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The real orthogonal group and special orthogonal group are, under matrix multiplication,An orthogonal matrix preserves Euclidean inner products; taking determinants of $Q^{T}Q=I$ shows that its determinant is $1$ or $-1$. Its inverse is $Q^T$, so $Q{Q}^T=I$ as well.

For $Q\in SO(3)$, use $Q-I=Q(I-Q^T)$. The determinant then satisfiesIt follows that $\det (Q-I)=0$. Thus the real matrix $Q-I$ has a nonzero vector in its kernel of a linear map, yieldingThis is the three-dimensional case of odd-dimensional special orthogonal transformation has a fixed vector; geometrically it supplies the fixed rotation axis, unless the transformation is the identity and every direction is fixed.

The claim is false for $O(3)$. The matrix $Q=-I_3$ is orthogonal with determinant $-1$ and has only the eigenvalue $-1$. It therefore has no nonzero eigenvector with eigenvalue $1$.

Introduce $v=\dot{x}$. The phase plane system for the damped pendulum isAt an equilibrium point, $v=0$ and $\sin x=0$, so all equilibrium points are $\boxed{(x,v)=(n\pi ,0),n\in \mathbb{Z}}$. Angles identified modulo $2\pi$ give one downward and one upward equilibrium point.

The linearization of a dynamical system at $(n\pi ,0)$ has Jacobian matrix and eigenvaluesWhen $n$ is even and $k>\omega$, both eigenvalues are real and strictly negative, since $0<\sqrt{k^2-{\omega }^2}<k$. These downward equilibrium points are stable nodes. When $n$ is even and $k<\omega$, the eigenvalues are $-k\pm i\sqrt{{\omega }^2-k^2}$, so the downward equilibrium points are stable foci, approached with decaying oscillations.

When $n$ is odd, $\sqrt{k^2+{\omega }^2}>k$, so one eigenvalue is positive and one is negative in either parameter range. Every upward equilibrium point is consequently an unstable saddle equilibrium. All these equilibrium points are hyperbolic equilibria in the two requested cases, so their linearization of a dynamical system gives their local nonlinear classification.

False. Every entry of $A=\left(\begin{array}{cc}1& 2\\ 2& 1\end{array}\right)$ is strictly positive, yet $A(1,-1{)}^T=-(1,-1{)}^T$. Its eigenvalues are $3$ and $-1$, so one has negative real part. Positivity of entries does not imply the positive-definite matrix property.

False. The matrix $A=\left(\begin{array}{cc}1& 1\\ 0& 2\end{array}\right)$ has real eigenvalues $1$ and $2$, because its characteristic polynomial is $(t-1)(t-2)$, but $A^{\ast }=\left(\begin{array}{cc}1& 0\\ 1& 2\end{array}\right)\ne A$. Real spectrum alone does not imply that it is a Hermitian matrix.

True. If $Av=\lambda v$ with $v\ne 0$ and $A=A^{\ast }$, thenSince $v^{\ast }v>0$, $\lambda =\bar{\lambda }$. Thus every eigenvalue of a Hermitian matrix is real.

False in general. Take $A=\mathrm{diag}(1,6,6)$ and $B=\mathrm{diag}(2,2,9)$. Both have trace $13$ and determinant $36$, but their eigenvalue multisets are different. For $n>3$, append equal identity blocks to preserve this counterexample. In dimensions one and two the assertion is true: in dimension two the characteristic polynomial is $t^2-(\mathrm{tr}A)t+\det A$, whose roots, with multiplicity, are determined by those two quantities.

For $n\ge 1$, an eigenvalue is a number $\lambda \in \mathbb{C}$ for which there is a nonzero vector $v$ satisfying $Av=\lambda v$. Equivalently, $\ker (A-\lambda I)\ne \{0\}$, or $\det (A-\lambda I)=0$. The characteristic polynomial $\det (tI-A)$ has degree $n$ and leading coefficient one. By the fundamental theorem of algebra it has a complex root $\lambda$. At that root $A-\lambda I$ is a singular matrix, so it has a nonzero kernel vector, proving that $A$ has an eigenvalue.

For the real Sturm-Liouville problem, multiply the equations for $y_n$ and $y_m$ by $y_m$ and $y_n$, respectively, and subtract. Integration by parts givesThe boundary term vanishes because both functions are zero at both endpoints. Distinct eigenvalues therefore imply weighted orthogonality. For equal indices the integral is the squared weighted norm, so

Let $c_n={\int }_0^{1}w{y}_n^{2}dx>0$. For $y=\sum a_n{y}_n$ in the operator domain, with convergence sufficient to evaluate its quadratic energy, weighted orthogonality givesFor finite sums these identities follow immediately; the corresponding convergent energy expansion follows by approximation in the operator or quadratic-form domain. For nonzero $y$, the Rayleigh quotient is a weighted average of the eigenvalues, soEquality occurs precisely when only a lowest-eigenvalue component is present.

Assume completeness as permitted and take a sufficiently regular solution of the diffusion equation. It belongs to the weighted expansion space at each positive time; its modal coefficients satisfy ${\dot{a}}_n=-{\lambda }_n{a}_n$. Alternatively differentiate its energy directly and use the preceding inequality:Thus the squared weighted norm is at most its initial value times $e^{-2{\lambda }_{1}t}$. This describes decay when ${\lambda }_1>0$; positivity of $p,w$ alone does not require ${\lambda }_1>0$ if the potential $q$ can be negative. The displayed inequality remains valid regardless of that sign.

A Hermitian matrix satisfies $A=A^{†}$, where $†$ means conjugate transpose. If $Av=\lambda v$ with $v\ne 0$, then $v^{†}Av=\lambda \Vert v{\Vert }^2$ is real, since it equals its complex conjugate $v^{†}{A}^{†}v$. Thus every eigenvalue is real. To prove the finite-dimensional spectral theorem, choose a unit eigenvector $v$; one exists because the complex characteristic polynomial has a root. Its orthogonal complement is invariant, since for $w\perp v$,The restriction to that complement is again Hermitian. Induction on dimension supplies an orthonormal basis of eigenvectors, so $A=U\mathrm{diag}({\lambda }_1,\dots ,{\lambda }_n)U^{†}$ for a unitary matrix $U$.

In this basis, $x^{†}Ax={\sum }_i{\lambda }_i|(U^{†}x{)}_i{|}^2$. Therefore $\boxed{A>0⟺{\lambda }_i>0\text{ for every }i}$. In that case the square root of a matrixis Hermitian, positive definite and satisfies $C^2=A$. For uniqueness, any positive-definite Hermitian $D$ satisfying $D^2=A$ commutes with $A$ and hence preserves each eigenspace of $A$. Its restriction to the $\lambda$-eigenspace is Hermitian, and its positive eigenvalues all solve ${\mu }^2=\lambda$. Thus that restriction is $\sqrt{\lambda }I$, forcing $D=C$. This is the principal square root of a positive semidefinite matrix, here strictly positive definite.

Now let $D=\sqrt{B}$ and $X=C-D$. Expanding without assuming that $C$ and $D$ commute givesThe Hermitian matrix $X$ has $X^2\ge 0$ because $v^{†}{X}^{2}v=\Vert Xv{\Vert }^2$, whereas $A-B>0$. Hence $CX+XC>0$. If $Xv=\mu v$, $v\ne 0$, thenSince $C>0$, every real eigenvalue $\mu$ of $X$ is positive. The positive-eigenvalue criterion now gives the strict finite-dimensional operator monotonicity of the square root:

A Jordan normal form of a complex $n\times n$ matrix $A$ is a similarity transformation $P^{-1}AP=J$ in which $J$ is a direct sum of Jordan blocksFor each eigenvalue, the multiset of block sizes is uniquely determined by $A$; only the order of the blocks is arbitrary. The change-of-basis matrix $P$ is generally not unique.

For the given matrix, expansion of the characteristic polynomial givesThe upper-left $2\times 2$ minor of $A-I$ is $-5$, so its rank is $2$ and the eigenspace for $1$ has dimension $1$. Thus the two occurrences of the eigenvalue $1$ belong to a single size-$2$ Jordan block. More explicitly, the Jordan chain and the remaining eigenvector can be chosen asConsequently, with $P=(v_1{v}_2{v}_3)$, for which $\det P=1$, the answer is

Use complex inner products linear in the first argument throughout. The spectral theorem for compact self-adjoint operators states that, for a compact operator $T=T^{\ast }$ on a Hilbert space, its nonzero spectral values are real eigenvalues, each of finite multiplicity, with no possible accumulation point except zero. Moreover,where $P_{\lambda }$ are the mutually orthogonal eigenspace projections. The operator series converges in operator norm when the nonzero eigenvalues are listed in decreasing absolute value. The sum is at most countable, even if the kernel is nonseparable. Zero need not be an eigenvalue.

First prove that a nonzero compact self-adjoint operator has an eigenvalue of magnitude its norm, without assuming a spectral decomposition. Put $r=\Vert T\Vert >0$ and choose unit vectors ${\xi }_n$ with $\Vert T{\xi }_n\Vert \to r$. Since $\Vert T^2\xi \Vert \le r\Vert T\xi \Vert$, expansion givesCompactness of $T^2$ supplies a subsequence for which $T^2{\xi }_n$ converges; the displayed estimate then forces ${\xi }_n$ itself to converge to a unit vector $\xi$ satisfying $T^2\xi =r^2\xi$. The vectors $\xi +T\xi /r$ and $\xi -T\xi /r$ are eigenvectors for $r$ and $-r$, respectively, whenever nonzero, and they cannot both vanish. This proves the claim.

Self-adjointness gives reality of every eigenvalue and orthogonality of eigenspaces for distinct eigenvalues. An eigenspace for $\lambda \ne 0$ cannot contain an infinite orthonormal sequence: its images under $T$ would have pairwise distance $\sqrt{2}|\lambda |$, contradicting compactness. The same argument proves that there are only finitely many mutually orthogonal eigenvectors with eigenvalues of magnitude at least any given $\epsilon >0$. Hence all nonzero eigenspaces together have a countable orthonormal basis, and their eigenvalues tend to zero if there are infinitely many.

Let $K$ be the orthogonal complement of the closed span of all these eigenspaces. It is invariant under $T$. If $T{|}_K$ were nonzero, the norm-attaining eigenvalue argument applied to that compact self-adjoint restriction would produce another nonzero eigenvector, a contradiction. Thus $K\subseteq \ker T$, and the converse inclusion follows from orthogonality. This establishes the asserted decomposition. On its orthogonal summands the tail of the operator series has norm equal to the supremum of the omitted eigenvalue magnitudes, proving norm convergence.

Finally, a nonzero complex number not among these eigenvalues has positive distance from their set and from zero. Inverting $T-zI$ separately on the eigenspaces and kernel therefore gives a bounded inverse. This proves that no other nonzero spectral values occur and completes the theorem.