Eigenvalues

Study Sheet

Eigenvalues

The directions a matrix merely stretches

Eigenvalues & Eigenvectors

Tip
Stretch directions

An eigenvector v\mathbf v satisfies Av=λvA\mathbf v = \lambda\mathbf v: the matrix only STRETCHES it by the eigenvalue λ\lambda. Find eigenvalues from det⁡(A−λI)=0\det(A - \lambda I) = 0; for 2×22\times2 this is λ2−(trA)λ+det⁡A=0\lambda^2 - (\mathrm{tr}A)\lambda + \det A = 0, so eigenvalues SUM to the trace and MULTIPLY to the determinant.

Concept
Powers become easy

If Av=λvA\mathbf v = \lambda\mathbf v then Akv=λkvA^k\mathbf v = \lambda^k\mathbf v, and tr(Ak)=∑λik\mathrm{tr}(A^k) = \sum\lambda_i^k. This is why eigenvalues run the long-term behavior of everything iterative — Fibonacci (QnQ^n holds FnF_n, and det⁡Q=−1\det Q = -1 gives Cassini’s identity in one line), Markov chains, and populations alike.

Example
A symmetric 2×2

(2112)\begin{pmatrix}2&1\\1&2\end{pmatrix}: trace 44, determinant 33, so λ2−4λ+3=0\lambda^2 - 4\lambda + 3 = 0 gives λ=1,3\lambda = 1, 3 — with perpendicular eigenvectors (1,−1)(1,-1) and (1,1)(1,1), as symmetry guarantees.

Side note
Significance: the axes the world actually uses

Google ranks pages by an eigenvector; bridges have resonant eigenfrequencies (soldiers break step crossing them); quantum energy levels ARE eigenvalues; population models stabilize along their dominant eigenvector. Whenever a system is applied repeatedly, its eigenvalues write its future.

Try it
Try it: eigenvalues at sight

Find the eigenvalues of the triangular matrix (4702)\begin{pmatrix}4&7\\0&2\end{pmatrix}. Work: for triangular matrices the eigenvalues ARE the diagonal entries: 44 and 22 (the characteristic determinant is (4−λ)(2−λ)(4-\lambda)(2-\lambda) regardless of the 77). Check the trace: 4+2=64 + 2 = 6 ✓ and det: 88 ✓.

Proofs & Why It Matters

Tip
Proof: eigenvalues sum to the trace, multiply to the determinant (2×2)

Expand the characteristic polynomial: det⁡(a−λbcd−λ)=λ2−(a+d)λ+(ad−bc)\det\begin{pmatrix}a-\lambda & b\\ c & d-\lambda\end{pmatrix} = \lambda^2 - (a+d)\lambda + (ad - bc). By Vieta, its roots λ1,λ2\lambda_1, \lambda_2 satisfy λ1+λ2=a+d=tr A\lambda_1 + \lambda_2 = a + d = \mathrm{tr}\,A and λ1λ2=ad−bc=det⁡A\lambda_1\lambda_2 = ad - bc = \det A.

Reminder — Vieta's formulas:  r1+r2=−ba,r1r2=car_1+r_2=-\frac{b}{a},\qquad r_1 r_2=\frac{c}{a}

The trace and determinant are the eigenvalues' fingerprints, readable without solving anything. ■\blacksquare

Tip
Proof: Aᵏv = λᵏv

Induction on kk. Base: Av=λvA\mathbf v = \lambda\mathbf v is given. Step: if Akv=λkvA^k\mathbf v = \lambda^k\mathbf v, apply AA once more: Ak+1v=A(λkv)=λk(Av)=λk+1vA^{k+1}\mathbf v = A(\lambda^k\mathbf v) = \lambda^k(A\mathbf v) = \lambda^{k+1}\mathbf v.

So along an eigenvector, iterating the matrix is just repeated scalar multiplication — the engine behind diagonalization, Fibonacci formulas, and Markov-chain limits. ■\blacksquare

Going Deeper: Explanations & Worked Problems

Concept
Finding eigenvalues and eigenvectors: the complete procedure

Step one: eigenvalues are the roots of det⁡(A−λI)=0\det(A - \lambda I) = 0 — the values of λ\lambda making A−λIA - \lambda I singular, because only a singular matrix can send a NONZERO vector to zero. For a 2×22\times2, skip straight to λ2−(trA)λ+det⁡A=0\lambda^2 - (\mathrm{tr}A)\lambda + \det A = 0.

Step two: for EACH root, solve (A−λI)v=0(A - \lambda I)\mathbf v = \mathbf 0 — the matrix is now singular by design, so its rows are dependent and one row determines v\mathbf v up to scale. Step three: use them — along eigenvectors the matrix acts as a number, so AkA^k acts as λk\lambda^k, systems decouple, and Akx0A^k\mathbf x_0 for large kk tilts toward the eigenvector of the LARGEST ∣λ∣|\lambda| (the power method, and the reason Markov chains forget their starting point). Two structural gifts: symmetric matrices always have real eigenvalues with orthogonal eigenvectors, and distinct eigenvalues always give independent eigenvectors.

Example
Worked: full eigen-analysis of a symmetric matrix

Analyze A=(2112)A = \begin{pmatrix}2&1\\1&2\end{pmatrix}.

Step 1 — characteristic equation via trace and determinant: tr=4\mathrm{tr} = 4, det⁡=3\det = 3, so λ2−4λ+3=0\lambda^2 - 4\lambda + 3 = 0, factoring as (λ−1)(λ−3)=0(\lambda - 1)(\lambda - 3) = 0: eigenvalues 11 and 33.

Step 2 — eigenvector for λ=3\lambda = 3: (A−3I)v=(−111−1)v=0(A - 3I)\mathbf v = \begin{pmatrix}-1&1\\1&-1\end{pmatrix}\mathbf v = \mathbf 0 gives v1=v2v_1 = v_2: take (1,1)(1, 1).

Step 3 — for λ=1\lambda = 1: (1111)v=0\begin{pmatrix}1&1\\1&1\end{pmatrix}\mathbf v = \mathbf 0 gives v1=−v2v_1 = -v_2: take (1,−1)(1, -1).

Step 4 — checks: A(1,1)=(3,3)=3(1,1)A(1,1) = (3,3) = 3(1,1) ✓; A(1,−1)=(1,−1)A(1,-1) = (1,-1) ✓; and (1,1)⋅(1,−1)=0(1,1)\cdot(1,-1) = 0 — orthogonal, as symmetry promised.

Step 5 — payoff: A10(1,0)=A10⋅12[(1,1)+(1,−1)]=12[310(1,1)+(1,−1)]A^{10}(1, 0) = A^{10}\cdot\tfrac12\left[(1,1) + (1,-1)\right] = \tfrac12\left[3^{10}(1,1) + (1,-1)\right] — a closed form with no repeated multiplication anywhere.

Example
Worked: eigenvalues running a dynamical system

A population splits between two sites, redistributing each year by pk+1=Mpk\mathbf p_{k+1} = M\mathbf p_k with M=(0.80.30.20.7)M = \begin{pmatrix}0.8&0.3\\0.2&0.7\end{pmatrix} (columns sum to 11: nobody is lost).

Step 1 — eigenvalues: tr=1.5\mathrm{tr} = 1.5, det⁡=0.56−0.06=0.5\det = 0.56 - 0.06 = 0.5: λ2−1.5λ+0.5=0\lambda^2 - 1.5\lambda + 0.5 = 0 factors as (λ−1)(λ−0.5)=0(\lambda - 1)(\lambda - 0.5) = 0.

Step 2 — the λ=1\lambda = 1 eigenvector: (M−I)v=0(M - I)\mathbf v = \mathbf 0 gives −0.2v1+0.3v2=0-0.2v_1 + 0.3v_2 = 0, so v∝(3,2)\mathbf v \propto (3, 2).

Step 3 — conclusion: the λ=0.5\lambda = 0.5 component HALVES every year and dies; every starting split converges to the ratio 3:23 : 2 — sixty percent at site one, forever, regardless of history.

Step 4 — check one iteration from (1,0)(1, 0): M(1,0)=(0.8,0.2)M(1,0) = (0.8, 0.2), then (0.7,0.3)(0.7, 0.3), then (0.65,0.35)(0.65, 0.35) — marching monotonically toward (0.6,0.4)(0.6, 0.4) ✓. The eigenvalue 11 holds the destination; the second eigenvalue sets the speed of forgetting.