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If $\alpha v=\lambda v$ with $v\ne 0$, then $m_{\alpha }(\alpha )v=m_{\alpha }(\lambda )v$. The left side vanishes, so $m_{\alpha }(\lambda )=0$.

Conversely, if $m_{\alpha }(\lambda )=0$, write $m_{\alpha }(X)=(X-\lambda )q(X)$. Then $(\alpha -\lambda I)q(\alpha )=0$. If $\alpha -\lambda I$ were invertible, multiplying by its inverse would imply $q(\alpha )=0$, contradicting the minimal degree. Thus $\alpha -\lambda I$ is not invertible. In finite dimension this means it has a nonzero kernel, furnishing an eigenvector with eigenvalue $\lambda$. Hence the eigenvalues are exactly the roots of the minimal polynomial, with the scalar-field convention stated in part (a).

By orthogonal diagonalization of the positive-definite symmetric matrix $A$, form its positive symmetric inverse square root $S=A^{-1/2}$. Then $S^{T}AS=I$, and $C=SBS$ is real symmetric. Let $Q$ have an orthonormal eigenbasis of $C$ as its columns, so $Q^{T}CQ=\mathrm{diag}({\lambda }_1,\dots ,{\lambda }_n)$. Taking $P=SQ$ givesBoth factors of $P$ are invertible. This proves the simultaneous congruence diagonalization with a positive form by an explicit construction; it is congruence rather than simultaneous similarity.

To find the entries without constructing $P$, take determinants inSince $\det P\ne 0$, the entries are precisely the roots, with multiplicity, of $\det (B-\lambda A)=0$. Equivalently they are the eigenvalues of $A^{-1}B$, although that matrix need not itself be symmetric.

For the given $A$, its leading principal minors are $2,3,4$, so it is positive definite. Expanding the determinant givesThus the diagonal entries, in any order, are

Relative to the coordinate basis, the linear map has matrixExpanding $\det (tI-F)$ in its first column gives the monic characteristic polynomialIts eigenvalues are $1$ and $(1\pm i\sqrt{7})/2$, three distinct complex numbers. Eigenvectors corresponding to distinct eigenvalues are linearly independent, so they form a basis of ${\mathbb{C}}^3$ and $f$ is diagonalizable.

Every annihilating polynomial must vanish at all three eigenvalues: applying $p(f)$ to a corresponding eigenvector gives $p(\lambda )v$. Conversely, in the eigenbasis the product of the three distinct linear factors annihilates $f$. Hence the minimal polynomial isIf $af+b{f}^2=0$ with scalars $a,b$ not both zero, the nonzero polynomial $at+b{t}^2$ of degree at most two would annihilate $f$, contradicting this minimal degree of three. Thus $f$ and $f^2$ are linearly independent endomorphisms.

Fix real $y$ and use a rectangle with vertical sides at real coordinates $\pm R$. The function $e^{-z^2}$ is entire, so its closed contour integral is zero by Cauchy's integral theorem. The bottom horizontal side contributes ${\int }_{-R}^R{e}^{-x^2}dx$ and the top contributes the negative of ${\int }_{-R}^R{e}^{-(x+iy{)}^2}dx$. This identity applies for either sign of $y$, with the corresponding oriented rectangle.

On a vertical side, $z=\pm R+is$ with $|s|\le |y|$, andThe sum of the two vertical integrals in absolute value is at most $2|y|e^{-R^2+y^2}\to 0$. Both horizontal improper integrals converge absolutely, since $|e^{-(x+iy{)}^2}|=e^{y^2}{e}^{-x^2}$. Letting $R\to \infty$ therefore provesComplete the square in the Fourier transform:The shifted-integral identity applies after scaling by $\sqrt{2}$. The Gaussian integral is $\sqrt{2\pi }$: the square of $\int e^{-u^2}du$ becomes $2\pi {\int }_0^{\infty }{e}^{-r^2}rdr=\pi$ in polar coordinates. Thus the Gaussian Fourier transform yieldsThe eigenvalue reflects the unnormalized transform convention in this question.

Under a Cartesian frame change $B_i^{\prime }=R_{ik}{B}_k$, where $R$ is an orthogonal matrix. Orthogonality gives $B_k^{\prime }{B}_k^{\prime }=B_k{B}_k$ and $R_{ik}{R}_{j\ell }{\delta }_{k\ell }={\delta }_{ij}$. Thereforewhich is the tensor component transformation law. Also $T_{ij}=T_{ji}$ directly, proving that it is a symmetric Cartesian second-rank tensor for every choice of the two scalars.

For $\mathbf{B}\ne 0$, the direction of $\mathbf{B}$ is an eigenvector with eigenvalue $(\alpha +\beta )|\mathbf{B}{|}^2$. Every vector perpendicular to $\mathbf{B}$ is an eigenvector with eigenvalue $\beta |\mathbf{B}{|}^2$, a two-dimensional eigenspace. If $\alpha =0$, these eigenvalues coincide and every nonzero vector is an eigenvector. If $\mathbf{B}=0$, the tensor is zero and again every nonzero vector is an eigenvector, now with eigenvalue zero.

For the position-dependent field with zero divergence, the product rule givesContraction of the two Levi-Civita symbols in the curl and cross product givesConsequentlyThese constants give the magnetic part of the Maxwell stress tensor, with its physical prefactor suppressed. They are forced if the identity is to hold for every divergence-free field: $\mathbf{B}=(y,0,0)$ forces $2\beta =-1$; at the origin $\mathbf{B}=(y,1,0)$ then forces $\alpha =1$.

Apply the divergence theorem separately to each component:Since the whole vector field vanishes on $S$, all components of $T$ vanish there, and hence the vector volume integral is zero. The argument assumes a differentiable field and a boundary for which the divergence theorem applies.

If $U$ is a unipotent matrix, $N=U-I$ is a nilpotent matrix. If $Uv=\lambda v$ for a nonzero eigenvector, then $N^{m}v=(\lambda -1{)}^{m}v=0$ for some $m$, forcing $\lambda =1$.

Conversely, suppose the only eigenvalue is $1$. Over $\mathbb{C}$, the characteristic polynomial is then $(t-1{)}^n$. The Cayley-Hamilton theorem gives $(U-I{)}^n=0$, so $U-I$ is nilpotent. Thus $U$ is unipotent if and only if its only eigenvalue is one.

For any complex $\lambda$, the shifted linear operator $T=\alpha -\lambda I$ is normal: expansion of $T^{\ast }T-T{T}^{\ast }$ leaves just ${\alpha }^{\ast }\alpha -\alpha {\alpha }^{\ast }=0$. Apply part (i) to this shifted normal operator:Hence a nonzero vector is an eigenvector of $\alpha$ with eigenvalue $\lambda$ if and only if it is an eigenvector of ${\alpha }^{\ast }$ with eigenvalue $\bar{\lambda }$. The eigenspaces coincide, with conjugate eigenvalues.

Proceed by induction on $\dim V$. The zero-dimensional case has the empty orthonormal basis. In positive dimension, the fundamental theorem of algebra gives a root of the characteristic polynomial, hence a unit eigenvector $e$ of $\alpha$, say with eigenvalue $\lambda$. Part (ii) makes it an eigenvector of ${\alpha }^{\ast }$ as well. Thus the line spanned by $e$ and its orthogonal complement are invariant under both operators by the introductory adjoint argument.

On $\{e{\}}^{\perp }$, the restriction of ${\alpha }^{\ast }$ is the adjoint of the restriction of $\alpha$, and the restrictions still commute. The restricted linear operator is therefore normal. By induction it has an orthonormal basis of eigenvectors. Adjoining $e$ proves an orthonormal eigenbasis exists for $V$, establishing the finite-dimensional spectral theorem by construction.

For the final deduction, use this orthonormal basis $e_j$ and write $\alpha e_j={\lambda }_j{e}_j$. DefineThe real nonnegative diagonal entries make $\beta$ a Hermitian operator, and the modulus-one diagonal entries make $\gamma$ a unitary operator. Both are diagonal in the same orthonormal basis, so they commute and $\beta \gamma e_j={\lambda }_j{e}_j$ even when ${\lambda }_j=0$. This proves the required factorization without assuming invertibility.

Conversely, if $\alpha =\beta \gamma$ with ${\beta }^{\ast }=\beta$, ${\gamma }^{\ast }\gamma =\gamma {\gamma }^{\ast }=I$, and $\beta \gamma =\gamma \beta$, then ${\alpha }^{\ast }={\gamma }^{\ast }\beta$ andThus $\alpha$ is normal. Normality is equivalent to a commuting Hermitian–unitary factorization, with the Hermitian factor even selectable nonnegative. This is the polar decomposition with a unitary extension on the kernel.

The minimal polynomial divides the annihilating polynomial $t^2(t-1)$, since $A^3-A^2=0$. Every nonconstant monic divisor can occur in dimension four: use Jordan blocks of sizes at most two at eigenvalue $0$ and of size one at eigenvalue $1$, adding scalar blocks as needed. The complete list isFor example the first four occur for $0$, $J_2(0)\oplus 0\oplus 0$, $I$, and $\mathrm{diag}(0,0,1,1)$, respectively; the final one occurs for $J_2(0)\oplus 0\oplus 1$.

Failure to be diagonalizable requires a size-two Jordan block at zero. The condition $A^2\ne 0$ requires at least one block at one, since every allowed zero block squares to zero. A size-two zero block and one scalar block at one leave only one dimension. It can carry zero or one. Thus, up to permutation of blocks, the only possible Jordan normal forms are

Use the weighted inner product $⟨u,v{⟩}_r={\int }_a^{b}r(x)u(x)v(x)dx$. For real eigenfunctions, integration by parts with the zero endpoint values givesThus $({\lambda }_n-{\lambda }_m)⟨y_m,y_n{⟩}_r=0$. The regular separated Sturm-Liouville problem has simple eigenvalues: for a fixed eigenvalue, prescribing $y(a)=0$ leaves only the initial flux $p(a)y^{\prime }(a)$ free, so uniqueness of the associated first-order system makes any two eigenfunctions proportional. Distinct eigenfunctions are therefore orthogonal, and choosing their signs and scales so that $\int r{y}_n^2=1$ givesThe positive energy form $\int (p{y}^{\prime 2}+q{y}^2)$ realizes $r^{-1}\mathcal{L}$ as a positive self-adjoint operator in this weighted space, with compact inverse on the bounded interval. The spectral theorem for compact self-adjoint operators supplies a complete eigenbasis. This is the standard regular Sturm-Liouville eigenfunction expansion, also valid with continuous $p$ by working with the continuous flux $p{y}^{\prime }$.

For a source $F$, expand a zero-endpoint solution of $(\mathcal{L}-\mu r)y=F$ in that basis. Multiplication by $y_n$ and unweighted integration giveSince $\mu$ is not an eigenvalue, every coefficient is determined. The spectral Green function of a regular Sturm-Liouville problem is consequentlyWeighted completeness means $r(x){\sum }_n{y}_n(x)y_n(\xi )=\delta (x-\xi )$ as a distribution, so applying $\mathcal{L}-\mu r$ to this series produces the required delta source. The source integration is with $d\xi$, explaining the absence of an additional factor $r(\xi )$ in the stated kernel.

For the subsequent operator $L=d^2/d{x}^2$ the sign is explicitly reversed: $y_n=\sqrt{2/\pi }\sin (nx)$ and ${\lambda }_n=-n^2$. To cover every real non-eigenvalue $\mu$, define the homogeneous solutionIt satisfies $S_{\mu }(0)=0$, $S_{\mu }^{\prime }(0)=1$ and $S_{\mu }^{\prime \prime }=\mu S_{\mu }$. Matching the two endpoint solutions continuously at $x=\xi$ with derivative jump $G_x({\xi }^{+},\xi )-G_x({\xi }^{-},\xi )=1$ yieldsThe denominator is nonzero precisely off the Dirichlet eigenvalues $\mu =-n^2$. In particular $G_0=-x_{<}(\pi -x_{>})/\pi$. At $x=\xi =\pi /2$, its spectral expansion givesThis odd reciprocal-square sum from a Dirichlet Green function proves $\boxed{\sum _{j=0}^{\infty }(2j+1{)}^{-2}={\pi }^2/8}$.