Linear Algebra › Eigenvalues & eigenvectors › full formula sheet

Av = λv,  v ≠ 0

The eigen-equation

The directions a matrix only stretches — never rotates.

Notation on this page: v is a nonzero vector, λ (lambda) is a scalar — just a number — A is a square matrix, and I is the identity matrix.

Where it comes from

A square matrix A is a machine that takes vectors and returns vectors — usually scrambling them: rotating, stretching, and skewing all at once. The problem eigenvalues solve: find the special directions the machine treats simply — vectors it only stretches (or shrinks, or flips), never rotates off their own line. Those directions are the matrix’s natural axes, and they show up everywhere: Google’s PageRank, the principal components of a dataset, quantum states, and whether a system of differential equations blows up or settles down.

The naive first attempt: just solve Av = λv directly for v. Try it on A = [[2, 1], [0, 3]] with v = (x, y):

2x + y
=
λx
First row of Av = λv.
3y
=
λy
Second row.

Two equations — but three unknowns: x, y, and λ. You are stuck in a circle: you can’t pin down v without knowing λ, and you can’t find λ without knowing v. Guessing is no way out either: try v = (0, 1) and you get Av = (1, 3), which is not a multiple of (0, 1) — dead end, and nothing in the equations tells you what to try next.

Here is the intuition that breaks the circle. Compare a scaling D = [[2, 0], [0, 3]] with a shear S = [[1, 2], [0, 1]]. Feed D the vector (1, 0): out comes (2, 0) — same line, stretched by 2. Feed it (0, 1): out comes (0, 3) — same line, stretched by 3. D is full of stretching directions. Now feed S the vector (0, 1): out comes (2, 1) — knocked off its line; S rotated it while stretching. Only the x-axis survives S untouched ((1, 0) → (1, 0), stretched by λ = 1).

So the eigen-equation Av = λv is a filter: out of all vectors, keep only the ones A treats as pure scaling. The number λ says how much: λ = 2 doubles, λ = 1/2 halves, λ = −1 flips and keeps the length, λ = 0 crushes to zero.

And the v ≠ 0 rule? Without it the equation is vacuous: A0 = 0 = λ0 holds for every λ, so every number would be an “eigenvalue” of every matrix — meaningless. The nonzero requirement is what makes eigenvalues well-defined. It is not a technicality; it is the whole game.

That leaves the circularity problem: how do you find λ without already knowing v? The trick — coming up in the derivation — is to stop solving for v first and instead ask: for which λ does this system have a nonzero solution at all? Rearranged that way, the circle breaks.

Eigen is German for “own” or “characteristic” (David Hilbert, ~1904; the ideas run back through Euler and Cauchy’s work on principal axes). Eigenvectors are the matrix’s own directions.

Derivation

From Av = λv to something solvable. Watch Step 3 — it is the move that breaks the circle.

Av
=
λv  (v ≠ 0)
Step 1 — the eigen-equation. v is stretched by the scalar λ, never rotated off its own line.
Av − λv
=
0
Step 2 — move everything to one side. Now the question reads: for which λ does the shifted problem send some nonzero v to zero?
Av − λIv
=
0
Step 3 — insert I. λv = λIv since Iv = v, and we may not write A − λ — you cannot subtract a scalar from a matrix. This is the move that makes factoring possible.
(A − λI)v
=
0
Step 4 — factor out v. The equation now says: the matrix (A − λI) sends the nonzero vector v to the zero vector.
det(A − λI)
=
0
Step 5 — the characteristic equation. An invertible matrix sends only 0 to 0, so a nonzero v exists exactly when (A − λI) is singular — determinant zero. For an n×n matrix this is a degree-n polynomial in λ; its roots are the eigenvalues. For each root, solve (A − λI)v = 0 to get the eigenvectors. ∎

Why does “nonzero solution” mean “determinant zero”? If (A − λI) were invertible, you could multiply both sides by its inverse and get v = 0 — the only solution. So a nonzero v exists exactly when the matrix is not invertible: singular, i.e. determinant zero. That single equivalence is the engine of the whole subject.

How to use it

The procedure, every time:

  1. Form A − λI — subtract λ down the diagonal only; off-diagonal entries are untouched. (Why only the diagonal? λI has λ’s on the diagonal and zeros elsewhere.)
  2. Characteristic polynomial: compute det(A − λI) and set it equal to zero.
  3. Solve for λ — the roots are the eigenvalues. For a 2×2 there is a shortcut (below).
  4. One eigenvalue at a time: plug each λ into (A − λI)v = 0 and row-reduce. The nonzero solutions are the eigenvectors — pick the simplest one (clear fractions, smallest integers).
  5. Sanity-check: multiply Av and confirm it equals λv. (Why: one arithmetic slip in row-reduction hands you a vector that isn’t an eigenvector at all — the check catches it in seconds.)

The 2×2 shortcut

λ² − tr(A)λ + det(A) = 0tr(A) = sum of diagonal entries (the trace); for [[a, b], [c, d]], det(A) = ad − bc

Why it works: for [[a, b], [c, d]], det(A − λI) = (a−λ)(d−λ) − bc = λ² − (a+d)λ + (ad−bc). Bonus memory hook: the trace is the sum of the eigenvalues and the determinant is their product — for [[4, 1], [2, 3]] below, 5 + 2 = 7 = tr(A) and 5·2 = 10 = det(A). ✓

Judgment calls

Eigenvectors are never unique. If v works, so does every nonzero multiple cv, since A(cv) = cλv = λ(cv). (1, 1), (2, 2), and (−3, −3) are all “the” eigenvector for λ = 5 — pick the tidiest.

λ = 0 means A is singular. Av = 0 with v ≠ 0 is a nontrivial null space, so det(A) = 0 and A has no inverse. (Consistent: plug λ = 0 into det(A − λI) = 0 and you get det(A) = 0 directly.)

Repeated eigenvalues need care. A double root can produce fewer independent eigenvectors than its multiplicity — then the matrix can’t be diagonalized (“defective”). One line, no deep dive: just don’t assume two equal eigenvalues means two eigenvectors.

Where this goes

Diagonalization: if an n×n matrix has n independent eigenvectors, stack them as the columns of P and put the eigenvalues on the diagonal of D — then A = PDP−1, and powers become trivial: Ak = PDkP−1. That is the payoff: in the eigenbasis, the matrix just scales coordinates.

Steady states: a Markov (stochastic) matrix always has λ = 1, and its eigenvector is the long-run distribution. Google’s PageRank is exactly this: the web’s link matrix, and the page ranks are its λ = 1 eigenvector.

Common mistake: solving (A − λI)v = 0, getting only v = 0, and calling it “the eigenvector.” The zero vector is never an eigenvector — if row-reduction leaves no free variable, your λ isn’t actually an eigenvalue (or the algebra slipped). Go back and check.

Worked examples

Four problems, easiest first. In each one, read every step — the why of each move is the lesson.

Example 1 — diagonal (the confidence builder): A = [[2, 0], [0, 3]]

  1. Form A − λI. = [[2−λ, 0], [0, 3−λ]]. (Why only the diagonal changes: λI = [[λ, 0], [0, λ]].)
  2. Characteristic polynomial. The determinant of a diagonal matrix is the product of the diagonal: (2−λ)(3−λ) = 0. (Why set it to zero: we need a nonzero v, so A − λI must be singular.)
  3. Solve. A product is zero exactly when a factor is: λ = 2 or λ = 3.
  4. Eigenvectors. For λ = 2: [[0, 0], [0, 1]]v = 0 gives y = 0 with x free → v = (1, 0). For λ = 3: v = (0, 1). (Why “x free”: a zero row leaves that variable unconstrained — that freedom is the eigenvector direction.)
  5. Check. A(1, 0) = (2, 0) = 2(1, 0) ✓; A(0, 1) = (0, 3) = 3(0, 1) ✓

The lesson: diagonal matrices confess immediately — eigenvalues on the diagonal, eigenvectors the standard basis vectors.

Example 2 — the full 2×2: A = [[4, 1], [2, 3]]

  1. A − λI = [[4−λ, 1], [2, 3−λ]].
  2. Characteristic polynomial: det = (4−λ)(3−λ) − 2·1 = λ² − 7λ + 12 − 2 = λ² − 7λ + 10 = 0. (Or the shortcut: λ² − tr(A)λ + det(A) = λ² − 7λ + 10.)
  3. Solve: λ² − 7λ + 10 = (λ−5)(λ−2) = 0 → λ = 5, 2.
  4. λ = 5: [[−1, 1], [2, −2]]v = 0. The rows are multiples of each other — expected, not an error: det = 0 means singular, so the rows must be dependent. One equation survives: −x + y = 0 → v = (1, 1).
  5. Check: A(1, 1) = (5, 5) = 5(1, 1) ✓
  6. λ = 2: [[2, 1], [2, 1]]v = 0 → 2x + y = 0. Set x = 1 (why: the variable is free — pick the value giving the smallest integers) → v = (1, −2).
  7. Check: A(1, −2) = (2, −4) = 2(1, −2) ✓
Common mistake: panicking when the rows come out as multiples of each other. That dependence is the proof you did it right (singular matrix = dependent rows). If your rows are independent, recheck your λ.

Example 3 — no real eigenvalues: rotation A = [[0, −1], [1, 0]]

  1. A − λI = [[−λ, −1], [1, −λ]].
  2. Characteristic polynomial: det = (−λ)(−λ) − (−1)(1) = λ² + 1 = 0.
  3. No real roots: λ² = −1 has no real solution. So over ℝ there are no eigenvalues.
  4. Why, geometrically: this matrix rotates every vector 90° counterclockwise — try (1, 0) → (0, 1): knocked off its line. Every nonzero vector gets rotated, so no direction is merely stretched. The algebra (λ² + 1 = 0) and the geometry agree.
Common mistake: “no real eigenvalues” is a complete, respectable answer — not a sign your work went wrong. (Over ℂ the eigenvalues are ±i, but that’s beyond this page.) The real mistake is forcing it: writing λ = 1 because “λ² = −1, close enough.” Never round away a minus sign.

Example 4 — error-spotting: the missing I

A student computes eigenvalues of A = [[4, 1], [2, 3]] but writes det(A − λ) = 0 — subtracting the scalar straight off the matrix:

  1. The wrong move: “A − λ” = [[4−λ, 1−λ], [2−λ, 3−λ]] — λ subtracted from every entry. (Why it’s nonsense: you can’t subtract a number from a matrix; it’s a type error, like subtracting 3 from a grocery list.)
  2. The wrong polynomial: det = (4−λ)(3−λ) − (1−λ)(2−λ) = (12 − 7λ + λ²) − (2 − 3λ + λ²) = 10 − 4λ. Setting it to zero gives λ = 2.5.
  3. Why 2.5 is garbage: a 2×2 has two eigenvalues (counting multiplicity) — a single root is already suspicious. And the check fails: A − 2.5I = [[1.5, 1], [2, 0.5]], whose determinant is 0.75 − 2 = −1.25 ≠ 0, so (A − 2.5I)v = 0 has only v = 0 — no eigenvector exists. The “answer” doesn’t even satisfy its own equation.
  4. The fix: A − λI = [[4−λ, 1], [2, 3−λ]] — λ only on the diagonal, because λI = [[λ, 0], [0, λ]]. Then det = λ² − 7λ + 10, and λ = 5, 2, as in Example 2. ✓
Common mistake: if your characteristic polynomial has lower degree than n (here: linear instead of quadratic), you almost certainly subtracted λ from the wrong places. Count your roots.

Memorization tips

  • Say the procedure in one breath: “A minus lambda-I, determinant zero, solve for lambda, null space for v.” If you can say it, you can do it.
  • Trace = sum, determinant = product. The eigenvalues of [[4, 1], [2, 3]] are 5 and 2 because 5 + 2 = 7 = tr(A) and 5·2 = 10 = det(A). Reconstruct eigenvalues from those two numbers as a self-check.
  • v ≠ 0 or it’s all lies: the zero vector “satisfies” every λ. The nonzero rule is what makes eigenvalues meaningful.
  • Order of operations: eigenvalues first (the characteristic equation — no v in sight), eigenvectors second (null space). Never the reverse.
  • Diagonal and triangular matrices confess: eigenvalues sit on the diagonal. [[5, 7], [0, −2]] has eigenvalues 5 and −2 — no determinant needed.
  • The collapse check: λ = 0 is an eigenvalue exactly when det(A) = 0 — i.e. exactly when A is singular. Eigenvalues detect invertibility.
  • Picture it: eigenvectors are the directions the matrix only stretches. Shear [[1, 2], [0, 1]] keeps just the x-axis; scaling [[2, 0], [0, 3]] keeps both axes; rotation [[0, −1], [1, 0]] keeps none (over ℝ).

Final challenge

Five mixed questions — basics, applications, and the traps, all in one. Score 5/5 and the eigen-equation is yours.

← Back to the Linear Algebra formula sheet

How to learn a formula here

  1. Read each section in order — every section ends with a short quiz. Take it before moving on; the questions test exactly what you just read.
  2. Work the examples with the answers covered, then uncover one step at a time and compare.
  3. Finish with the final challenge — five mixed questions including the classic traps.
  4. Retake what you miss — every quiz reshuffles each attempt, and every answer explains itself.

Frequently asked questions

What is an eigenvector?

A nonzero vector v that a matrix A only stretches (or flips and stretches), never rotates: Av = λv. The stretch factor λ is the eigenvalue.

Why must the eigenvector v be nonzero?

Because A0 = λ0 holds for every λ, so the zero vector would make every number an eigenvalue — meaningless. The nonzero requirement is what makes eigenvalues well-defined.

How do I find eigenvalues?

Solve the characteristic equation det(A − λI) = 0. Its roots are the eigenvalues. Then solve (A − λI)v = 0 for each λ to get eigenvectors.

Can a matrix have no (real) eigenvalues?

Yes. A rotation matrix like [[0,−1],[1,0]] has characteristic polynomial λ² + 1 = 0, with no real roots — every nonzero vector gets rotated off its own line.

What does ‘eigen’ mean?

German for ‘own’ or ‘characteristic’ (David Hilbert, ~1904). Eigenvectors are the matrix’s own natural axes.

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