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Proportional-integral control is the proportional and integral part of a PID controller, with the derivative term omitted. Proportional-integral control combines proportional control with integral action. For the ideal controlled plant, a constant reference signal, exact measurement and a constant control disturbance, the ordinary differential equations giveThe characteristic polynomial is ${\lambda }^2+k_p\lambda +k_i$. Positive $k_p$ and $k_i$ imply asymptotic stability of the equilibrium point: real roots of a polynomial have negative sum and positive product, hence are both negative; a nonreal complex conjugate pair has real part $-k_p/2<0$. This is a short stability check for this ideal controlled plant, rather than a rule for every dynamical system.

At the equilibrium point, $e_{\ast }=0$, $x_{\ast }=r$, and $z_{\ast }=-w/k_i$. For $w=0.2$ and $k_i=1$, the stored value $z_{\ast }=-0.2$ supplies $u_{\ast }=-0.2$ even though the tracking error is zero. This separates the eventual accuracy from the transient response: integral action cancels the lasting offset but can make the plant output pass beyond the goal. The authors discuss practical limits on integral action in Åström and Murray, Feedback Systems, Chapter 11.

For $k_p=2$, $k_i=1$, $x(0)=z(0)=0$ and $w=0$, the characteristic polynomial is $(\lambda +1{)}^2$. The initial conditions are $e(0)=1$ and $\dot{e}(0)=-2$, givingThe tracking error changes sign at $t=1$. Since $\dot{x}=(2-t)e^{-t}$, the largest plant output occurs at $t=2$ and is $1+e^{-2}\simeq 1.135335$, about $13.53\%$ above the reference signal.

For a separate comparison, start the control disturbance $w=0.2$ at $t=4$. For $t\ge 4$, write $s=t-4$. The exact responses areThe proportional control response tends to $1.1$, while the proportional-integral control response returns to $1$. At $t=8$, the values are approximately $1.099966$ and $1.017001$. The added term in the proportional-integral control response eventually vanishes, exposing how memory removes a constant control disturbance's lasting effect.

Let an invertible matrix $T$ have monic invariant factors of a linear operator $f_j$ of degrees $d_j$. Their constant coefficients are nonzero, since their product is the characteristic polynomial and $\det T\ne 0$. On a cyclic block, $f(T)=0$ implies $T^{-d}f(T)=0$, so $f^{\vee }(T^{-1})=0$, where $f^{\vee }(t)=t^{d}f(1/t)/f(0)$ is monic of degree $d$. The vector $v$ remains cyclic for $T^{-1}$: multiplying $v,T^{-1}v,\dots ,T^{-(d-1)}v$ by the invertible map $T^{d-1}$ gives the reversed original cyclic basis. Thus this polynomial is the minimal polynomial of the inverse block.

Normalized reciprocation preserves divisibility because $(fg{)}^{\vee }=f^{\vee }{g}^{\vee }$ when both constant coefficients are nonzero. Consequently the rational canonical form of $T^{-1}$ is the direct sum of the companion matrices $C(f_j^{\vee })$. The normalization by $f_j(0)$ is essential to make each reciprocal polynomial monic.

The Rayleigh quotient measures the quadratic form of a self-adjoint operator per unit squared norm. For a nonzero vector $x$ and a finite-dimensional Hermitian matrix $A$,For a real symmetric matrix, replace the conjugate transpose $x^{\ast }$ by $x^T$. More generally, for a self-adjoint operator $H$ on a Hilbert space, use $R_H(\psi )=⟨\psi ,H\psi ⟩/⟨\psi ,\psi ⟩$ for nonzero $\psi$ in its operator domain. On an eigenvector it returns the exact eigenvalue. Multiplying the vector by any nonzero scalar leaves the quotient unchanged, so it measures a direction rather than a chosen length.

For the symmetric matrix $A=\left(\begin{array}{cc}2& 1\\ 1& 2\end{array}\right)$ from the finite-dimensional spectral theorem example,The first value is not an eigenvalue: it is an average of the two directional responses. To see every direction, write the unit vector $q(\theta )=\cos \theta e_{+}+\sin \theta e_{-}$ in the orthonormal eigenbasis. ThenThe largest value is $3$ along $e_{+}$; the smallest is $1$ along $e_{-}$. A direction halfway between them has quotient $2$. In the left panel, the color on the unit circle represents the quotient; in the right panel, the same values are plotted against the angle measured from $e_{+}$.

The Hermitian Rayleigh quotient range explains why this is useful: evaluating one trial vector gives an upper bound on the smallest eigenvalue and a lower bound on the largest one, without solving a characteristic polynomial. At an extreme eigenvector, a small angular error changes the quotient only quadratically. Near $e_{+}$ in this example, $3-R_A(q(\theta ))=2{\sin }^2\theta =2{\theta }^2+O({\theta }^4)$. This is the geometric reason for quadratic Rayleigh-quotient improvement for the power method and motivates the Rayleigh-Ritz method.

There is a concrete historical link with normal modes. In The Theory of Sound (1877), Volume I, Sections 88–89, Lord Rayleigh studies the period of a constrained trial vibration. He shows that the period has a stationary point at a normal mode and that a trial type bounds the longest natural period. These energy ratios are a mechanical setting for the modern Rayleigh quotient and generalized Rayleigh quotient; the book describes the result through vibration periods, rather than today's matrix notation.^[ref]

The real case of the finite-dimensional spectral theorem says that a self-adjoint operator on a finite-dimensional real inner product space has a real orthonormal eigenbasis. In orthonormal coordinates this is the spectral theorem for real symmetric matrices, $A=Q\Lambda Q^T$. The worked ellipse under finite-dimensional spectral theorem shows why both orthogonal diagonalization and orthogonality matter: the new coordinates separate the quadratic form into squares without distorting lengths or angles.

A real normal matrix need not have real eigenvalues. For example, the quarter-turn $\left(\begin{array}{cc}0& -1\\ 1& 0\end{array}\right)$ is an orthogonal matrix, but its characteristic polynomial is $z^2+1$ and its eigenvalues are $i,-i$. Over the complex numbers it has an orthonormal eigenbasis; over the real numbers it has no eigenvector. Self-adjointness supplies the reality conclusion in the real spectral theorem.

For a symplectic matrix $M$ of size $2m$ with determinant one, the defining identity $M^{T}JM=J$ gives $M^{-1}=J^{-1}{M}^{T}J$. Since similarity and transposition preserve the characteristic polynomial, ${\chi }_{M^{-1}}={\chi }_M$. Factoring $xI-M=M(x{M}^{-1}-I)$ then proves the displayed reciprocal identity for $x\ne 0$. If ${\chi }_M(z)=(z-\lambda {)}^{d}R(z)$ with $R(\lambda )\ne 0$, this identity gives ${\chi }_M(x)=x^{2m-d}(1-\lambda x{)}^{d}R(1/x)$ near $x={\lambda }^{-1}$. Thus reciprocal eigenvalues have the same algebraic multiplicity. The fixed points $\pm 1$ require the stronger parity argument in even multiplicities at fixed points of reciprocal root pairing.

Take a different real matrix,Its only eigenvalue is $2$, with algebraic multiplicity two. However, the polynomial $q(t)=t-2$, which vanishes at that eigenvalue, gives $q(B)=N\ne 0$. Merely checking values at eigenvalues cannot prove a matrix identity. The full characteristic polynomial does work:The matrix is a Jordan block, and its minimal polynomial is also $(t-2{)}^2$.

The binomial theorem gives, for positive integers $n$,Terms containing $N^2$ or a higher matrix power vanish. After removing the common growth factor $2^n$, the remaining linear map is the shear mapping $I+(n/2)N$. It moves $(x,y)$ to $(x+ny/2,y)$. The single eigenvalue controls the exponential scale, while the nilpotent matrix adds a factor growing linearly in $n$.

In terms of polynomial interpolation, reduction modulo $(t-2{)}^2$ retains both a value and a derivative:Indeed, polynomial division gives $f(t)=f(2)+f^{\prime }(2)(t-2)+(t-2{)}^{2}h(t)$. This is a simple Hermite interpolation condition. For $f(t)=t^n$, it gives exactly the displayed matrix power. Repeated eigenvalues retain information about how a polynomial changes, not just where it vanishes.

A right singular vector $v_j$ is a unit vector in the input space satisfying $A^{\ast }A{v}_j={\sigma }_j^2{v}_j$. For positive singular values, $A{v}_j={\sigma }_j{u}_j$ determines the matching left singular vector. In the ellipse example, $A{v}_1=(5,0{)}^T$ is not parallel to $v_1$: a right singular vector need not be an eigenvector of $A$. Indeed, the characteristic polynomial of $A$ is ${\lambda }^2-(18/5)\lambda +5$, with eigenvalues $(9\pm i\sqrt{44})/5$ and no real eigenvectors. Nevertheless, its real right singular vectors give a complete orthonormal basis. Eigenvectors describe preserved directions; right singular vectors describe orthogonal input channels of stretching.^[ref]

An arbitrary input mixes these channels. For $x=c_1{v}_1+c_2{v}_2$, orthogonality gives $\Vert Ax{\Vert }_2^2=25|c_1{|}^2+|c_2{|}^2$, while $\Vert x{\Vert }_2^2=|c_1{|}^2+|c_2{|}^2$. Equal contributions, $c_1=c_2=1/\sqrt{2}$, produce output Euclidean norm $\sqrt{13}$ from a unit vector. Concentrating the entire input along $v_1$ produces the maximum $5$. This is why the right singular vectors give a direct reading of how any input will behave. For a compact operator, the same definition describes the positive terms of its singular value system.

The Routh-Hurwitz criterion tests whether all roots of a real polynomial have negative real part, by sign conditions on determinants formed from its coefficients. Applied to a characteristic polynomial, it gives an algebraic test for strict linear asymptotic stability.

Consider the diagonal matrix $D=2I_2$ and a Jordan blockBoth have characteristic polynomial $(t-2{)}^2$, trace $4$, determinant $4$, and eigenvalue $2$ with algebraic multiplicity two. But $\ker (D-2I)$ is the whole plane, whereas $(J-2I)(a,b{)}^T=(b,0{)}^T$ vanishes only when $b=0$. Their eigenspaces have different dimensions, so eigenvectors under similarity rules out matrix similarity. There is an even shorter proof: $S^{-1}(2I)S=2I$ for every invertible matrix $S$.

The missing information is how the linear operator acts beyond its eigenvectors. Here $D-2I=0$, while $J-2I\ne 0$ but $(J-2I{)}^2=0$. Their minimal polynomials are respectively $t-2$ and $(t-2{)}^2$. Over a field where the characteristic polynomial splits, the full sizes of the Jordan blocks, encoded by the Jordan normal form, determine the matrix similarity class.

For $u=(1,\dots ,1{)}^T$, $J_n=u{u}^T$ maps $v$ to $({\sum }_i{v}_i)u$. Therefore $u$ is an eigenvector with eigenvalue $n$, while the zero-sum vector subspace is its kernel. For $n\ge 2$ a basis of the kernel is $e_1-e_n,\dots ,e_{n-1}-e_n$. These vectors and $u$ form an eigenbasis, proving diagonalization and characteristic polynomial $t^{n-1}(t-n)$. Both algebraic multiplicity and geometric multiplicity are one at $n$ and $n-1$ at zero. At $n=1$ the sole eigenvalue is 1; the zero-sum space has dimension zero and contributes no additional root.

ConsiderIts row sums vanish, so $(1,1,1{)}^T$ is an eigenvector for the zero eigenvalue. Its characteristic polynomial iswith sum of principal minors of order twoThe other eigenvalues are $(-S\pm \sqrt{S^2-4T})/2$. No symmetry is required for this formula; positive diagonal symmetrization of a matrix is a sufficient condition for real roots.

The model matrix $B=\left(\begin{array}{cc}0& 1\\ ϵ& d\end{array}\right)$ has a defective double eigenvalue when $d=ϵ=0$. Its characteristic polynomial is ${\lambda }^2-d\lambda -ϵ$, giving the displayed roots. For fixed nonzero $d$ the small root is $-ϵ/d+O({ϵ}^2)$ and the other is $d+ϵ/d+O({ϵ}^2)$. These expansions cease to be ordered at $d=O(\sqrt{ϵ})$, the distinguished limit in which both roots split on a square-root scale. This is the generic leading behavior when a perturbation closes a two-dimensional Jordan block through a nonzero lower-left entry; it need not occur for a perturbation that leaves that entry zero.

The steady-state tracking error iswhen this finite limit exists. A finite steady-state tracking error does not require the reference signal, the plant output, or the control input to become constant: all can grow while their difference settles. If the tracking error grows without bound or keeps oscillating, report that behavior rather than assigning a finite steady-state tracking error. Eventual accuracy and asymptotic stability are different questions.

For the first-order low-pass filter $\tau \dot{y}+y=Ku+d$ introduced in reference signal, assume exact measurement, constant control disturbance $d$, and no actuator saturation or feedback delay. Under proportional control $u=k_p(r-y)$ with $k_p>0$, a constant reference signal $r=R$ givesThe decaying homogeneous solution has rate $(1+K{k}_p)/\tau >0$, so this equilibrium point is approached from any finite initial condition. A nonzero tracking error is needed to maintain the control input that balances the model's tendency to relax toward $y=d$. This differs from the accumulating controlled plant $\dot{x}=u+w$, for which proportional control already tracks a constant reference signal exactly when $w=0$. The presence or absence of a step offset depends on both the controlled plant and the controller.

For a ramp reference signal $r=vt$, the same proportional control givesIts long-time particular solution grows with slope $v/(1+K{k}_p)$ when $v\ne 0$. Thus the controlled plant and homogeneous closed-loop response are stable, but the tracking error has no finite steady-state tracking error. A changing goal requires a different performance test from a fixed destination.

Now use proportional-integral control $u=k_{p}e+k_{i}z$, $\dot{z}=e$, with $k_i>0$. For the same controlled plant, differentiate the error equation on intervals where the inputs are smooth:The characteristic polynomial $\tau {\lambda }^2+(1+K{k}_p)\lambda +K{k}_i$ has both roots of a polynomial in the open left half-plane: real roots have negative sum and positive product, and a nonreal complex conjugate pair has negative real part. This proves decay of the homogeneous error for this model. It does not prove stability of an arbitrary controlled plant controlled by positive gains. At a reference jump, use the original ordinary differential equations to set the new initial conditions, rather than treating the reference as smooth at that jump.

With a constant reference signal and constant control disturbance, proportional-integral control therefore gives $e_{\mathrm{ss}}=0$. At its equilibrium point, $y_{\ast }=R$, $u_{\ast }=(R-d)/K$, and $z_{\ast }=(R-d)/(K{k}_i)$. The stored integral maintains the required command while the current tracking error vanishes. For a ramp reference signal, however, the forcing above is $v$, andThe tracking error settles even though the accumulated integral and the required control input grow. This finite offset is the error needed to keep increasing that command. For $r=a{t}^2/2$ with $a\ne 0$, the forcing is $\tau a+at$ and even this proportional-integral control leaves a tracking error growing with slope $a/(K{k}_i)$. Integral action is therefore a precise mechanism for selected input classes, not a promise of perfect tracking for every reference signal. ^[ref]

Here are completely explicit synthetic examples with $y(0)=z(0)=0$, $d=0$, and no sensor bias. Choose $\tau =2\mathrm{s}$, $K=1{}^{\circ }\mathrm{C}$ per percentage point, $k_p=1$ percentage point per degree, and $k_i=0.5$ percentage points per degree-second. Thus $K{k}_p=1$ is dimensionless, and $K{k}_i=0.5{\mathrm{s}}^{-1}$. In the following formulas, $t$ denotes the numerical time in seconds and $y,e,r$ denote numerical temperature rises in degrees. For the unit-step reference signal $r=1$, proportional control and proportional-integral control respectively giveFor the ramp reference signal $r=0.2t$, the corresponding solutions areThe control inputs here are numerical heater-command percentage points. Substitution into $2\dot{y}+y=u$ and into the controller equations checks these solutions without invoking an unverified limiting rule. The proportional control step offset is $0.5$ degrees; the proportional-integral control step offset is zero. The proportional control ramp error grows; the proportional-integral control ramp offset is $0.4$ degrees, exactly $v/(K{k}_i)$.

The conclusions have specific limits. With constant sensor bias $b$, stable proportional-integral control for a feasible constant reference signal drives the measured tracking error to zero but leaves the physical discrepancy ${\epsilon }_{\mathrm{ss}}=b$. Under proportional control, the measured offset instead becomes $(R-b-d)/(1+K{k}_p)$. If actuator saturation prevents the required control input, zero step offset is not achievable by these formulas. A changing control disturbance contributes $-\dot{d}$ to the error equation, while feedback delay can change stability; neither is covered by the constant-disturbance result. These restrictions are what make steady-state tracking error useful: it identifies the input, measurement convention, and model for which an accuracy claim is actually justified.

Three nonnegative amounts convert in the directed cycle $A\to B\to C\to A$ at positive first-order rates $a,b,c$. The linear system of ordinary differential equations conserves $A+B+C=N$. Its characteristic polynomial is $\lambda [{\lambda }^2+(a+b+c)\lambda +ab+bc+ca]$. The two nonzero eigenvalues have negative real part, so every trajectory converges to the unique steady state with that total:This is also a column-convention continuous-time Markov chain after normalizing by $N>0$: every component loses to one successor and receives from one predecessor. A complex-conjugate pair produces damped oscillations about equilibrium even though the amounts remain nonnegative. A repeated nonzero eigenvalue is defective and requires a Jordan chain.

If $A$ is diagonalizable, the anticommutator map $B\mapsto AB+BA$ has eigenvalues $2{\lambda }_1$, ${\lambda }_1+{\lambda }_2$ twice, and $2{\lambda }_2$. Their product gives the displayed characteristic polynomial. The density of diagonalizable complex matrices and continuity of polynomial coefficients prove the formula also for defective matrices.

Consider the real matrixThe Cayley-Hamilton theorem gives $A^2=3A-I$. ConsequentlyThusAll four entries come from just two scalar coefficients. If $A^n=a_{n}A+b_{n}I$, multiplying once by $A$ and replacing $A^2$ gives the linear recurrence relationEquivalently $a_{n+2}=3a_{n+1}-a_n$, with $a_0=0,a_1=1$. The appearance of Fibonacci numbers is explained by $A=Q^2$, where $Q=\left(\begin{array}{cc}1& 1\\ 1& 0\end{array}\right)$ advances the Fibonacci recurrence. No eigenvectors need to be computed.

The same identity computes an inverse. Rearrange it as $A(3I-A)=I$, givingThis works because the constant term of this characteristic polynomial is nonzero and the coefficients lie in a field. Generally, when $c_0\ne 0$ over a field, rearranging the Cayley-Hamilton theorem expresses $A^{-1}$ as a polynomial evaluated at a matrix. Over a general commutative ring, a nonzero constant need not be a unit; invertibility requires that extra condition.

The nonzero-sum two-cycles of the cubic swap map exist for $a>3$. Their two-step characteristic polynomial has trace $T$ and determinant $D$ as displayed. Since $1-T+D=-2(a-3)(a+1)<0$, one periodic-orbit multiplier exceeds one, proving instability throughout. The other crosses minus one at $a=(1+\sqrt{29})/2$; the cycles are saddles before that threshold and repelling afterwards. At $a=3$ they meet the fixed branches in an unstable subcritical-flip branch.

Let $L$ be the symmetric zero-diagonal Jacobi matrix with $L_{j,j+1}=L_{j+1,j}=a_j$, and let the skew-symmetric matrix $B$ haveThen $\dot{L}=[B,L]$ is equivalent to the Finite Volterra lattice. The corresponding similarity flow makes the eigenvalues of $L$, the coefficients of its characteristic polynomial, and its trace invariants first integrals.