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An eigenvalue of $C$ is a scalar $\lambda$ for which a nonzero vector $v$ satisfies $Cv=\lambda v$. Equivalently, $\det (\lambda I-C)=0$, because the homogeneous linear system has a nontrivial solution exactly when its coefficient matrix is singular. This characteristic polynomial is the monic quadraticA nonzero quadratic has at most two distinct roots, proving the eigenvalue bound. Eigenvalues and diagonalization are taken over $\mathbb{C}$, also for a matrix with real entries.

For the matrix commutator, the cyclic trace identity follows directly from its entries:Hence $\mathrm{tr}[A,B]=0$. Factoring the characteristic polynomial as $(\lambda -{\lambda }_1)(\lambda -{\lambda }_2)$ and comparing the coefficient of $\lambda$ gives $\mathrm{tr}C={\lambda }_1+{\lambda }_2$, with the eigenvalues counted with algebraic multiplicity. If both eigenvalues are zero, trace and determinant are zero. The Cayley-Hamilton theorem then gives $C^2=0$; no diagonalizability assumption is needed.

Now put $M=[A,B]$. Since its trace is zero, its characteristic polynomial is ${\lambda }^2+\det M$ and the scalar square of a traceless two-by-two matrix identity givesThis proves the scalar-square assertion for both real and complex entries. If $\det M=0$, it gives $M^2=0$. If $\det M\ne 0$, the two eigenvalues are the distinct nonzero numbers $\pm \sqrt{-\det M}$. Their eigenvectors are linearly independent: a nonzero vector cannot satisfy two different eigenvalue equations. They form a basis, making $M$ a diagonalizable matrix. Thus either the commutator is diagonalizable over $\mathbb{C}$ or its square is zero. These alternatives need not be disjoint, since the zero matrix has both properties. Over $\mathbb{R}$ the diagonalization assertion would require real eigenvalues; complex diagonalization is essential to the unrestricted statement.

At a critical point $x_0$ of a twice continuously differentiable real function, the Taylor expansion isThe Hessian matrix is real symmetric, so an orthogonal eigenbasis diagonalizes its quadratic form. If every eigenvalue is positive, that form is bounded below by a positive multiple of $|h{|}^2$, which dominates the remainder and gives a strict local minimum. If every eigenvalue is negative, the same argument gives a strict local maximum. If both signs occur, displacement along the corresponding eigenvectors produces values above and below $f(x_0)$, so the point is a saddle point of a scalar function. These exhaust a nonsingular Hessian matrix.

A singular Hessian matrix does not always make the classification impossible: nonzero eigenvalues of both signs still give a saddle. If the Hessian matrix is positive or negative semidefinite, higher-order terms in its null directions must be examined. For example, $x^2+y^4$ and $x^2-y^4$ have the same Hessian matrix $\mathrm{diag}(2,0)$ at zero, but the first has a strict minimum and the second a saddle. The function $x^2$ has a non-strict minimum along a whole line.

For the given polynomial,A stationary point has $y=3x^2/2$ and then $x(2\alpha -x^2)=0$. Hence the origin is always critical, and two additional points exist exactly for $\alpha >0$:The Hessian matrix isAt the origin it is $\mathrm{diag}(2\alpha ,2)$, giving a strict local minimum for $\alpha >0$ and a saddle for $\alpha <0$. At either extra critical point it has determinant $64\alpha -36(2\alpha )=-8\alpha <0$, so both extra points are saddles.

At $\alpha =0$, complete the square:Along $x=0$ it is positive away from zero, whereas along $y=3x^2/2$ it is negative away from zero. Thus the origin remains a degenerate saddle despite its nonnegative Hessian matrix. This is the parabolic quartic critical-point classification. The complete result isFor $\alpha >0$ the local minimum is not global, since along the same parabola $f=\alpha x^2-x^4/4\to -\infty$.

Write $J_k(\lambda )$ for the Jordan block with diagonal $\lambda$ and ones on the superdiagonal. Its characteristic polynomial and minimal polynomial are both $(x-\lambda {)}^k$. Therefore one answer to the first request isFor the second request, takeThe algebraic multiplicities are five and two, and the largest block at each eigenvalue has size two, so both prescribed polynomials hold. They are not similar matrices: the eigenspace at three has dimensions three and four respectively, and similarity preserves eigenspace dimension.

There is no third similarity class. At eigenvalue one, the total size two and maximal block size two force the single block $J_2(1)$. At eigenvalue three, the blocks partition five, each has size at most two, and at least one has size two. The only partitions are $2+2+1$ and $2+1+1+1$. The uniqueness of Jordan normal form up to block order therefore leaves exactly the two classes already exhibited.

Let $I$ denote the identity endomorphism. The two-root minimal polynomial gives $(\alpha -{\lambda }_{1}I)(\alpha -{\lambda }_{2}I)=0$. Every vector has the decompositionThe first summand lies in the eigenspace $E_1=\ker (\alpha -{\lambda }_{1}I)$, and the second in $E_2$. If a vector belongs to both, subtracting its two eigenvalue equations gives $({\lambda }_1-{\lambda }_2)v=0$, hence $v=0$. Thus $V=E_1\oplus E_2$.

For distinct real roots ${\lambda }_1,\dots ,{\lambda }_m$, define polynomial projectorsThe Lagrange interpolation polynomials in this expression sum to one: their sum has degree at most $m-1$ and equals one at the $m$ distinct roots. Hence ${\sum }_i{P}_i=I$. Also $(\alpha -{\lambda }_{i}I)P_i=0$, because its numerator is the minimal polynomial evaluated at $\alpha$. Thus every vector is a sum of eigenvectors, $v={\sum }_i{P}_{i}v$. On $E_j$, $P_i$ acts as ${\delta }_{ij}I$, so applying $P_i$ to any zero sum of vectors from the eigenspaces shows that every summand is zero. ThereforeChoosing a basis in each eigenspace proves diagonalizability over the reals. Conversely, if a real matrix is similar over the reals to a diagonal matrix, a polynomial annihilates it exactly when it vanishes at each distinct real diagonal entry. Its minimal polynomial is therefore the product of those distinct linear factors. We have proved real diagonalizability is equivalent to a minimal polynomial with distinct real roots.

First work over a field of characteristic different from two. If a nonzero symmetric bilinear form had $\varphi (v,v)=0$ for every vector, polarization would givea contradiction. Hence choose $e$ with $\varphi (e,e)\ne 0$. Every $v$ decomposes uniquely aswith the second summand orthogonal to $e$. Induct on the orthogonal complement to obtain a diagonal basis. If the restricted form becomes zero, any basis of that remaining space completes the diagonalization. In this basis the rank of a quadratic form is exactly the number of nonzero diagonal coefficients; the remaining basis vectors span its radical of a bilinear form.

Over the reals, rescale the nonzero basis vectors so the diagonal entries are $p$ copies of $+1$, $q$ copies of $-1$ and $z=n-p-q$ zeros. Define the signature of a quadratic form by $\sigma =p-q$, so $r=p+q$. To prove it is basis independent, note that $p$ is the largest dimension of a linear subspace on which the form is positive-definite. The positive coordinate linear subspace attains dimension $p$; on any other positive linear subspace, projection to those positive coordinates is injective, since its kernel has nonpositive quadratic value. Thus its dimension is at most $p$. The same argument for the negative form characterizes $q$. This proves Sylvester's law of inertia and that the signature is well-defined.

Let $e_i$ and $f_i$ be positive and negative normalized basis vectors, and let $Z$ be the radical of a bilinear form. The span of $Z$ and $e_i+f_i$ for $1\le i\le \min (p,q)$ is a totally isotropic subspace: all mutual bilinear pairings vanish. Its dimension is $z+\min (p,q)$. Conversely, project any totally isotropic subspace $U$ to the nondegenerate positive-plus-negative quotient. The projected linear subspace is still null, and its projections to both the positive and negative coordinate spaces are injective: a vector with one component zero would have strictly signed quadratic value unless it vanished. Its dimension is therefore at most $\min (p,q)$, while the kernel of the quotient map on $U$ has dimension at most $z$. HenceThis proves both construction and sharpness of the maximum dimension of a totally isotropic subspace.

For the five-variable example, the symmetric coefficient matrix isIts cyclic structure makes the vectors $(1,{\zeta }^k,{\zeta }^{2k},{\zeta }^{3k},{\zeta }^{4k})$, $\zeta =e^{2\pi i/5}$, eigenvectors with eigenvalues ${\zeta }^k+{\zeta }^{-k}=2\cos (2\pi k/5)$. Thus the eigenvalues areTheir product is $2$, and three are positive and two negative. ThereforeThe real and imaginary parts of each conjugate pair of eigenvectors give the corresponding real two-dimensional eigenspaces, consistent with the real inertia calculation.

Let $d_n(x)=\det (xI-A_n)$, and set $d_0=1$, $d_1=x-a_1$. Expansion of the tridiagonal determinant along its last row givesFor the second term, the last-row entry is $-b_n$, its cofactor has negative sign, and the remaining minor has last-column entry $-b_n$ multiplying $d_{n-2}$; their product is $-b_n^2{d}_{n-2}$. Thus $d_n$ and $p_n$ have identical initial values and recurrence, and induction provesThe eigenvalues of the Jacobi matrix $A_n$ are precisely the roots of this characteristic polynomial. Part(a) therefore gives all $n$ eigenvalues simple and lying strictly in $(a,b)$. This conclusion follows from orthogonality as well as symmetry; symmetry alone would establish real eigenvalues but not their interval location.

Work in the finite-dimensional vector space $V$ implicit in the requested matrix representation and $n=\dim V$. Put $K_j=\ker {\alpha }^j$, with $K_0=0$. These form an increasing sequence ending in $K_m=V$, and $\alpha (K_j)\subseteq K_{j-1}$. Choose a basis of $K_1$, extend it to a basis of $K_2$, and continue until a basis of $V$ is obtained. Each basis vector introduced at stage $j$ maps into the span of vectors chosen at earlier stages. Its matrix column can therefore have nonzero entries only in earlier rows. This proves that the matrix is strictly upper triangular.

For a strictly upper-triangular $n\times n$ matrix $N$, a nonzero contribution to $(N^r{)}_{ij}$ requires an index chain $i<i_1<\cdots <i_{r-1}<j$. No such chain has $n$ links among $n$ indices. Consequently $\boxed{{\alpha }^n=0}$. An example with exact index four is the Nilpotent Jordan blockHere $M^3$ has entry one in position $(1,4)$ and all others zero, while $M^4=0$.

In the adapted basis, $I+A$ is upper triangular with diagonal entries one, so all its eigenvalues are one. This does not persist for the product of two such matrices with unrelated adapted bases. TakeBoth square to zero, buthas characteristic polynomial $t^2-3t+1$ and eigenvalues $(3\pm \sqrt{5})/2$, neither equal to one. The obstruction is that the two nilpotent matrices need not admit a common triangularizing basis.