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Suppose a diagonalizable matrix has a simple eigenvalue ${\lambda }_{\ast }$ uniquely closest to a shift $\mu \notin \mathrm{spec}(A)$. If the starting vector has a nonzero coefficient $c_{\ast }$ in the target eigenbasis direction, fixed-shift inverse iteration converges to that eigenvector line. In fact, the eigenvalues of $(A-\mu I{)}^{-1}$ are $({\lambda }_j-\mu {)}^{-1}$, so after $k$ inverse applications, division by $c_{\ast }({\lambda }_{\ast }-\mu {)}^{-k}$ leaves the target eigenvector plusThis proves direction error $O(q_{\mu }^k)$ in Big O notation. The multiplicative constant depends on the eigenbasis and starting coefficients, and no uniform reduction at the first step follows for a nonsymmetric matrix. Over the real numbers, ${\lambda }_{\ast }<\mu$ produces alternating signs. A nonzero ordinary inner product with a right eigenvector is not generally the same as a nonzero coefficient in the eigenbasis; the latter is the required condition.

A nonsymmetric matrix is a square matrix unequal to its transpose. Its right eigenvectors need not be orthogonal, and it need not be a diagonalizable matrix. For instance, $A=\left(\begin{array}{cc}1& 1\\ 0& 2\end{array}\right)$ has eigenvectors $(1,0{)}^T$ and $(1,1{)}^T$ for its distinct eigenvalues. A starting vector $(1,0{)}^T$ has nonzero inner product with $(1,1{)}^T$ but zero coefficient in that direction in the eigenbasis. This is why power iteration and inverse iteration require a nonzero spectral component rather than an ordinary nonorthogonality test in the general case.

For a fixed shift $s$ that is not an eigenvalue, inverse iteration is the power method applied to $(A-sI{)}^{-1}$. Before normalization, the $k$th vector is proportional toIf $s<{\lambda }_1$, factor out the first coefficient:Every ratio in parentheses lies strictly between zero and one. The bracket therefore tends to $w_1$, and normalization in the Euclidean norm givesThis proves the stated convergence to $w_1$ or $-w_1$ and exhibits the convergence of fixed-shift inverse iteration rate.

Now suppose ${\lambda }_m<s<{\lambda }_{m+1}$. The closest eigenvalue to $s$ is one of ${\lambda }_m$ and ${\lambda }_{m+1}$.

Finally, if ${\lambda }_n<s$, the closest eigenvalue is ${\lambda }_n$ and its inverse eigenvalue is negative. Thus the direction approaches the line spanned by $w_n$, while normalized vectors alternate between the two signs. These cases are the sign behavior of fixed-shift inverse iteration.

Fix a shift $s$ that is not an eigenvalue of $A$. Inverse iteration computesand may use $r({\mathbf{x}}_k)$ as the eigenvalue estimate.

For a real symmetric matrix, suppose that one eigenvalue ${\lambda }_{\ast }$ is uniquely closest to $s$ and that ${\mathbf{x}}_0$ has a nonzero component in its eigenspace. Since the iteration is the power method applied to $(A-sI{)}^{-1}$, the normalized vectors converge, up to sign, to an eigenvector for ${\lambda }_{\ast }$. If ${\lambda }_{\mathrm{next}}$ is the second-closest eigenvalue to the shift, the asymptotic direction-error ratio isThus a shift close to the desired eigenvalue gives rapid convergence. The Rayleigh quotients converge to ${\lambda }_{\ast }$ and, for a simple eigenvalue, their errors are of second order in the vector error. This is convergence of fixed-shift inverse iteration.

The power method starts from $q_0\ne 0$ and iteratesIf the eigenvalue of largest modulus is separated by a spectral gap and the initial vector is nonorthogonal to a corresponding eigenvector, the iterates converge in direction to that eigenvector.

Inverse iteration with shift $s$ instead solvesIt applies the power method to $(A-sI{)}^{-1}$ and therefore finds an eigenvector whose eigenvalue is closest to $s$.

Let $q_1$ and $q_n$ be the first and last columns of ${\tilde{Q}}_k$. Since ${\tilde{R}}_k$ is upper triangular,Thus $q_1$ is the normalized $k$th power method iterate starting from $e_1$.

Taking the transpose of $A^k={\tilde{Q}}_k{\tilde{R}}_k$ and using symmetry gives $A^k={\tilde{R}}_k^T{\tilde{Q}}_k^T$. ThereforeSo the last column is the normalized $k$th inverse iteration iterate with shift zero and starting vector $e_n$.

Choose a shift $\mu$ which is not an eigenvalue. Inverse iteration applies power iteration to $B=(A-\mu I{)}^{-1}$, implemented by a linear system solve rather than by forming an explicit matrix inverse:A single LU decomposition of $A-\mu I$ can be reused for all the solves at this fixed shift. An eigenpair $A{u}_j={\lambda }_j{u}_j$ transforms intoThus an eigenvalue ${\lambda }_{\ast }$ uniquely closest to $\mu$ becomes the dominant eigenvalue of $B$. For a diagonalizable matrix, the argument in the preceding subpart gives convergence of diagonalizable fixed-shift inverse iteration, provided the starting coefficient in the target eigenspace is nonzero, with power iteration direction error $O(q_{\mu }^k)$, whereFor any prescribed simple eigenvalue, taking a sufficiently close but unequal shift makes the numerator small while the denominator stays close to its positive separation from the remaining eigenvalues. This makes $q_{\mu }$ small and gives very fast convergence per iteration. The Rayleigh quotient of $v_k$ estimates ${\lambda }_{\ast }$ directly. If ${\beta }_k$ estimates the associated eigenvalue of $B$, the corresponding estimate is $\mu +{\beta }_k^{-1}$.

The selection principle also applies when the other eigenvalues have nontrivial blocks in the Jordan normal form: their contributions acquire powers of $k$, multiplying strictly smaller geometric factors. A simple uniquely closest eigenvalue still attracts the direction when the Riesz projection of $v_0$ onto its generalized eigenspace is nonzero. For nonreal target eigenvalues, allow complex shifts and arithmetic; a real shift cannot distinguish a conjugate pair of equal distances. Setting $\mu ={\lambda }_{\ast }$ makes the solve singular, and an exactly zero starting target component prevents convergence to that target in exact arithmetic. When ${\lambda }_{\ast }<\mu$ is real, account for the sign behavior of fixed-shift inverse iteration.

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.