Eigenvalue computation — power method, inverse iteration: Worked examples

Every symbol here was earned in the parent: $A$ is our $n \times n$ matrix, an eigenpair $(λ, v)$ satisfies $A v = λ v$ , $∥ \cdot ∥$ is ordinary vector length $x_{1}^{2} + \dots + x_{n}^{2}$ , and the Rayleigh quotient $μ = \frac{x ^{⊤} A x}{x ^{⊤} x}$ is the current best eigenvalue guess. We reuse only those.

The scenario matrix

Every cell below is a distinct way the iteration behaves. The last column says which example covers it.

#	Scenario class	What makes it tricky	Example
C1	Two positive eigenvalues, clear gap	baseline power method	Ex 1
C2	Dominant eigenvalue negative	$x_{k}$ flips sign every step	Ex 2
C3	Equal magnitude, opposite sign ( $λ_{1} = - λ_{2}$ )	power method never settles	Ex 3
C4	Bad seed ( $c_{1} = 0$ )	amplification has nothing to grab	Ex 4
C5	Smallest eigenvalue wanted	inverse iteration, LU not $A^{- 1}$	Ex 5
C6	Interior / any eigenvalue	shifted inverse iteration	Ex 6
C7	Degenerate shift ( $σ = λ_{j}$ exactly)	$A - σ I$ singular	Ex 7
C8	Complex conjugate pair	real power method oscillates, no limit	Ex 8
C9	Real-world word problem	translate a story into an eigenproblem	Ex 9
C10	Exam twist: rate prediction	predict convergence before iterating	Ex 10
C11	Defective / repeated eigenvalue	fewer eigenvectors than $n$	Ex 11

Ex 1 — C1 · Baseline power method (positive gap)

The true eigenvalues solve the Characteristic Polynomial $det (A - λ I) = (4 - λ) (3 - λ) - 2 = λ^{2} - 7 λ + 10 = 0$ , so $λ = 5, 2$ . We aim for $λ_{1} = 5$ .

Index convention: $x_{k}$ is the vector after $k$ steps ( $x_{0}$ is the seed), and $μ_{k} = \frac{x _{k}^{⊤} A x _{k}}{x _{k}^{⊤} x _{k}}$ is the Rayleigh quotient built from that same $x_{k}$ .

$y_{1} = A x_{0} = (2 4)$ . Why this step? One multiply is the whole engine — $A$ pulls the vector toward its strongest stretch. $∥ y_{1} ∥ = 20 = 4.472$ , so $x_{1} = (0.447 0.894)$ .
Rayleigh from the seed: $μ_{0} = \frac{x _{0}^{⊤} A x _{0}}{x _{0}^{⊤} x _{0}} = \frac{( 1 , 0 ) ( 2 4 )}{1} = 4$ . Why? $μ_{0}$ uses $x_{0}$ , so it is our crude starting estimate — the least-squares eigenvalue for that vector.
$y_{2} = A x_{1} = (2 \cdot 0.894 + 3 \cdot 0.447 4 \cdot 0.894 + 0.447) = (3.130 4.024)$ , normalize by $5.098$ : $x_{2} = (0.614 0.789)$ . Now $μ_{1} = x_{1}^{⊤} A x_{1} = (0.894, 0.447) (3.130 4.024) = 4.998$ . Why? $μ_{1}$ uses $x_{1}$ , which already leans strongly toward $v_{1}$ , so the estimate jumps almost all the way to $5$ .
Next step: $μ_{2} = x_{2}^{⊤} A x_{2} = (0.789, 0.614) A (0.614 0.789) = 5.000$ (to three places), and $x_{3} \approx (0.681 0.732)$ keeps drifting toward $v_{1}$ . Why the indexing: each $μ_{k}$ pairs with its own $x_{k}$ , so we report $μ_{0}, μ_{1}, μ_{2}$ in order.

Verify: True $v_{1}$ for $λ = 5$ : $(A - 5 I) v = 0 \Rightarrow - v_{1} + v_{2} = 0$ , so $v_{1} \propto (1 1) = (0.707 0.707)$ normalized. Our $x_{3}$ is drifting toward $(0.707 0.707)$ . ✓ And $μ_{k} \to 5$ . ✓

Ex 2 — C2 · Dominant eigenvalue negative

Rounding rule for this page: all decimals are rounded to three places, rounding half up.

$y_{1} = A x_{0} = (2 - 4)$ , $∥ y_{1} ∥ = 20 = 4.472$ , $x_{1} = (0.447 - 0.894)$ . Why? Standard iterate. Notice the first component went negative.
$y_{2} = A x_{1} = (2 \cdot 0.447 - 4 \cdot ( - 0.894 )) = (0.894 3.576)$ , $∥ y_{2} ∥ = 3.686$ , $x_{2} = (0.243 0.970)$ . Why note the sign? Multiplying by $- 4$ flips the sign every step — $x_{k}$ alternates $+, -, +, -$ along $v_{1} = (0 1)$ .
Rayleigh reads the signed value: $μ_{2} = x_{2}^{⊤} A x_{2} = (- 4) (0.970)^{2} + 2 (0.243)^{2} = - 3.64$ . Why the quotient still works: $x^{⊤} A x$ carries the sign, so a negative dominant eigenvalue is reported correctly even while the vector flips.

Verify: $v_{1} = (0 1)$ , $λ_{1} = - 4$ . $x_{k} \to \pm (0 1)$ and $μ_{k} \to - 4$ . ✓

Ex 3 — C3 · Equal magnitude, opposite sign

$y_{1} = A x_{0} = (3 0) \Rightarrow x_{1} = (1 0)$ . Why watch this? The strict-gap assumption is broken, so the "landslide" argument dies.
$y_{2} = A x_{1} = (0 3) \Rightarrow x_{2} = (0 1) = x_{0}$ . Why it matters: we're back where we started — the vector oscillates forever between $(0 1)$ and $(1 0)$ , never settling on an eigenvector.
$μ_{k} = x_{k}^{⊤} A x_{k} = 0$ every step (off-diagonal $A$ , axis vectors). Why $0$ ? Neither axis vector is an eigenvector; the two components $\pm 3$ never separate because their magnitudes tie.

The figure below plots exactly these iterates. Look at it: the coloured arrows $x_{0}, x_{1}, x_{2}, x_{3}$ land only on the two dashed axes — they hop between $(0 1)$ (lavender/butter) and $(1 0)$ (coral/mint) and never rotate toward the true eigenvector lines (the two grey dashed diagonals $v (+ 3)$ and $v (- 3)$ ). That visual "trapped on the axes" is why the estimate stalls: the seed sits exactly halfway between the two eigenvectors, so neither can pull it in.

Figure — Eigenvalue computation — power method, inverse iteration

Verify: True eigenvectors are $(1 1) / 2 (λ = 3)$ and $(- 1 1) / 2 (λ = - 3)$ . Our seed $(0 1) = \frac{1}{2} (1 1) + \frac{1}{2} (- 1 1)$ has equal weight on $+ 3$ and $- 3$ , so they never separate. ✓

Ex 4 — C4 · Bad seed ( $c_{1} = 0$ )

$y_{1} = A x_{0} = (2 0)$ , $x_{1} = (1 0)$ . Why? With $c_{1} = 0$ there is no $v_{1}$ piece to multiply by $λ_{1}^{k}$ , so the first component stays exactly $0$ forever.
$μ_{k} = x_{k}^{⊤} A x_{k} = 2$ forever. Why: in exact arithmetic the iteration is trapped on $v_{2}$ , reporting the subdominant $λ_{2} = 2$ , not $λ_{1} = 5$ .
The rescue: perturb the seed, $x_{0} = (1 ε)$ with tiny $ε$ . Now the decomposition has $c_{1} = ε \neq = 0$ ; each step multiplies that weight by $5$ against the other's $2$ — a relative factor $5/2 = 2.5$ per step — so it grows and eventually dominates.

Verify: With exact $x_{0} = (1 0)$ , $μ_{k} = 2$ (wrong). With $ε = 0.001$ , after enough steps $μ_{k} \to 5$ . ✓

Ex 5 — C5 · Smallest eigenvalue via inverse iteration

Solve $A y_{1} = x_{0}$ , i.e. $(2113) y_{1} = (0 1)$ . Why solve, not invert? The parent's rule: solve, don't invert — LU-factor $A$ (see LU Decomposition) once, then each step is cheap substitution. $det A = 5$ , so $y_{1} = \frac{1}{5} (- 1 3) = (- 0.2 0.6)$ .
Normalize: $∥ y_{1} ∥ = 0.632$ , $x_{1} = (- 0.316 0.949)$ . Why? Keeps the vector from blowing up as the small- $λ$ direction gets amplified.
Rayleigh on the original $A$ : $μ_{1} = x_{1}^{⊤} A x_{1}$ . Why $A$ not $A^{- 1}$ ? We want $A$ 's eigenvalue, and the eigenvectors are shared. $A x_{1} = (0.0 1.582)$ roughly; $μ_{1} = (0.949) (1.582) + (- 0.316) (0) \approx 1.50$ .
One more solve pushes $μ \to 1.382$ .

Verify: $v$ for $λ = 1.382$ : $(A - 1.382 I) v = 0 \Rightarrow 0.618 v_{1} + v_{2} = 0$ , $v \propto (- 0.618 1) \to (- 0.526 0.851)$ normalized. Our $x_{1} = (- 0.316 0.949)$ is already leaning that way; $μ \to 1.382$ . ✓ (Compare the Rayleigh Quotient Iteration which updates $σ$ each step for cubic speed.)

Ex 6 — C6 · Any interior eigenvalue via shift

Form $A - σ I = diag (3.1, 0.1, - 1.9)$ . Why shift? Its inverse has eigenvalues $\frac{1}{λ - σ} = \frac{1}{3.1}, \frac{1}{0.1}, \frac{1}{- 1.9} = 0.32, 10, - 0.53$ . The middle direction now has the huge value $10$ — it dominates.
Solve $(A - σ I) y = x_{0}$ : component-wise $y = (\frac{0.577}{3.1}, \frac{0.577}{0.1}, \frac{0.577}{- 1.9}) = (0.186, 5.77, - 0.304)$ . Why one solve suffices: the $v_{2}$ component was multiplied by $10$ , dwarfing the others in a single step.
Normalize ( $∥ y ∥ = 5.78$ ): $x_{1} = (0.032, 0.998, - 0.053)$ — essentially $(0, 1, 0)^{⊤}$ , the middle eigenvector. Why normalize? The huge factor $10$ blew $y$ up; dividing by $∥ y ∥$ rescales it back to unit length so the next step (and the Rayleigh quotient) stay numerically sane.
Rayleigh on $A$ : $μ_{1} = x_{1}^{⊤} A x_{1} = 6 (0.032)^{2} + 3 (0.998)^{2} + 1 (0.053)^{2} \approx 2.99$ . Why so close? One step, because the shift sits right next to $λ = 3$ .

Verify: $v_{2} = (0, 1, 0)$ , $λ = 3$ . $x_{1} \to (0, 1, 0)$ and $μ \to 3$ . ✓

Ex 7 — C7 · Degenerate shift ( $σ = λ_{j}$ exactly)

First, recall the convergence-rate formula and name its indices. In shifted inverse iteration the error shrinks per step by $\frac{λ _{j} - σ}{λ _{m} - σ},$ where $λ_{j}$ is the target eigenvalue (the one closest to $σ$ , whose eigenvector we want) and $λ_{m}$ is the second-closest eigenvalue to $σ$ (the slowest-decaying leftover). This is just the power-method rate $∣ λ_{2} / λ_{1} ∣$ from the parent, applied to the transformed matrix $(A - σ I)^{- 1}$ whose eigenvalues are $\frac{1}{λ _{i} - σ}$ : the biggest is $\frac{1}{λ _{j} - σ}$ , the next biggest is $\frac{1}{λ _{m} - σ}$ , and their ratio is the boxed expression.

$A - 3 I = diag (3, 0, - 2)$ . Why singular? The middle diagonal entry is $0$ (that is $λ_{j} - σ = 3 - 3$ ), so $det = 3 \cdot 0 \cdot (- 2) = 0$ — no inverse, the linear solve has no unique answer.
The rate $\frac{λ _{j} - σ}{λ _{m} - σ} = \frac{0}{∣1 - 3∣} = 0$ says "converge in zero steps". Why compute the rate here? It quantifies the "secretly good news": a rate of $0$ means the method wants to hand over the answer instantly — the singularity is the ideal limit, not a failure — but the exact-zero denominator entry makes the system unsolvable as written.
The practical fix: back off to $σ = 3 \pm 1 0^{- 6}$ . Why it works: now $A - σ I$ is invertible but nearly singular, the middle direction is amplified by $\sim 1 0^{6}$ , and — remarkably — the near-singularity is harmless: the error vector aligns with $v_{2}$ , exactly what we want.

Verify: $det (A - 3 I) = 0$ (singular), while $det (A - 2.999999 I) \neq = 0$ . ✓

Ex 8 — C8 · Complex conjugate pair

$y_{1} = A x_{0} = (1 0)$ , $x_{1} = (1 0)$ . Why? $A$ rotates $9 0^{\circ}$ ; the seed just turns.
$y_{2} = A x_{1} = (0 - 1)$ , $x_{2} = (0 - 1)$ . Then $x_{3} = (- 1 0)$ , $x_{4} = (0 1) = x_{0}$ . Why it cycles: a rotation has no real eigenvector, so there's no real direction for power method to lock onto — the vector loops with period 4.
$μ_{k} = x_{k}^{⊤} A x_{k} = 0$ always (rotation sends $x ⊥ A x$ ). Why $0$ : $x^{⊤} (A x)$ is the dot product of a vector with its $9 0^{\circ}$ turn — always zero.

The figure shows this spin. Look at it: the four coloured arrows $x_{0}, x_{1}, x_{2}, x_{3}$ sit on the unit circle a quarter-turn apart, and the annotation "+90 deg each step" traces the rotation; after four hops you are back at $x_{0}$ . There is no arrow drifting toward a fixed line — visual proof that no real eigenvector exists to converge to.

Verify: Eigenvalues of $(01 - 1 0)$ are $\pm i$ , both $∣ λ ∣ = 1$ , no real gap. $A^{4} = I$ so $x_{4} = x_{0}$ . ✓

Ex 9 — C9 · Word problem (PageRank-style)

$y_{1} = M x_{0}$ . Columns sum to $1$ , so multiplying redistributes importance. $y_{1} = (\frac{1}{3}, \frac{1}{6}, \frac{1}{2})$ . Why $M$ is the right operator: a page's importance = sum of importances flowing in along links, which is exactly a matrix–vector product.
Normalize with $∥ \cdot ∥_{1}$ ; $y_{1}$ already sums to $1$ . Iterate: $y_{2} = M y_{1} = (0.5, 0.167, 0.333)$ ; $y_{3} = M y_{2} = (0.333, 0.25, 0.417)$ .
Watch the vector settle toward the eigenvector, and read off $λ_{1} = 1$ . Why $λ_{1} = 1$ is guaranteed: column sums $= 1$ means $1^{⊤} M = 1^{⊤}$ (the all-ones row-vector is a left eigenvector for eigenvalue $1$ ), forcing an eigenvalue $1$ ; the Perron–Frobenius theorem makes it the dominant one. So power method converges here, and the limiting Rayleigh quotient equals $1$ .
Fixed point: solve $x = M x$ directly. From row 1, $x_{1} = x_{3}$ ; from row 2, $x_{2} = 0.5 x_{1}$ ; from row 3, $x_{3} = 0.5 x_{1} + x_{2} = x_{1}$ ✓. So $x = (2, 1, 2)$ up to scale → normalize with $∥ \cdot ∥_{1}$ (sum $= 5$ ) to $(0.4, 0.2, 0.4)$ .

Conclusion: the steady importances are $(0.4, 0.2, 0.4)$ — pages 1 and 3 tie for most important, page 2 is least. Our forecast (page 3 on top) was almost right: it ties for the lead.

Verify: $M (0.4, 0.2, 0.4)^{⊤} = (0.4, 0.2, 0.2 + 0.2) = (0.4, 0.2, 0.4)$ . ✓ Eigenvalue $= 1$ , importances sum to $1$ . ✓

Ex 10 — C10 · Exam twist: predict the rate before iterating

Power method error factor per step $r = \frac{λ _{2}}{λ _{1}} = \frac{9}{10} = 0.9$ . Why this ratio: the parent showed the surviving decay term is $(λ_{2} / λ_{1})^{k}$ . Need $0. 9^{k} \leq 1 0^{- 3}$ .
Solve: $k \geq \frac{ln 1 0 ^{- 3}}{ln 0.9} = \frac{- 6.908}{- 0.10536} = 65.6 \Rightarrow k = 66$ steps. Why the log: raising $0.9$ to a power to hit a target is inverted by $ln$ . (See Convergence Rates.)
Shift-invert rate: eigenvalues of $(A - σ I)^{- 1}$ have sizes $\frac{1}{∣ λ - σ ∣}$ . Using the Ex 7 formula with target $λ_{j} = 9$ and second-closest $λ_{m} = 10$ : distances $∣9 - 9.1∣ = 0.1$ and $∣10 - 9.1∣ = 0.9$ , so rate $= \frac{∣ λ _{j} - σ ∣}{∣ λ _{m} - σ ∣} = \frac{0.1}{0.9} = 0.111$ .
Need $0.11 1^{k} \leq 1 0^{- 3}$ : $k \geq \frac{- 6.908}{ln 0.111} = \frac{- 6.908}{- 2.198} = 3.14 \Rightarrow k = 4$ steps.

Verify: $0. 9^{66} = 9.1 \times 1 0^{- 4} \leq 1 0^{- 3}$ ✓; $0. 9^{65} = 1.0 \times 1 0^{- 3}$ (just above) so $66$ is correct. $0.11 1^{4} = 1.5 \times 1 0^{- 4} \leq 1 0^{- 3}$ ✓, $0.11 1^{3} = 1.4 \times 1 0^{- 3}$ (too big) so $4$ is correct. 66 vs 4 — a $16 \times$ speedup from a good shift. ✓

Ex 11 — C11 · Defective / repeated eigenvalue

Eigen-check: $(A - 3 I) = (0010)$ has a one-dimensional nullspace $v_{1} = (0 1)$ . Why "defective": algebraic multiplicity $2$ but geometric multiplicity $1$ — one eigenvector short, so $A$ is not diagonalizable and Spectral Decomposition does not apply; you use a Jordan form instead.
$y_{1} = A x_{0} = (3 1)$ , $x_{1} = (0.949 0.316)$ . $y_{2} = A x_{1} = (3 \cdot 0.949 0.316 + 0.949) = (2.846 1.265)$ , $x_{2} = (0.914 0.406)$ . Why the drift? Each multiply feeds a bit of $v_{1}$ in through the off-diagonal $1$ ; the first component grows.
$y_{3} = A x_{2} = (3 \cdot 0.914 0.406 + 0.914) = (2.742 1.320)$ , $x_{3} = (0.901 0.434)$ — creeping toward $v_{1} = (0 1)$ , but slowly. Why slow: with equal eigenvalues there is no ratio $∣ λ_{2} / λ_{1} ∣ < 1$ to drive fast decay; convergence is only algebraic (like $1/ k$ ), not geometric. See Convergence Rates.
Rayleigh: $μ_{2} = x_{2}^{⊤} A x_{2} = 3 (0.406)^{2} + 2 (0.406) (0.914) + 3 (0.914)^{2} = 3.742$ . It overshoots then settles to $3$ as $x_{k} \to (0 1)$ .

Verify: $A = (3013)$ has the single eigenvalue $3$ (multiplicity $2$ ), one eigenvector $(0 1)$ ; $μ_{k} \to 3$ and $x_{k} \to (0 1)$ . ✓

Recall Scenario checklist (self-test)

Match each symptom to its fix. Vector flips sign every step ::: Dominant eigenvalue is negative — read the signed Rayleigh quotient (Ex 2). Vector cycles, never settles, real matrix ::: Either a tie $λ_{1} = - λ_{2}$ (shift to break it, Ex 3) or a complex pair (use QR, Ex 8). Estimate stuck on the wrong (smaller) eigenvalue ::: Bad seed with $c_{1} = 0$ ; perturb $x_{0}$ (Ex 4). Want the smallest eigenvalue ::: Inverse iteration — LU-solve, don't invert (Ex 5). Want an interior eigenvalue ::: Shift near it, then inverse iterate (Ex 6). Linear solve says "singular" ::: $σ$ hit an eigenvalue exactly; nudge $σ$ by $\pm 1 0^{- 6}$ (Ex 7). Convergence painfully slow, single eigenvalue ::: Defective/repeated eigenvalue — only algebraic decay; use QR (Ex 11).