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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.

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.

For a diagonalizable matrix with distinct eigenvalues $\lambda ,\mu ,\dots$, this polynomial is the projection operator onto the eigenspace for $\lambda$ along the direct sum of the other eigenspaces. It is one on its own eigenspace and zero on the others. Consequently $P_{\lambda }{P}_{\mu }=0$ for different eigenvalues, $\sum P_{\lambda }=I$ and $A=\sum \lambda P_{\lambda }$. These projections need not be orthogonal projections; that stronger property holds for a normal matrix. This is the polynomial interpolation form of the spectral decomposition.