the diagonal and imaginary eigenvalues.  The exponential of a Hermitian matrix is positive definite and has real positive eigenvalues, while the exponential of an anti-Hermitian matrix is unitary with  unit modulus eigenvalues and therefore determinant.


{\bf 2.} [{\it bonus}: We saw in H04\#1 that the {\it trace} and {\it determinant} are matrix {\bf invariants} under similarity transformation.  In general, an $n\times n$ matrix has $n$ independent invariants, including the trace and determinant.

{\bf \quad a)} Show that the {\it characteristic equation} $|A-\lambda I| = \lambda^{n}+a_{n-1}\lambda^{n-1}+\ldots+a_{1}\lambda+a_{0}=0$, substituting $\lambda\to A$ in the second equality, is invariant under similarity transforms.  Thus $A$ has $n$ independent invariants: either $a_{0}\ldots a_{n-1}$ of this equation, its $n$ eigenvalue roots, or the traces of $A^{1},A^{2},\ldots, A^{n}$.  The first and last coefficients $a_{n-1}$ and $a_{0}$ are $\mathrm{tr}(A)$  and $\mathrm{det}(A)$, respectively, while the others are $k$-dimensional `perimeters', for example, {\it surface area}, of the transformation.

{\bf \quad b)} Prove the \href{https://www.cuemath.com/algebra/cayley-hamilton-theorem/}{Cayley-Hamilton Theorem}, that any matrix $A$ satisfies its own characteristic equation, for  the case of diagonal matrices.  The full theorem follows from the invariance of the characteristic equation.  Show that $A^{-1}=-(A^{n-1}+a_{n-1}A^{n-2}+\ldots+a_{1}I)/a_{0}$.]


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