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Place equal masses at the six vertices $\pm {\mathbf{e}}_i$ of a regular octahedron and identical unstressed axial springs on its twelve edges. For the invariant axial displacement family ${\mathbf{u}}_{\pm i}=\pm q_i{\mathbf{e}}_i$, the quadratic potential and kinetic energies areHence $m\ddot{\mathbf{q}}=-k(I+11^{\mathsf{T}})\mathbf{q}$. The breathing vector $1=(1,1,1)$ has squared frequency $4k/m$, while every traceless axial vector with $q_1+q_2+q_3=0$ has squared frequency $k/m$. Orthogonal projection onto these subspaces solves any initial condition in this family without an eighteen-dimensional characteristic polynomial.

For a real consistent zero-stable linear multistep method, transform its first characteristic polynomial by $z=(1+x)/(1-x)$ to $(1-x{)}^s\rho (z)=xP(x)$. The root condition for a multistep method puts all finite roots of $P$ in the closed left half-plane; real linear factors and conjugate quadratic factors consequently give nonnegative coefficients, after choosing $P(0)>0$. The order criterion requires a polynomial of degree at most $s$ to agree through degree $p-1$ with $P(x)x/(2\mathrm{arctanh}x)$. The latter quotient has strictly negative nonconstant even coefficients. Its first even coefficient beyond degree $s$ includes a strictly negative multiple of $P(0)$ and cannot vanish. This gives the stated bound.

Every square matrix satisfies its own characteristic polynomial. More explicitly, for an $n$-by-$n$ matrix $A$ over a commutative ring with identity, formThenHere $0$ is the zero matrix, and the constant term multiplies the identity matrix. This is a polynomial evaluated at a matrix; it is stronger than saying that each eigenvalue is a root of a polynomial. Neither diagonalizability nor distinct eigenvalues is required.^[ref]

The practical payoff is that arbitrarily high matrix powers can be reduced to a combination of $I,A,\dots ,A^{n-1}$. These powers arise when repeatedly applying a linear map, solving a linear recurrence relation, or evolving a finite state space model. The characteristic polynomial turns that repeated action into a fixed-order scalar recurrence. The minimal polynomial may reduce the number of required powers further.

Over a field, a cyclic vector gives a companion matrix, whose characteristic relation annihilates its generator and hence every iterate. If there is no cyclic vector spanning the whole space, a nonzero proper cyclic subspace is invariant. Induction annihilates the restriction and the quotient; the quotient polynomial first sends the whole space into the invariant subspace, and the restriction polynomial then kills it. The block-triangular characteristic polynomial is the product of these two polynomials.

On two-by-two complex matrices let ${\rho }_A(X)=XA$. In row-entry coordinates, its matrix is $\mathrm{diag}(A^T,A^T)$, so its characteristic polynomial is the square of that of $A$. For every polynomial $p$, $p({\rho }_A)(X)=Xp(A)$. Thus $p(A)=0$ implies $p({\rho }_A)=0$, and the reverse implication follows by taking $X=I$. The operators have exactly the same annihilating polynomials and hence the same minimal polynomial. In general, right multiplication on $r$-by-$n$ matrices has characteristic polynomial ${\chi }_A^r$ and the same minimal polynomial.

For a complex $n\times n$ matrix $A$, $\mathrm{tr}(A^k)$ is the sum of the $k$th powers of its eigenvalues, counted with algebraic multiplicity. This does not require diagonalizability: a similar upper triangular matrix has the eigenvalues on its diagonal, and taking powers raises those diagonal entries to their corresponding powers. Newton's identities therefore recover its characteristic polynomial from the first $n$ such traces.

For the Jordan block $A=\left(\begin{array}{cc}2& 1\\ 0& 2\end{array}\right)$, $\mathrm{tr}A=4$ and $\mathrm{tr}(A^2)=8$. Hence $e_1=4$ and $e_2=(4^2-8)/2=4$, giving $\det (zI-A)=z^2-4z+4=(z-2{)}^2$. The repeated eigenvalue is recovered even though this matrix is not diagonalizable.^[ref]

A matrix and its transpose have the same characteristic polynomial, allowing two independent Gershgorin eigenvalue counting arguments for one spectrum. If isolated row discs certify simple eigenvalues near one group of centres and isolated column discs certify the remaining ones, pairwise separation of all the certified small discs establishes a simple spectrum even if neither full family has pairwise disjoint discs.

For example, let $n=2m$, the diagonal centres be $1,\dots ,n$, and all adjacent off-diagonal entries have magnitude strictly below $1/4$. Add a corner entry $a_{n1}=c$. The row radii are below $1/4,1/2,\dots ,1/2,|c|+1/4$; the column radii reverse the two endpoint bounds. If $|c|\le m-3/4$, rows isolate centres $1,\dots ,m$ and columns isolate centres $m+1,\dots ,n$. The nearest potentially enlarged-disc contact has centre separation $m$, while the radius sum is strictly less than $|c|+3/4\le m$. Each of the $n$ certified small discs contains one simple eigenvalue. For real entries these eigenvalues are real, since a nonreal conjugate pair would put two roots in the same conjugation-invariant isolated disc.

This is a sufficient uniform certificate. Overlap beyond this bound does not prove repeated eigenvalues; individual matrices can have simple real spectra well beyond a Gershgorin disc separation bound.

For $C_n=⟨g:g^n=e⟩$, the complex irreducible representations are precisely the $n$ one-dimensional maps $g\mapsto e^{2\pi ij/n}$, with $0\le j<n$. Their characters have the displayed values, forming the discrete Fourier character table. Indeed a representing matrix $A$ satisfies $A^n=I$. Applying the fundamental theorem of algebra to its characteristic polynomial gives an eigenvalue and an eigenvector over the complex numbers. Its line is an invariant subspace under every power of $A$; irreducibility therefore forces the entire space to be that line. Its eigenvalue is an $n$th root of unity. Different roots give inequivalent one-dimensional representations, and there are exactly $n$ possibilities.

All conjugacy classes are singletons because the group is Abelian. The number of the displayed irreducibles equals the group order, and their squared dimensions sum to $n$, in agreement with the sum of squares of irreducible representation dimensions. Over the real numbers, conjugate nonreal characters instead combine into the planes described by the irreducible real representations of a finite cyclic group.

A matrix is defective when an eigenvalue has geometric multiplicity smaller than its algebraic multiplicity. Over a field in which the characteristic polynomial splits, this is precisely the obstruction to an eigenbasis: at least one Jordan block has size greater than one, and generalized eigenvectors are needed to complete a basis. The shear mapping in diagonalizable matrix is the simplest example.

For the constant-coefficient model $x^{\prime }(t)=-kx(t-\tau )$ with $k>0$ and $\tau \ge 0$, the zero equilibrium point is asymptotically stable if and only if $k\tau <\pi /2$. For positive $\tau$, perturbations are measured by the supremum norm of the initial history of delayed feedback; in the stable regime, every continuous history produces a solution tending to zero. At equality, nondecaying sinusoids prevent asymptotic stability. For $k\tau >\pi /2$, growing oscillatory solutions exist. These statements concern this ideal linear time-invariant system, rather than an arbitrary controlled plant. The corresponding linear equation and its threshold are developed in Cain and Reynolds, section 6.4, pages 174–177.

Substitute the exponential function $x(t)=e^{st}$, allowing a complex number $s$. The characteristic equation isFor positive feedback delay this is not a characteristic polynomial: it has infinitely many roots. A root's real part determines growth or decay of its exponential mode, and its imaginary part determines the angular frequency. To find the first boundary, set $s=i\omega$ with $\omega >0$ and use Euler's formula:The smallest positive crossing is $\omega =k$ and $\omega \tau =\pi /2$. Thus at ${\tau }_c=\pi /(2k)$ the complex conjugate roots are $s=\pm ik$. In fact, with the matching initial history of delayed feedback,solves the delay differential equation exactly: a quarter-cycle feedback delay turns the past displacement into precisely the current derivative. Its cycle lasts $2\pi /k=4{\tau }_c$. The amplitude is arbitrary; this linear model does not select a finite-amplitude oscillation.

The roots cross toward growth as feedback delay increases. Differentiating the characteristic equation at fixed $k$ givesThis calculation identifies the direction of the first destabilization. The full threshold also requires ruling out earlier unstable roots: if $s=\alpha +i\beta$ with $\alpha \ge 0$ and $\beta >0$, the characteristic equation implies $\cos (\beta \tau )\le 0$ and $0<\beta \le k$, hence $k\tau \ge \beta \tau \ge \pi /2$. A real nonnegative root is impossible because $s+k{e}^{-s\tau }>0$ there. This supplies the geometric reason the threshold is a quarter-cycle rather than an arbitrary fitting constant.

The tradeoff is concrete. Without feedback delay, increasing $k$ shortens the exponential decay time constant $1/k$. With known $\tau >0$, the same proportional control must satisfy $k<\pi /(2\tau )$. For $\tau =0.2\mathrm{s}$, the threshold is $k\approx 7.854{\mathrm{s}}^{-1}$: $k=5{\mathrm{s}}^{-1}$ is stable and $k=8{\mathrm{s}}^{-1}$ is unstable. A design close to the boundary leaves little room for additional feedback delay; reducing $k$ or reducing measurement latency increases that room. The zero-delay time constant is only a comparison scale, not the decay time of the delayed response. Frank discusses the gain–delay compromise in Control Theory Tutorial, section 13.4 and figure 13.3.

In the Leibniz formula for a determinant of a block diagonal matrix, a term is zero unless its permutation maps every block's row indices onto that same block's column indices. The surviving permutations are precisely independent permutations within the blocks; their signs and products of entries factor over the blocks. Summing therefore gives the product of the individual determinants. Applied to $tI-T$, this also proves that the characteristic polynomial of a block diagonal linear operator is the product of the characteristic polynomials of its blocks.

For a matrix over a field $F$, diagonalizability requires an eigenbasis with entries in $F$ and eigenvalues in $F$. Passing to a larger field can supply missing roots of the characteristic polynomial. For example,is not a diagonalizable matrix over the rational numbers, since the quadratic factor has no rational root. It is a diagonalizable matrix over the real numbers and complex numbers, since its three eigenvalues are distinct. More generally, diagonalizability is equivalent to the minimal polynomial splitting into distinct linear factors over the specified field.

For an integer matrix, reduction modulo a prime number is a different operation from extension of its coefficient field, and can merge eigenvalues. In the same example modulo $5$, ${\chi }_A(t)=(t-1)(t-3{)}^2$, while $\ker (A-3I)$ is spanned by $(1,3,4{)}^T$. Thus the geometric multiplicity of $3$ is one and its algebraic multiplicity is two, preventing diagonalizability even though all eigenvalues belong to the finite field.

A finite-dimensional linear operator is diagonalizable over a specified field when it has an eigenbasis. Equivalently, its matrix $A$ admitswhere $P$ is an invertible matrix whose columns are eigenvectors, and $D$ is a diagonal matrix of the corresponding eigenvalues. This change of basis reveals a linear map as independent one-dimensional scalings. The matrix may look coupled in the original coordinates, but each coordinate in the eigenbasis evolves on its own.

For the symmetric matrix $A=\left(\begin{array}{cc}2& 1\\ 1& 2\end{array}\right)$ computed in eigenvector, takeThen $AP=PD$, hence $P^{-1}AP=D$. The orthonormal basis version is $Q=P/\sqrt{2}$, an orthogonal matrix satisfying $Q^{T}AQ=D$. Every real symmetric matrix has an orthonormal eigenbasis by the spectral theorem; a general diagonalizable matrix need not have orthogonal eigenvectors.

The immediate payoff is repeated application: adjacent factors $P^{-1}P$ cancel, giving $A^n=P{D}^n{P}^{-1}$. In this example,The component in $E_3$ grows while the component in $E_1$ stays fixed. After division by its Euclidean norm, this particular sequence approaches $(1,1{)}^T/\sqrt{2}$. This is the mechanism behind the power method; an initial vector entirely in $E_1$ would stay there and never find the dominant eigenvalue $3$.

Two different obstructions deserve comparison. The planar rotation $R$ in eigenvector is diagonalizable over $\mathbb{C}$ but not over $\mathbb{R}$, because its eigenvalues are $i$ and $-i$. By contrast, the shear mappinghas only $\lambda =1$, of algebraic multiplicity two, and $E_1=\mathrm{span}\{(1,0{)}^T\}$, of geometric multiplicity one. There is no second independent eigenvector, even over $\mathbb{C}$. This is a defective matrix, already a two-dimensional Jordan block. Its powers areThus even though both eigenvalues are one, repeated application can produce growth. Eigenvalues alone miss the coupling recorded by a Jordan block. A repeated eigenvalue is harmless when it has enough independent eigenvectors, as $I$ demonstrates. More generally, a matrix is diagonalizable exactly when its characteristic polynomial splits over the chosen field and every eigenvalue has equal geometric multiplicity and algebraic multiplicity. The proof outline is to join bases of the distinct eigenspaces; their direct sum fills the space exactly under these conditions.

A physical motivation is the separation of normal modes. Consider two equal masses $m$ between three identical springs of stiffness $k$, with the outer ends fixed. By Hooke's law and Newton's second law, the displacement vector $q=(q_1,q_2{)}^T$ obeysThe eigenvectors $(1,1{)}^T$ and $(1,-1{)}^T$ have eigenvalues $1$ and $3$: the masses move together in the first normal mode and oppositely in the second. Writing $q=a(t)(1,1{)}^T+b(t)(1,-1{)}^T$ separates the coupled equations into two harmonic oscillators,Their angular frequencies are $\sqrt{k/m}$ and $\sqrt{3k/m}$; the eigenvalues give squared angular frequencies after multiplication by $k/m$. This is the beauty of diagonalization: finding the right directions turns interacting motions into independent patterns. For a physical treatment, see Yen-Jie Lee's "Coupled Oscillators, Normal Modes" (MIT 8.03SC, 2016), and the local article normal modes of two equal masses between three springs.^[ref]

Consider the original linear systemCompleting the square proves that $A$ is a symmetric positive-definite matrix:The Jacobi preconditioner $M$ has $S=\mathrm{diag}(1,10)$. Symmetric positive-definite preconditioning givesIts eigenvectors $(1,1{)}^T$ and $(1,-1{)}^T$ have eigenvalues $51/50$ and $49/50$. HenceThe characteristic polynomial of $A$ is ${\lambda }^2-101\lambda +2499/25$. Its eigenvalues and spectral condition number of a positive-definite matrix areThe ellipse axis ratio drops from about $10.0020$ to $1.0202$. The transformation multiplies the second coordinate by ten, bringing the two energy scales together.

Both panels show the same energy level sets, expressed in error coordinates $e=x-x_{\ast }$ and $\eta =M^{1/2}e$. Their coordinate axes have equal scales. The marked displacement $(1,0.1{)}^T$ becomes $(1,1{)}^T$; its energy remains $e^{T}Ae/2=1.02$. These are exact contours of the specified quadratic form.

Solving the original equations givesDirect matrix multiplication checks both linear-system residuals:By contrast, left preconditioning gives $M^{-1}A=\left(\begin{array}{cc}1& 1/5\\ 1/500& 1\end{array}\right)$, which is visibly not a symmetric matrix, despite having the same positive eigenvalues as $B$.

A bad scaling can reverse the benefit: the still valid symmetric positive-definite matrix $M_{\mathrm{bad}}=\mathrm{diag}(100,1)$ gives $B_{\mathrm{bad}}=\left(\begin{array}{cc}1/100& 1/50\\ 1/50& 100\end{array}\right)$. Its Rayleigh quotients on the coordinate unit vectors are $0.01$ and $100$, so ${\kappa }_2(B_{\mathrm{bad}})\ge 10000$. Cheapness and positive definiteness do not guarantee improved conditioning.

A positive diagonal matrix $D=\mathrm{diag}(d_1,\dots ,d_n)$, with $d_i>0$, is an invertible matrix and changes the coordinate scales through the similarity transformation $B=D^{-1}AD$. Its diagonal entries are still $a_{ii}$, while $b_{ij}=a_{ij}{d}_j/d_i$. Thus the Gershgorin discs have the same centres but different radii. Matrix similarity preserves the characteristic polynomial and eigenvalues, so the original and scaled disk unions are both valid bounds and may be intersected. Suitable scaling can compensate for badly unbalanced off-diagonal entries; it need not shrink every radius.^[ref]

For example,The original Gershgorin discs are $|z-2|\le 100$ and $|z-3|\le 0.01$. Their union is simply the huge first disk; the small second disk contains no eigenvalue. After scaling, both radii are $1$, giving $|z-2|\le 1$ or $|z-3|\le 1$. The two Gershgorin discs overlap, so their individual eigenvalue counts are not certified. The scaled matrix is a real symmetric matrix, however, so matrix similarity now proves that the original matrix has real eigenvalues in $[1,4]$ and is an invertible matrix. Directly, both matrices have characteristic polynomial ${\lambda }^2-5\lambda +5$, henceThe scaling changes the available bounds, while these eigenvalues stay fixed.

A scalar $\lambda$ is an eigenvalue of a linear operator $A:V\to V$ whenfor some nonzero vector $v$, called an eigenvector. The scalar must belong to the field over which $V$ is a vector space. Geometrically, a real eigenvalue tells us what $A$ does along a special line: every vector on that line is multiplied by the same number. The eigenvector gives the line and the eigenvalue gives its scaling.

For a real eigenvalue, $\lambda >0$ preserves the orientation along the line, $\lambda <0$ reverses it, and $\lambda =0$ collapses the line to zero. The length is multiplied by the modulus $|\lambda |$, so $|\lambda |>1$ stretches and $0<|\lambda |<1$ contracts. For example, the diagonal matrix $\mathrm{diag}(2,-1/2,0)$ doubles one coordinate direction, reverses and halves another, and annihilates the third. Its eigenvalues are $2,-1/2,0$; the last direction lies in its kernel.

For a finite square matrix, the eigenvalues are the roots of a polynomial $p_A$, its characteristic polynomial:Here $I$ is the identity matrix. The reason is that $Av=\lambda v$ means $(A-\lambda I)v=0$ with $v\ne 0$: the shifted matrix must have a nontrivial kernel, equivalently a zero determinant. The worked symmetric matrix example in eigenvector shows both stages: find the eigenvalues, then find their eigenspaces.

Eigenvalues are especially useful for repeated application of a linear map. If $Av=\lambda v$, thenOne multiplication proves the next step, so a matrix power becomes an ordinary power on each eigenvector. This explains growth, decay, and alternating signs along these directions. Gilbert Strang develops this connection, and its extension to the matrix exponential, in his lecture "Eigenvalues and Eigenvectors" (MIT 18.065, 2018).^[ref]

A documented historical connection is C. G. J. Jacobi's "Über ein leichtes Verfahren die in der Theorie der Säcularstörungen vorkommenden Gleichungen numerisch aufzulösen" (1846, Journal für die reine und angewandte Mathematik 30, pp. 51–94). Its opening discusses symmetric matrices arising from small oscillations, with a common unknown subtracted from every diagonal entry. In modern notation this is $(A-\lambda I)v=0$; Jacobi seeks the parameter and the ratios of the components of $v$. This is evidence of an early physical and computational setting for the eigenvalue problem, without assigning its discovery to a single person.^[ref][ref]

If a $k$-dimensional invariant subspace is mapped to the span of the first $k$ coordinate vectors by a similarity transformation, the transformed matrix has block formIts characteristic polynomial factors as $\det (\lambda I-B)\det (\lambda I-D)$, so its eigenvalues are the eigenvalues of the two diagonal blocks, counted with algebraic multiplicity.

Geometric multiplicity is the dimension of an eigenspace; algebraic multiplicity is its multiplicity as a root of the characteristic polynomial. For a real symmetric matrix they coincide because it has an orthonormal eigenbasis.

A nonzero vector $v$ is an eigenvector of a linear operator $A$ when $Av=\lambda v$ for some eigenvalue $\lambda$. For a real vector space and $\lambda \ne 0$, the line through $v$ is preserved: the image can point forward or backward, but cannot turn away from that line. When $\lambda =0$, the image is zero and has no direction. The zero vector is excluded because it satisfies the equation for every scalar and identifies no special direction.

Every nonzero scalar multiple of an eigenvector is an eigenvector for the same eigenvalue, since $A(cv)=cAv=\lambda (cv)$. Thus there is usually no preferred length or sign; choosing a unit vector is a convenient normalization. Eigenvectors for a fixed eigenvalue, together with zero, form its eigenspace.

For a fully computed example, consider the real symmetric matrixIts characteristic polynomial isFor $\lambda =3$, $(A-3I)v=0$ gives $-x+y=0$, so $v_3=(1,1{)}^T$ is an eigenvector. For $\lambda =1$, $(A-I)v=0$ gives $x+y=0$, so $v_1=(1,-1{)}^T$ is an eigenvector. Direct multiplication checks both:The map stretches the line $y=x$ threefold and fixes the line $y=-x$ pointwise. These eigenvectors are orthogonal because $v_3^T{v}_1=0$; dividing each by $\sqrt{2}$ gives an orthonormal basis. A general vector turns because its components along these two lines receive different scalings. For example, $(1,0{)}^T$ maps to $(2,1{)}^T$ and is not an eigenvector.

A contrasting example is the planar rotation through $90$ degrees:Every nonzero real vector turns to a perpendicular direction, so there are no real eigenvectors. Its characteristic polynomial is ${\lambda }^2+1$, which has no real roots. Over the complex numbers, however, $R(1,-i{)}^T=i(1,-i{)}^T$ and $R(1,i{)}^T=-i(1,i{)}^T$. The same matrix therefore has an eigenbasis in ${\mathbb{C}}^2$ even though it has none in ${\mathbb{R}}^2$: the choice of field matters.

If the derivative characteristic polynomial $\sigma$ of a linear multistep method has a simple root outside the unit disk and keeps its degree in the limit, then its absolute stability region is bounded. Otherwise a stable sequence with $|z|\to \infty$ would give amplification polynomials $\sigma -\rho /z$ tending coefficientwise to $\sigma$. A root near the exterior root would persist by continuity, contradicting the root condition for a multistep method. This argument excludes stability at every sufficiently large complex $z$, not only along the negative real axis.