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The strictly positive case states that a real square matrix $W$ with $W_{ij}>0$ has a positive simple eigenvalue $\lambda$ and positive left and right eigenvectors, with all other eigenvalues strictly smaller in modulus. The Brouwer fixed-point theorem applied to $v\mapsto Wv/(1^{T}Wv)$ on the nonnegative unit simplex gives $v>0$ and $Wv=\lambda v$. For another eigenvector $z$, maximize $|z_i|/v_i$; the triangle inequality gives $|\mu |\le \lambda$. Equality forces every modulus ratio and complex argument to agree because every entry is positive, so $z$ is proportional to $v$ and $\mu =\lambda$. Apply the same argument to $W^T$ for a positive left eigenvector; its positive pairing with $v$ rules out a generalized eigenvector at $\lambda$, establishing algebraic simplicity. For merely nonnegative matrices a nonnegative leading eigenvector exists. An irreducible nonnegative matrix has a positive leading eigenvector and simple Perron eigenvalue; a primitive nonnegative matrix has the strict modulus gap.

For a bounded linear operator $T:X\to X$ on a complex Banach space, the point spectrum is precisely its set of eigenvalues:Here $I$ is the identity operator. Thus membership requires an actual nonzero eigenvector, whereas the spectrum of a bounded operator only requires failure of a bounded inverse. Every eigenvalue belongs to the spectrum, but the inclusion can be strict. The multiplication operator discussed below has an entire interval in its spectrum and empty point spectrum.^[ref]

The strict positivity of a quadratic form on every nonzero vector. A real symmetric matrix has positive definiteness exactly when its associated quadratic form does, making it a positive-definite matrix. A Hermitian form is positive definite when its value on $(v,v)$ is real and strictly positive for every nonzero vector.

Think of a positive-definite quadratic form as an energy cost for a displacement: every direction costs something, and zero cost identifies the origin. For example, the real symmetric matrixis a positive-definite matrix because the two squares can vanish together only at $(0,0)$. The mixed term couples the coordinates, but cannot overwhelm their positive contributions: $q(1,1)=8$ and $q(1,-1)=4$. The graph is a bowl, and its positive level sets are ellipses with axes along the eigenvectors, rather than along the original coordinates.

The real and complex settings require different symmetry assumptions. For a real symmetric matrix use $q(x)=x^{T}Ax$ with $x\in {\mathbb{R}}^n$; for a complex Hermitian matrix use $q(z)=z^{\ast }Az$ with $z\in {\mathbb{C}}^n$, where $z^{\ast }$ is the conjugate transpose. The latter expression is real, so strict positivity is meaningful. A complex quadratic form such as $z_1^2+z_2^2$ has no such positivity: it vanishes at $(1,i)$ and takes negative real values at $(i,0)$. A Hermitian form uses complex conjugation, which changes this example to $|z_1{|}^2+|z_2{|}^2$.

For a real matrix $M$ without a symmetry hypothesis, its quadratic form depends only on its symmetric part of a matrix:Positive eigenvalues of $M$ alone do not establish positive definiteness of this quadratic form. For example, $M=\left(\begin{array}{cc}1& 4\\ 0& 1\end{array}\right)$ has both eigenvalues equal to $1$, yet $q(1,-1)=1-4+1=-2$. The relevant test is positivity of the symmetric part of a matrix, not the spectrum of an arbitrary matrix.^[ref]

Vary the coupling in the opening example. The eigenvectors $e_{+}=(1,1{)}^T/\sqrt{2}$ and $e_{-}=(1,-1{)}^T/\sqrt{2}$ form a fixed orthonormal eigenbasis, with eigenvalues $3+t$ and $3-t$. Writing a vector as $u{e}_{+}+v{e}_{-}$ givesFor $|t|<3$, both directions cost energy and $A_t$ is a positive-definite matrix. At $t=3$, the quadratic form is $6u^2$: the entire line spanned by $e_{-}$ has zero energy, and $A_t$ is a singular positive semidefinite matrix. At $t=-3$, the flat line is instead spanned by $e_{+}$. For $|t|>3$, the quadratic form is an indefinite quadratic form; one direction lowers the energy without bound as its displacement grows. At $t=4$, for instance, $q_t(e_{+})=7$ while $q_t(e_{-})=-1$.

The geometric transition is precise. For $|t|<3$, $q_t=1$ is an ellipse. At $t=3$, it consists of the two parallel lines $u=\pm 1/\sqrt{6}$, extending freely in the zero-energy direction. For $t>3$, it is a hyperbola. The boundary case retains nonnegative energy but loses both the bounded positive level sets and the unique zero-energy vector.

For a strictly positive real matrix $A$, the map $x\mapsto Ax/{\sum }_i(Ax{)}_i$ is a continuous self-map of the probability simplex. A Brouwer fixed-point theorem fixed point satisfies $Ax=\lambda x$ with $\lambda >0$. The image lies in the interior of the simplex, so every entry of this eigenvector is strictly positive. Nonnegative entries and nonzero columns alone do not imply this strict conclusion.

For a two-level Fourier recurrence ${\hat{u}}^{n+1}=a(\theta ){\hat{u}}^n+b(\theta ){\hat{u}}^{n-1}$, the amplification matrix isStability requires its powers to be uniformly bounded in the time index and the mesh frequencies. If its two amplification roots have modulus at most one and their separation has a positive mesh-independent lower bound, the eigenvectors $({\xi }_j,1{)}^T$ give a uniformly bounded diagonalization of a matrix, proving stability. A repeated unit-modulus root of this companion matrix instead produces a Jordan block and linear growth in time. Thus checking only the moduli of the roots is insufficient.

Power iteration finds a dominant eigenvalue and its eigenvector by repeatedly applying a matrix and rescaling the resulting vector. The essential idea is selective amplification: components along different eigenvectors grow at different rates, so one direction can eventually dominate. It is the computational form of the power method and is useful when applying a large sparse matrix to a vector is cheap but computing its whole spectral decomposition is expensive.

For a square matrix over the real numbers or complex numbers $A$, choose a unit vector $x_0$ and repeatHere $\Vert \cdot {\Vert }_2$ is the Euclidean norm. If $y_{k+1}=0$, stop before division: the current vector is an eigenvector for the zero eigenvalue, and this iteration cannot continue. Otherwise, in exact arithmetic,Repeated normalization keeps the stored vectors at manageable size without changing their directions; it does not guarantee that an individual multiplication cannot exceed the range of floating-point arithmetic. The dominant eigenvalue means largest modulus, which need not mean largest signed real value.^[ref]

Suppose $A\in {\mathbb{C}}^{n\times n}$ is a diagonalizable matrix, $n\ge 2$, with an eigenbasis $v_1,\dots ,v_n$ and spectral decompositionAssume the dominant eigenvalue ${\lambda }_1$ is a simple eigenvalue andThen the normalized power iteration approaches the one-dimensional eigenspace of ${\lambda }_1$. Its asymptotic direction-error bound has factor $\gamma =|{\lambda }_2/{\lambda }_1|<1$. This spectral gap is a gap in modulus: if $g=|{\lambda }_1|-|{\lambda }_2|$, then $\gamma =1-g/|{\lambda }_1|$. A small relative gap gives slow convergence.^[ref]

The short proof is the exact eigenbasis identityThe remaining sum has norm bounded by a constant times ${\gamma }^k$. This proves convergence of direction, allowing a changing sign or complex factor. A tiny $c_1$ or a poorly conditioned, nonorthogonal eigenbasis can delay visible convergence: the condition number of $V$ affects constants, while $\gamma$ controls the eventual rate. The bound need not be an equality; if the initial coefficient of the second-largest eigenvalue vanishes, the active spectrum of the power method gives a faster rate. When $\gamma =0$, all unwanted components disappear after one multiplication.^[ref]

The hypothesis is $c_1=(V^{-1}{x}_0{)}_1\ne 0$, not an ordinary nonzero inner product with the right eigenvector. Let $w_1^{\ast }$ be the first row of $V^{-1}$, a left eigenvector normalized so $w_1^{\ast }{v}_j$ is $1$ for $j=1$ and $0$ otherwise; $\ast$ denotes conjugate transpose. Then $c_1=w_1^{\ast }{x}_0$. For a Hermitian matrix, choose an orthonormal eigenbasis and the criterion becomes $v_1^{\ast }{x}_0\ne 0$: the starting vector must be nonorthogonal to the dominant eigenvector. For a general diagonalizable matrix, the nonorthogonal eigenbasis makes these two tests different, as illustrated under power iteration failure cases.

For a unit vector $u$ spanning the target eigenspace and a normalized iterate $x_k$, defineThis is the norm of the component in the orthogonal complement of $u$, equivalently the sine of the acute angle to its line. It ignores the arbitrary sign or complex unit factor of an eigenvector. Under power iteration convergence for a diagonalizable matrix, ${\epsilon }_k=O({\gamma }^k)$ in Big O notation. The orthogonal projection defines how we measure direction; it does not assume that the eigenbasis of $A$ is orthogonal. In computation, the projection form avoids subtracting nearly equal numbers inside the square root.

A tied largest modulus can prevent even direction convergence. With $D=\mathrm{diag}(2,-2)$ and $x_0=(1,1{)}^T/\sqrt{2}$,The two directions form a cycle; neither is an eigenvector. There is no dominant-modulus spectral gap. A tied modulus is not a universal failure: an initial eigenvector remains on its line, and $A=2I$ preserves every direction. Subspace iteration addresses a separated cluster by targeting a dominant invariant subspace instead of one eigenvector.

An orthogonal start can fail despite a strict spectral gap. For $A=\left(\begin{array}{cc}2& 1\\ 1& 2\end{array}\right)$ and $x_0=(1,-1{)}^T/\sqrt{2}$, the dominant component is zero, so $A{x}_0=x_0$ and every iterate equals $x_0$. The Rayleigh quotient is exactly $1$ and the eigenpair residual is zero, although the dominant eigenvalue is $3$. Exact arithmetic never creates a missing component.^[ref]

The third panel contrasts these failures with $N=\mathrm{diag}(-3,1)$, $x_0=(1,1{)}^T/\sqrt{2}$, for whichHere the horizontal eigenvector direction converges with a strict spectral gap, while the normalized vectors alternate signs. This is the successful direction convergence described under sign and phase in power iteration.

For a nonorthogonal eigenbasis, even the word “orthogonal” must be used carefully. The diagonalizable matrix $C=\left(\begin{array}{cc}3& 2\\ 0& 1\end{array}\right)$ has right eigenvectors $v_1=(1,0{)}^T$, $v_2=(-1,1{)}^T$, and dominant left eigenvector $w_1=(1,1{)}^T$. For $x_0=(a,b{)}^T$, the dominant coefficient is $c_1=w_1^T{x}_0=a+b$. The starting vector $v_2$ is nonorthogonal to $v_1$ but stays at the smaller eigenvalue $1$; $(0,1{)}^T$ is orthogonal to $v_1$ but has $c_1=1$ and converges to the dominant direction. The left eigenvector test, not the right-eigenvector inner product, controls the outcome.

A random vector drawn from a continuous probability distribution with a probability density function on the full real or complex vector space has $c_1\ne 0$ with probability one. This avoids the exact missing-component hyperplane, but does not guarantee a large component or fast convergence. Floating-point arithmetic may seed a missing component, yet a structurally preserved zero can remain zero, so accidental perturbations are not a convergence guarantee.

Iterate $q^{(k+1)}=A{q}^{(k)}/\Vert A{q}^{(k)}\Vert$. For a self-adjoint operator with a unique dominant eigenvalue separated in modulus from the others, a starting vector that is nonorthogonal to its dominant eigenvector gives convergence of the eigendirection at the spectral-ratio rate. For a general diagonalizable matrix, the condition is a nonzero coefficient in the dominant component of its eigenbasis, detected by the corresponding left eigenvector; an ordinary dot product with the right eigenvector does not generally detect that coefficient.

The power method is power iteration; its convergence depends on the dominant spectral component of the starting vector.

For example, $B=\left(\begin{array}{cc}2& 1\\ 0& 1\end{array}\right)$ has dominant right eigenvector $(1,0{)}^T$ and a dominant left eigenvector $(1,1{)}^T$. The starting vector $q^{(0)}=(-1,1{)}^T/\sqrt{2}$ has nonzero dot product with the right eigenvector, but $B{q}^{(0)}=q^{(0)}$: it stays forever in the eigenspace for the smaller eigenvalue $1$. Its coefficient along the dominant direction is zero, as the left eigenvector detects. An orthonormal eigenbasis makes the two tests agree in the self-adjoint case.