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Applying the quadratic formula to the characteristic polynomial gives the two eigenvalues

Commuting inverse. Assume $I-M$ is invertible, as required by the displayed inverse. Multiplication givesMultiply on both sides by $(I-M{)}^{-1}$ to obtainMore generally, a matrix commuting with an invertible matrix also commutes with its inverse.

Orthogonality. For a real skew-symmetric matrix $A$, $x^{T}Ax=0$. If $(I-A)x=0$ for a real vector $x$, multiply by $x^T$ to obtain $\Vert x{\Vert }^2=0$. Thus $I-A$ is invertible; the same argument applies to $I+A$. For the Cayley transform of a real skew-symmetric matrix,The factors commute by the first result, soTherefore $\boxed{N\text{ is an orthogonal matrix}.}$

Eigenvector mapping. If $Av=av$, then $(I-A{)}^{-1}v=(1-a{)}^{-1}v$. ConsequentlyThe denominator cannot vanish. In fact the eigenvalues of a real skew-symmetric matrix are purely imaginary: $v^{†}Av$ is the negative of its complex conjugate, so $a=-\overline{a}$. For $a=i\beta$ the corresponding eigenvalue of $N$ is $(1+i\beta )/(1-i\beta )=e^{2i\arctan \beta }$, consistent with orthogonality.

Spectrum in three dimensions. Let $Q$ be any real three-by-three orthogonal matrix. For a complex eigenvector $v$, $\Vert Qv\Vert =\Vert v\Vert$ implies $|\lambda |=1$. A real cubic characteristic polynomial has a real root, necessarily $+1$ or $-1$. Any nonreal roots form a complex-conjugate pair of unit modulus. If all three roots are real, two of the three signs agree and can be grouped as a pair with angle zero or $\pi$. Thus in every case the eigenvalues can be writtenTheir product is $\epsilon =\det Q$.

Determinant and fixed axis for the Cayley transform. Since $A^T=-A$,so $\boxed{\det N=1}$. This determinant argument works in every dimension. In dimension three the preceding spectrum result then guarantees an eigenvector with $Nx=x$. Set $z=(I-A{)}^{-1}x$. The fixed-vector equation says $(I+A)z=(I-A)z$, so $Az=0$ and $x=(I-A)z=z$. HenceThis is the axis fixed by the three-dimensional rotation. Equivalently, a real odd-dimensional skew-symmetric matrix has zero determinant, because $\det A=\det A^T=\det (-A)=-\det A$.

Nilpotence and eigenvalues. If $Av=\lambda v$ with $v\ne 0$, then $A^{n}v={\lambda }^{n}v$. Thus $A^n=0$ forces $\lambda =0$ for every eigenvalue. Conversely, if all $n$ complex eigenvalues, counted with algebraic multiplicity, vanish, the monic characteristic polynomial is ${\chi }_A(t)=t^n$. The Cayley-Hamilton theorem then gives $A^n=0$. HenceThis is the criterion for a nilpotent matrix.

Symmetric case. Here a symmetric matrix is understood to be real. The spectral theorem gives $A=QD{Q}^T$ with $Q$ orthogonal and $D$ real diagonal. The previous result makes every diagonal entry zero, so $A=0$. Alternatively, choose $2^r\ge n$, so $A^{2^r}=0$. For every real vector $x$,It follows that $A^{2^{r-1}}=0$. Repeat until $A=0$. This uses positivity of the quadratic form $x^T{M}^{T}Mx$. The restriction to real symmetric, or complex Hermitian matrices, matters: a complex matrix satisfying only $A^T=A$ need not have this positivity.

The parameters and powers. The first two columns of $T$ coincide, so $\det T=0$ for all $\alpha ,\beta$. The sum of its principal two-by-two determinants, each a minor of a matrix, isTogether with $\mathrm{tr}T=2+\beta$, this givesFor every eigenvalue to vanish both remaining coefficients must be zero. ThereforeDirect multiplication at these values givesIndeed the first row of $T^2$ multiplied by $T$ is $3{\mathrm{row}}_1(T)+3{\mathrm{row}}_2(T)-6{\mathrm{row}}_3(T)=0$, and the second row is its negative.

Geometry of repeated application. For any parameters the image of a linear map lies in the span of $a=(1,1,1{)}^T$ and $b=(1,\alpha ,\beta {)}^T$, since $Tx=(x_1+x_2)a+x_{3}b$. Thus it lies in a fixed plane through the origin, independently of the input. When $a,b$ are independent its normal is $a\times b=(\beta -\alpha ,1-\beta ,\alpha -1)$; if they coincide, the image is a line and is still contained in a plane.

For $\alpha =-5,\beta =-2$, put $s=x_1+x_2$ and $t=x_3$. ThenThe columns span this entire plane, so $T$ has rank two. The next two steps areThus the image of a linear map contracts from a plane to a line and then to the origin under successive applications. The line from $T^2$ is precisely the kernel of $T$.

An eigenvalue $\lambda$ and associated eigenvector $e\ne 0$ satisfy $Ae=\lambda e$. The characteristic polynomial $\det (A-\lambda I)$ has degree equal to the positive matrix size. The fundamental theorem of algebra gives a complex root, at which $A-\lambda I$ is singular and therefore has a nontrivial kernel. Hence every nonempty square complex matrix has an eigenvector.

A nondiagonalizable example is the Jordan block $\left(\begin{array}{cc}0& 1\\ 0& 0\end{array}\right)$: its only eigenvalue is zero, and its eigenspace is one-dimensional. A Hermitian matrix satisfies $A^{†}=A$ and has an orthonormal eigenbasis by the spectral theorem. One proof finds an eigenvector, notes that its orthogonal complement is invariant, and repeats. Its eigenvalues are real becausewhere $e^{†}e>0$.

With all eigenvalues negative, $A$ is invertible. Its same complete eigenbasis givesSome powered eigenvalues can coincide, but the original eigenvectors remain a complete basis. Writing $A=U\mathrm{diag}({\lambda }_i)U^{†}$, the convergent matrix exponential series givesThus the matrix inverse from a decaying matrix exponential isThe identity printed in the original PDF is missing a minus sign. Already for the one-by-one matrix $A=(-1)$ its integral is $1$, whereas its inverse is $-1$. Equivalently, integrating $d(e^{tA})/dt=A{e}^{tA}$ gives $A{\int }_0^{\infty }{e}^{tA}dt=-I$, confirming the correction without diagonalization.

Over the chosen scalar field, an $n$-by-$n$ matrix is diagonalizable if and only if it has $n$ linearly independent eigenvectors. Repeated eigenvalues are allowed provided all eigenspaces together supply a basis.

For the given matrix, the characteristic polynomial is ${\lambda }^2-4\lambda -5=(\lambda -5)(\lambda +1)$. One may chooseTaking these as columns gives the change of basisThis is not a rotation matrix. The PDF's terminology cannot hold literally: $A$ is real and nonsymmetric, whereas a real matrix orthogonally similar to a real diagonal matrix is symmetric. Also $V_1\cdot V_2=-4\ne 0$, so scaling these eigenvectors cannot make an orthonormal rotation basis. A general invertible similarity is the valid construction.

Let $D=P^{-1}AP$. Inserting $P{P}^{-1}$ between successive factors gives $A^n=P{D}^n{P}^{-1}$, and henceTo write $A^n={\alpha }_{n}A+{\beta }_{n}I$, match its action on the two eigenspaces:Solving gives the matrix powers for a diagonalizable matrix with two eigenvaluesThe cases $n=0,1$ recover $I,A$, respectively. Equivalently the spectral projections of a diagonalizable matrix are $(A+I)/6$ and $(5I-A)/6$, and the power multiplies them by $5^n$ and $(-1{)}^n$.

Let $Q$ be the real orthogonal matrix. If $Qv=\lambda v$ for a nonzero complex eigenvector, then preservation of the complex Euclidean norm givesso $|\lambda |=1$. Its characteristic polynomial has real coefficients, hence nonreal eigenvalues occur in conjugate pairs. Such a pair is $e^{i\alpha },e^{-i\alpha }$, and the remaining real eigenvalue is $+1$ or $-1$.

If all three eigenvalues are real, each is $\pm 1$; two have the same sign and can be paired as $e^{\pm i\alpha }$ with $\alpha =0$ or $\pi$. Thus in every case the spectrum has the formTheir product shows that $\epsilon =\det Q$.

Use the full $3\times 3$ matrix $B$ in the original PDF; the transcription has dropped its middle column. For either matrix, the three-dimensional characteristic polynomial is ${\chi }_M(t)=t^3-(\mathrm{tr}M)t^2+[(\mathrm{tr}M{)}^2-\mathrm{tr}(M^2)]t/2-\det M$. Both have zero trace and zero determinant, while $\mathrm{tr}(B^2)=0$ and $\mathrm{tr}(C^2)=72$. ConsequentlyFor $B$, a useful check isfor every value of $\alpha$. Thus $B$ is a nonzero nilpotent matrix for all $\alpha$; for example its second column never vanishes. By the preceding result, $\boxed{B\text{ is not diagonalizable for any }\alpha }$.

For $C$, the three eigenvalues $-6,0,6$ are distinct and real, so their eigenvectors are linearly independent. An explicit eigenbasis isWith $P=(v_{-},v_0,v_{+})$, $\det P=18\ne 0$ and $P^{-1}CP=\mathrm{diag}(-6,0,6)$. Hence $\boxed{C\text{ is diagonalizable over }\mathbb{R}\text{ and }\mathbb{C}}$.

A repeated root of the characteristic polynomial alone would not settle diagonalizability; for $B$ it is the nonzero nilpotence that completes the argument.

Keep the constraint from part (b). For equal masses and spring constants, the generalized eigenvalue problem for small oscillations reduces toThe horizontal coordinate decouples, givingIn this normal mode, mass 1 oscillates horizontally, mass 2 has its horizontal coordinate fixed, and neither mass has a vertical displacement.

For the vertical block, write $\lambda =m{\omega }^2/k$. Its characteristic polynomial is $(2-\lambda )(1-\lambda )-1={\lambda }^2-3\lambda +1$. Thus ${\lambda }_{\pm }=(3\pm \sqrt{5})/2$, and the first vertical equation gives $y_2/y_1=2-{\lambda }_{\pm }$. If $\phi =(1+\sqrt{5})/2$ is the golden ratio, the two vertical modes areIn the lower-frequency mode, the masses move vertically in phase, with the lower mass having the larger amplitude. In the higher-frequency mode they move oppositely, with lower-to-upper amplitude ratio $1/\phi$. Their Euclidean inner product is zero because $1+\phi (-1/\phi )=0$; since $M=mI$, this is also mass-matrix orthogonality of normal modes.

The three frequencies satisfy ${\omega }_{-}<{\omega }_x<{\omega }_{+}$. Each mode has a scalar coordinate $A\cos \omega t+B\sin \omega t$, and their superposition gives the general constrained motion. No zero-frequency mode remains because the first spring anchors the system. Gravity fixes the equilibrium extensions but drops out of these frequencies for zero-natural-length springs.

The $z$ direction is fixed, giving the eigenvalue one. The planar block has characteristic polynomialThus the full list of eigenvalues isThe last two are real precisely when $\sin \varphi =0$, so all three are real exactly for $\boxed{\varphi \in \pi \mathbb{Z}}$. At even multiples of $\pi$ the matrix is the identity and every vector is fixed. At odd multiples it is a half-turn: the axial eigenvalue remains $+1$, while the whole perpendicular plane has eigenvalue $-1$.

For other angles, the only real eigenvector direction is the rotation axis; the perpendicular plane is rotated rather than preserving an individual real direction. Its complex eigenvectors can be taken as $(1,-i,0{)}^T$ and $(1,i,0{)}^T$ for $e^{i\varphi }$ and $e^{-i\varphi }$, respectively. These represent the two complex conjugate planar eigenvalues.

Put $J_0=4\sigma D^4/3=M{D}^2/3$. The normal to the lamina is already an eigenvector, with moment $3J_0$. The planar block has characteristic polynomialso the principal moments of inertia areA corresponding orthonormal set of principal axes isThe sign of each axis vector is immaterial. The smaller-moment axis makes angle $\frac{1}{2}\arctan (2)$ with the positive $x_1$ direction, approximately $31.7^{\circ }$; the other planar axis is perpendicular. As a check, $I_{-}+I_{+}=3J_0=I_3$, the perpendicular axis theorem for a planar lamina.

Let $z={\omega }_1^2$, so ${\omega }_2^2=4z$. The sum and product of roots of the characteristic polynomial in (i) giveEliminating $z$ yieldsTherefore the octave tuning in a two-mass complete spring graph has two positive solutions:The two values are reciprocals. Both meet the requested frequency ratio; substituting either into the expressions in (i) gives ${\omega }_2/{\omega }_1=2$.

The quadratic form is ${\mathbf{x}}^{T}Q\mathbf{x}$ for the real symmetric matrixThe $y$ axis has eigenvalue $-3$. On the $xz$ plane, the characteristic polynomial isThus the orthogonal diagonalization from part (a) yieldsFor later use, one possible orthonormal basis isThen $u=(3x+z)/\sqrt{10}$, $v=y$, $w=(x-3z)/\sqrt{10}$. This explicit choice also confirms that the Euclidean norm is preserved: $x^2+y^2+z^2=u^2+v^2+w^2$.

Expanding the characteristic polynomial givesTherefore $a=-2z$ and $b=z^2-x^2-y^2$. Both are polynomial functions of the three real coordinates, so they are continuous functions, including on the open subset $X$ of invertible matrices.

On $X$, $b=\det A\ne 0$. Its image contains positive values, for example $b(I)=1$, and negative ones, for example $b(\mathrm{diag}(1,-1))=-1$. If $X$ were connected, its continuous real image would be an interval containing both signs and hence zero. That is impossible. Thus $X$ is not connected. More precisely its determinant image is $\mathbb{R}\setminus \{0\}$: take $x=y=0$ for a positive value and $z=y=0$ for a negative one.

The Jacobian matrix at the origin and its characteristic polynomial areThe fixed-point multipliers areBy the Jury stability criterion, both lie strictly inside the unit circle exactly whenThe last condition always holds, since ${\mu }^2+\mu +2=(\mu +1/2{)}^2+7/4$. Therefore the origin is asymptotically stable exactly for $-1<\mu <0$ among the hyperbolic parameter values.

For completeness, its linear stability throughout the parameter line is as follows. When $\mu <-1$, the complex-conjugate Floquet multipliers have modulus $|\mu |>1$, so the origin repels. When $-1<\mu <-1/3$, their modulus is $|\mu |<1$, so it attracts with oscillation. At $\mu =-1/3$ the repeated Floquet multiplier is $1/3$, still attracting. For $-1/3<\mu <0$, both Floquet multipliers are real and in $(0,1)$. For $0<\mu <1$, $p(1)<0$ and $p(0)>0$, so one Floquet multiplier lies in $(0,1)$ and the other exceeds one: the origin is a saddle fixed point of a map. For $\mu >1$, the complex-conjugate Floquet multipliers have modulus $\mu >1$, so it repels.

Possible changes at the unit circle are exhausted by $p(1)=0$, $p(-1)=0$, or a nonreal pair with product one. They giveAt $\mu =0$ the Floquet multipliers are $0,1$; the nonzero fixed point branches in part (a) meet the origin, and part (c) identifies a subcritical pitchfork bifurcation. Indeed at $\mu =0$ the exact restricted map is $y\mapsto y+y^3$, which makes the origin unstable despite its neutral linear direction.

At $\mu =-1$ the Floquet multipliers are $i,-i$: this is a 1:4 strong resonance of a planar map, coinciding with the change from repelling to attracting as $\mu$ increases. A generic nonresonant Neimark–Sacker bifurcation conclusion would be unjustified. At $\mu =1$ there is a double Floquet multiplier $+1$; the matrix has a nontrivial Jordan block, and the two further nonzero fixed point branches meet the origin. Linear stability alone is inconclusive at these unit-modulus boundary values. The repeated Floquet multiplier at $\mu =-1/3$ is strictly inside the unit circle and gives no bifurcation.

The determinant is invariant under transpose. Therefore the characteristic polynomials coincide:Their polynomial roots, including their algebraic multiplicities, are the same, so $A$ and $A^T$ have the same eigenvalues. This allows column sums as well as row sums to supply Gershgorin discs for the same spectrum.

With $f(x,y)=-x{y}^2-y^5$, write $t=x_{n+1}=x_n+h$ and $z=y_{n+1}$. The Backward Euler method requiresor the monic quintic polynomial equationConstruct its $5\times 5$ companion matrixIts characteristic polynomial is $\det (zI-C_n)=p(z)$. Thus the polynomial roots needed for this update are exactly the eigenvalues of $C_n$.

At each step, assemble $C_n$, compute its eigenvalues, for example with the QR algorithm, and choose its unique positive real eigenvalue as $y_{n+1}$. To justify this selection, letIf $y_n>0$, then $F(0)=-y_n<0$, $F(y_n)=h{y}_n^5+ht{y}_n^2>0$, andThe intermediate value theorem and strict monotonicity prove there is exactly one positive polynomial root, located in $(0,y_n)$. Starting at $y_0=1$ therefore makes every update well defined, positive and decreasing. In numerical arithmetic, select the nearly real eigenvalue in this interval and check the polynomial residual $|F(z)|$; a final Newton root-finding iteration can refine it. This reduces each update to the requested eigenvalue problem.

As appropriate for a genuine $k$-step linear multistep method, take ${\alpha }_k\ne 0$. Suppose the characteristic polynomial has a polynomial root $\xi$ with modulus $|\xi |>1$. From $\rho (\xi )=0$ and the triangle inequality,Dividing by the positive number $|{\alpha }_k||\xi {|}^{k-1}$ gives $|\xi |\le 1$, a contradiction. HenceThe nonzero leading coefficient matters: if ${\alpha }_k=0$, the assumed inequality forces all the ${\alpha }_j$ to vanish, so $\rho$ is the zero polynomial rather than a characteristic polynomial with $k$ roots. This degenerate case does not define the method in the question.

The four row Gershgorin discs have centres $9,3,-4,-5$ and radii $1,2,2,1$, respectively. Their real-axis intersections areNone of the Gershgorin discs contains zero, so the Gershgorin circle theorem givesThe discs centred at $9$ and $3$ are each isolated. By Gershgorin eigenvalue counting, they contain exactly one eigenvalue apiece, and both eigenvalues are simple. Each must be real: the characteristic polynomial is real, so a nonreal root would bring a distinct conjugate root into the same disc. Denote these eigenvalues by $\beta \in [8,10]$ and $\alpha \in [1,5]$. The remaining two eigenvalues lie in the union of the negative-centred discs.

For the shift $4$, the distances satisfyfor either of the other eigenvalues $\gamma$, since its real part is at most $-2$. Thus $\alpha$ is uniquely closest to $4$. Once $4$ is checked not to be an eigenvalue, the natural full target-disc interval enclosure and a conservative asymptotic direction factor areAs always, inverse iteration requires a nonzero initial target component. The following calculation verifies the shift and gives sharper information.

For the tridiagonal matrix, let $p_j(t)$ be the determinant of the leading $j\times j$ block of $tI-A$. Expansion along the last row givesIn particular, $\det A=577$, $p_4(4)=-239$ and $p_4(5)=-583$. Both shifted linear systems are nonsingular. Further exact evaluations areThe intermediate value theorem gives a real root in each of $(-6,-5)$, $(-4,-3)$, $(3,4)$ and $(8,9)$. As $p_4$ has polynomial degree four, these are all its roots and each is simple. In particular, $A$ is a diagonalizable matrix, $\alpha \in (3,4)$ and $\beta \in (8,9)$. The closest competitor to $\alpha$ at the shift $4$ is $\beta$, so this refinement yieldsFor the shift $5$, the original Gershgorin discs only give $|\alpha -5|\le 4$ and $|\beta -5|\ge 3$; those bounds do not decide which is closer. The refined interval enclosures do:for the negative eigenvalues. Therefore the answer for this particular matrix is yes:Both shifts select the same eigenvector line and alternate signs because $\alpha <\mu$. The factor bounds concern asymptotic directional convergence, rather than a universal error reduction from the first iterate for a nonsymmetric matrix. The wider interval $[1,5]$ remains a valid full-disc enclosure for both shifts; $[3,4]$ records the stronger information obtained from the characteristic polynomial.

The characteristic polynomial of the given Jones matrix isso a convenient choice of eigenvalue labels isDirect multiplication gives $J{\mathbf{u}}_{\pm }={\lambda }_{\pm }{\mathbf{u}}_{\pm }$, and ${\mathbf{u}}_{+}^{†}{\mathbf{u}}_{-}=0$. They represent opposite circular polarizations. Since $a,b$ are real, $J$ is a Hermitian matrix and both eigenvalues are real, but they need not be positive. A negative eigenvalue adds a wave phase reversal to that mode. The labels are algebraic: ${\lambda }_{+}$ is the larger numerical eigenvalue only when $b\ge 0$. If $b=0$, the eigenvalue is degenerate and every nonzero Jones vector is an eigenvector; the circular pair remains a valid basis.

For a complex matrix $C=(c_{ij})$, Gershgorin's theorem puts every eigenvalue in the union of the row discs $D_i=\{z:|z-c_{ii}|\le R_i\}$, where $R_i={\sum }_{j\ne i}|c_{ij}|$. To prove it, choose an eigenvector $v$ for an eigenvalue $\lambda$ and an index $i$ maximizing $|v_i|>0$. Its $i$th equation givesDivision proves $\lambda \in D_i$.

The usual counting refinement says that a connected component made of $k$ discs contains exactly $k$ eigenvalues, with algebraic multiplicity. For completeness, deform the matrix from its diagonal part to the original by multiplying all off-diagonal entries by $t\in [0,1]$. Its discs have radii $t{R}_i$ and stay inside the original discs. Continuous variation of the roots of the characteristic polynomial keeps the number inside any contour separating one original component from the others constant. At $t=0$, the diagonal entries give exactly $k$ roots there, proving the refinement.

For the application put $s=-\mu >0$. The matrix isIts three row discs have positive real centres and leftmost real coordinatesAll are strictly positive, so zero lies in none of the discs. It cannot be an eigenvalue. Thus $A-\mu B$ is nonsingular for every real $\mu <0$.