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In the two-dimensional span of the first two energy eigenstates, the measured observable has matrix $S=\left(\begin{array}{cc}7& 24\\ 24& -7\end{array}\right)$. Its square is $625I$, and its trace is zero, giving eigenvalues $25,-25$. The orthogonal complement has eigenvalue zero. Thus the full spectrum is $\boxed{\{-25,0,25\}}$.

The normalized eigenstates for $-25$ and $25$ are respectivelyThe lowest measurement outcome prepares ${\chi }_{-}$ up to an irrelevant overall phase. The subsequent unitary time evolution givesBy the Born rule, the probability of measuring the lowest eigenvalue again is the squared projection onto its one-dimensional eigenspace:This is the two-level quantum return probability with $q=9/25$. At $t=0$ the probability is one; if the two energies happen to be equal it remains one for every time.

Use completing the square in the Hermitian form $x^{\ast }Ax$, rather than unitary diagonalization of a normal matrix. The first pivot is one, soA second square completion givesTherefore one answer isThe diagonal entries of $D$ are rational, $\det T=1$, and the displayed identity for every $x$ proves $T^{\ast }DT=A$.

If $T$ were a unitary matrix, this matrix congruence would also be a similarity transformation, so the diagonal entries of $D$ would be the eigenvalues of $A$. Its characteristic polynomial isThus its eigenvalues are $3,3+\sqrt{10},3-\sqrt{10}$, two of which are irrational. A unitary factor cannot give a rational diagonal here. Alternatively, the permitted rational root theorem implies that three rational eigenvalues would all be integers, whereas the quadratic factor has no integer roots.

Put $C=A+yI$. For every $y$ except the finitely many roots of $\det (A+yI)$, $C$ is invertible and $CB$ is similar to $BC$, since $C^{-1}(CB)C=BC$. HenceBoth sides are polynomials in $y$. Agreement for infinitely many $y$ makes them identical, so evaluating at $y=0$ proves Characteristic polynomials of AB and BA:Their minimal polynomials need not agree. Forwe have $AB=B\ne 0$, $B^2=0$, and $BA=0$. The two minimal polynomials are respectively $t^2$ and $t$, although both characteristic polynomials are $t^2$.

For any $p(t)={\sum }_j{p}_j{t}^j$, the identity $(BA{)}^{j+1}=B(AB{)}^{j}A$ givesTaking $p=m_{AB}$ makes the right-hand side zero. Every annihilating polynomial is divisible by the minimal polynomial, soIf $AB$ is diagonalizable, $p$ has only simple roots. Therefore each nonzero eigenvalue of $BA$ has only size-one Jordan blocks, and the zero-eigenvalue blocks have size at most two. Squaring kills the latter nilpotent blocks, so $(BA{)}^2$ is diagonalizable, as is $(AB{)}^2$.

By their common characteristic polynomial, $AB$ and $BA$ have the same eigenvalues counted with algebraic multiplicity. Squaring produces the same multiset of squared eigenvalues, including combined multiplicities if two eigenvalues have the same square. The two diagonalizable squares therefore have the same diagonal normal form up to permutation. This proves similarity of squared matrix products with one diagonalizable product:It does not imply that $AB$ and $BA$ themselves are conjugate: the possible size-two nilpotent blocks at zero are exactly what squaring removes.

Using ${\chi }_M(t)=\det (tI-M)$, expansion or triangular reordering givesThe eigenspace for $1$ is spanned by $e_1$, so its algebraic multiplicity two requires a size-two Jordan block. A convenient Jordan chain isFor eigenvalue $2$, the eigenspace has basis $w_1=e_1+e_3$, $w_2=e_4$, hence both corresponding blocks have size one. Therefore a Jordan normal form and the minimal polynomial areThe ordered basis $(v_1,v_2,w_1,w_2)$ supplies the change-of-basis matrixIts determinant is one, and the chain equations give $MP=PJ$, proving $P^{-1}MP=J$. The square at $t=1$ in the minimal polynomial is essential: $M$ is not diagonalizable.

Put $U=\varphi (A(t))$. Applying the group homomorphism $\varphi$ to the identity just proved shows that $U$ is conjugate to $U^{m^2}$ for every positive integer $m$. Thus the multiset of its eigenvalues is unchanged by taking their $m^2$th powers.

Take $m=2$. The fourth-power operation permutes this finite multiset. For each eigenvalue $z$, iterating its permutation cycle gives $z^{4^r}=z$ for some positive integer $r$. Since $U$ is unitary, $z\ne 0$, so $z^{4^r-1}=1$: every eigenvalue is a root of unity. Choose a positive integer $d$ divisible by all their orders. Every eigenvalue of $U^{d^2}$ is then one. Because its eigenvalue multiset is also the original one, we obtainThis is square-power rigidity of a finite nonzero spectrum. Finiteness of the dimension is essential to the permutation-cycle argument.

Every unitary matrix is unitarily diagonalizable. Part (ii) therefore proves the stronger matrix statement $\varphi (A(t))=I$, not merely a statement about its eigenvalues. The kernel of a group homomorphism is a normal subgroup, and it contains every upper unipotent. By part (iii) their normal closure is the entire group. ThusThis proves the triviality of finite-dimensional unitary representations of SL2R. In fact the argument used only the algebraic homomorphism property and finite-dimensional unitarity; it did not need continuity.

For a compact Lie group $G$, the complex representation ring of a compact group $R(G)$ is the Grothendieck group of finite-dimensional continuous complex group representations, with the relations $[M\oplus N]=[M]+[N]$. Multiplication is $[M][N]=[M\otimes N]$, and the unit is the trivial one-dimensional representation. The ring of class functions iswith pointwise addition and multiplication. The character of $M$ is ${\chi }_M(g)=\mathrm{tr}{\rho }_M(g)$. The trace identities for direct sums and tensor products makea well-defined ring homomorphism $\chi :R(G)\to c\ell (G)$.

Normalize Haar measure by ${\int }_{G}dg=1$. Averaging a positive definite Hermitian inner product giveswhich is positive definite and $G$-invariant. This proves unitarization of a compact-group representation. In a unitary representation the orthogonal complement of an invariant subspace is invariant. Induction on dimension therefore gives complete reducibility of compact-group representations. Hence every element of $R(G)$ is a finite integral linear combination of irreducible classes.

We establish the needed character orthogonality for compact groups carefully. For irreducible unitary representations $V,W$, let $G$ act on ${\mathrm{Hom}}_{\mathbb{C}}(W,V)$ byIts average $P={\int }_G(g\cdot )dg$ is a projection onto ${\mathrm{Hom}}_G(W,V)$: averaging makes every image invariant, and it fixes every intertwiner. The trace of this action is ${\chi }_V(g)\overline{{\chi }_W(g)}$, since the inverse of a unitary matrix has conjugate trace. Taking the trace of the projection givesFor completeness, Schur's lemma follows here from invariance of the kernel and image of an intertwiner. A nonzero intertwiner between irreducibles is an isomorphism. An endomorphism of an irreducible complex representation has an eigenvalue $\lambda$; the noninvertible intertwiner $A-\lambda I$ must vanish. Thus the last dimension is one for isomorphic irreducibles and zero for inequivalent ones.

If a virtual character ${\sum }_V{n}_V{\chi }_V$ vanishes, integrating it against $\overline{{\chi }_W}$ gives $n_W=0$ for every irreducible $W$. The irreducible multiplicities therefore distinguish the classes, andThis also proves that two finite-dimensional representations with the same character are isomorphic.

For SU(2), the irreducibles are $V_k={\mathrm{Sym}}^k({\mathbb{C}}^2)$, $k\ge 0$, of dimension $k+1$. One can see this classification through the usual ${\mathfrak{sl}}_2$ operators $H,E,F$: a highest-weight vector $v$ has $Hv=kv$, $Ev=0$, and the relations $[H,E]=2E$, $[H,F]=-2F$, $[E,F]=H$ give, by induction,The torus weights are integers. If $F^{r}v\ne 0$ but $F^{r+1}v=0$, the identity at $j=r+1$ forces $k=r\ge 0$. The vectors $v,Fv,\dots ,F^{k}v$ span an invariant irreducible module, which in an irreducible representation is the whole space. This is the symmetric-power module. Conversely its successive monomial weight vectors are linked by $E,F$ with nonzero coefficients, proving its irreducibility. Thus the classification of finite-dimensional representations of SU2 gives every irreducible, rather than just a list of examples.

On the maximal torus $T=\{\mathrm{diag}(z,z^{-1}):|z|=1\}$, their characters areWith $x={\chi }_1=z+z^{-1}$, multiplication gives ${\chi }_{k+1}=x{\chi }_k-{\chi }_{k-1}$, starting with ${\chi }_0=1$. Hence each ${\chi }_k$ is a monic integral polynomial of degree $k$ in $x$. For example ${\chi }_2=x^2-1$ and ${\chi }_3=x^3-2x$. They form a basis over $\mathbb{Z}$, soThe character map realizes this ring as the finite integral polynomials in $z+z^{-1}$, interpreted as continuous class functions. Its image is not the whole infinite-dimensional ring of continuous class functions; injectivity is the assertion being proved.

Let each observation have dimension $p$, and define the sample mean and the divisor-$n$ sample covariance matrix byThe multivariate normal distribution gives the log-likelihoodWriting $y_i-\mu =(y_i-\bar{y})+(\bar{y}-\mu )$ makes the cross terms vanish, and the last sum becomes $n\mathrm{tr}(V^{-1}S)+n(\bar{y}-\mu {)}^T{V}^{-1}(\bar{y}-\mu )$. Since $V$ is a positive-definite matrix, its minimum over the mean is at $\hat{\mu }=\bar{y}$. ConsequentlyWhen $S$ is a positive-definite matrix, $\log \det (V^{-1}S)=\log \det S-\log \det V$. Thuswhere $C^{\prime }$ depends on the data but not on $V$.

For the maximum-likelihood estimator, consider the positive-definite matrix $Q=S^{1/2}{V}^{-1}{S}^{1/2}$, with positive eigenvalues $a_1,\dots ,a_p$. The part of the profile likelihood depending on $V$ is $\frac{n}{2}{\sum }_j(\log a_j-a_j)$. The elementary inequality $\log a-a\le -1$, with equality exactly at $a=1$, shows that the unique maximizing matrix is $Q=I$. HenceThe divisor is $n$, not the unbiased covariance divisor $n-1$. The regularity qualification matters: if $S$ is singular, the displayed logarithm of its determinant is not finite, and no positive-definite covariance maximizes the likelihood. Sending the fitted variance in a null direction of $S$ to zero makes the likelihood unbounded. For a nonsingular normal population, $S$ is positive definite almost surely when $n>p$, which is the setting of the large-sample test.

Principal component analysis replaces correlated measurements by orthogonal linear coordinates ordered by their variance. For centered observations collected in an $n\times p$ matrix $Z$, write the sample covariance matrix as $S=Z^{T}Z/(n-1)$. A unit loading vector $a$ gives score vector $Za$ with sample variance $a^{T}Sa$. The Rayleigh quotient is maximized by a unit eigenvector of $S$ for its largest eigenvalue.

The subsequent principal components maximize the same quadratic form subject to orthogonality to the earlier loading vectors. By the spectral theorem, if $S=Q\Lambda Q^T$ with descending eigenvalues, the scores $ZQ$ have diagonal covariance $\Lambda$. Their fractions of explained variance are ${\lambda }_j/{\sum }_i{\lambda }_i$. Repeated eigenvalues identify an eigenspace rather than a unique axis.

Keeping the first $q$ principal components gives the rank-$q$ reconstruction $\hat{Z}=Z{Q}_q{Q}_q^T$. Its squared reconstruction error isFor any rank-$q$ orthogonal projection $P$, retained variance is $\mathrm{tr}(SP)={\sum }_j{\lambda }_j{q}_j^{T}P{q}_j$. The weights $q_j^{T}P{q}_j$ lie in $[0,1]$ and sum to $q$, so this is at most the sum of the largest $q$ eigenvalues. Projecting onto any candidate reconstruction subspace is its best least-squares reconstruction; this proves the minimum-error property and agrees with the singular value decomposition. Thus PCA finds the linear subspace preserving the greatest variance, equivalently minimizing squared reconstruction error. The upper-left sketch shows the first axis aligned with the elongated cloud; the second is perpendicular. A scree plot and substantive interpretability help choose $q$.

These methods depend on measurement scale. A variable measured in large numerical units can dominate covariance-based principal component analysis; standardizing variables gives principal component analysis on a correlation matrix. Scores summarize patterns, but the method neither establishes causation nor guarantees that the largest-variance directions best predict a separate response.

Classical multidimensional scaling begins with a symmetric dissimilarity matrix $D$, rather than necessarily with measured coordinates. Its aim is to find a low-dimensional Euclidean configuration representing those dissimilarities. Square the entries, not the matrix product, and defineFor centered coordinates $X$, squared Euclidean distances satisfy $D_{ij}^2=B_{ii}+B_{jj}-2B_{ij}$ with $B=X{X}^T$. Multiplication on both sides by $J$ annihilates the two single-index terms, proving the double-centering formula.

If $B$ is positive semidefinite, use its spectral decomposition $B=U\Lambda U^T$ to constructKeeping all positive eigenvalues reproduces the original Euclidean distances exactly; their number is the minimal embedding dimension. Keeping fewer gives a low-rank approximation to the Gram matrix. It minimizes squared Gram-matrix error, often called strain, rather than necessarily the sum of squared errors in raw distances. Negative eigenvalues signal that the dissimilarities cannot be represented exactly by Euclidean coordinates; discarding them produces an approximation whose adequacy should be inspected.

In the upper-right sketch, the supplied distances come from four points in a rectangle, and the two-dimensional reconstruction recovers that shape. The overall translation, rotation and reflection of the configuration are not identifiable from distances. Classical scaling reconstructs centered geometry from pairwise distances. If the input distances are computed from centered multivariate observations, classical scaling and principal component analysis give the same score configuration up to an orthogonal transformation: their matrices $Z{Z}^T$ and $Z^{T}Z$ share the nonzero squared singular values.