σμimplied vol · KKmax(S−K,0)udβρᵢⱼfat tailsPnL · drawdownyield · TdSₜ = μSₜ dt + σSₜ dWₜC = S·N(d₁) − K·e⁻ʳᵀ·N(d₂)σₚ² = wᵀΣwd₁ = [ln(S/K) + (r + σ²⁄2)T] ⁄ σ√T∂V/∂t + ½σ²S²·∂²V/∂S² + rS·∂V/∂S − rV = 0Sharpe = (Rₚ − R_f) ⁄ σₚVaR₉₉ = μ − 2.33σρ = Cov(X,Y) ⁄ σₓσᵧE[Rₚ] = Σ wᵢ·E[Rᵢ]φ(x) = e^(−x²/2) ⁄ √2πE[Rᵢ] = R_f + βᵢ(E[Rₘ] − R_f)dXₜ = θ(μ − Xₜ)dt + σ dWₜσ²ₜ = ω + α·ε²ₜ₋₁ + β·σ²ₜ₋₁C − P = S − K·e⁻ʳᵀΔ = ∂V/∂S Γ = ∂²V/∂S²p = (e^(rΔt) − d) ⁄ (u − d)rₜ = ln(Pₜ ⁄ Pₜ₋₁)E[Xₜ₊₁ | ℱₜ] = Xₜf* = (bp − q) ⁄ bz = (x − μ) ⁄ σP(N=k) = λᵏe^(−λ) ⁄ k!Θ = −∂V/∂Tν = S·φ(d₁)·√Tmin wᵀΣw s.t. wᵀμ = μ*∫Σ∂∞λ
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σμimplied vol · KKmax(S−K,0)udβρᵢⱼfat tailsPnL · drawdownyield · TdSₜ = μSₜ dt + σSₜ dWₜC = S·N(d₁) − K·e⁻ʳᵀ·N(d₂)σₚ² = wᵀΣwd₁ = [ln(S/K) + (r + σ²⁄2)T] ⁄ σ√T∂V/∂t + ½σ²S²·∂²V/∂S² + rS·∂V/∂S − rV = 0Sharpe = (Rₚ − R_f) ⁄ σₚVaR₉₉ = μ − 2.33σρ = Cov(X,Y) ⁄ σₓσᵧE[Rₚ] = Σ wᵢ·E[Rᵢ]φ(x) = e^(−x²/2) ⁄ √2πE[Rᵢ] = R_f + βᵢ(E[Rₘ] − R_f)dXₜ = θ(μ − Xₜ)dt + σ dWₜσ²ₜ = ω + α·ε²ₜ₋₁ + β·σ²ₜ₋₁C − P = S − K·e⁻ʳᵀΔ = ∂V/∂S Γ = ∂²V/∂S²p = (e^(rΔt) − d) ⁄ (u − d)rₜ = ln(Pₜ ⁄ Pₜ₋₁)E[Xₜ₊₁ | ℱₜ] = Xₜf* = (bp − q) ⁄ bz = (x − μ) ⁄ σP(N=k) = λᵏe^(−λ) ⁄ k!Θ = −∂V/∂Tν = S·φ(d₁)·√Tmin wᵀΣw s.t. wᵀμ = μ*∫Σ∂∞λ
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Matrix Decompositions skill›Module 1 · Eigenvalues and spectra›Topic 1.1·~55 minIntermediate

Eigenvalues and eigenvectors

Skill positionModule 1 of 5 · Topic 1 of 3
1.1 Eigenvalues and eigenvectors
—Lab score— / 100Total XP—Streak—

Directions a matrix leaves alone

A square matrix AAA moves most vectors both in length and in direction. A few special directions are only stretched: Av=λvA\mathbf v=\lambda\mathbf vAv=λv with v≠0\mathbf v\neq\mathbf 0v=0. Then v\mathbf vv is an eigenvector and the stretch factor λ\lambdaλ is its eigenvalue. If λ=2\lambda=2λ=2 the vector doubles, if λ=−1\lambda=-1λ=−1 it flips, if λ=0\lambda=0λ=0 it collapses to zero. Everything a matrix does to its eigenvectors is therefore one number.

Av=λv  ⟺  det⁡(A−λI)=0,∑iλi=tr⁡A,∏iλi=det⁡AA\mathbf v=\lambda\mathbf v\iff\det(A-\lambda I)=0,\qquad \sum_i\lambda_i=\operatorname{tr}A,\qquad \prod_i\lambda_i=\det AAv=λv⟺det(A−λI)=0,i∑​λi​=trA,i∏​λi​=detA(1)

To find the eigenvalues, note that (A−λI)v=0(A-\lambda I)\mathbf v=\mathbf 0(A−λI)v=0 has a non-zero solution only when A−λIA-\lambda IA−λI is singular, so det⁡(A−λI)=0\det(A-\lambda I)=0det(A−λI)=0. For a 2×22\times22×2 matrix this is λ2−(tr⁡A)λ+det⁡A=0\lambda^2-(\operatorname{tr}A)\lambda+\det A=0λ2−(trA)λ+detA=0. Then for each root solve (A−λI)v=0(A-\lambda I)\mathbf v=\mathbf 0(A−λI)v=0 for the direction. Two facts give quick checks: the eigenvalues add up to the trace and multiply to the determinant. A matrix is singular exactly when one eigenvalue is zero, and AkA^kAk has eigenvalues λk\lambda^kλk with the same eigenvectors.

A worked 2 by 2 exampleTake A=(4123)A=\begin{pmatrix}4&1\\2&3\end{pmatrix}A=(42​13​). Its trace is 7 and its determinant is 12−2=1012-2=1012−2=10, so λ2−7λ+10=0\lambda^2-7\lambda+10=0λ2−7λ+10=0 and (λ−5)(λ−2)=0(\lambda-5)(\lambda-2)=0(λ−5)(λ−2)=0: the eigenvalues are 5 and 2. For λ=5\lambda=5λ=5 solve (−112−2)v=0\begin{pmatrix}-1&1\\2&-2\end{pmatrix}\mathbf v=\mathbf 0(−12​1−2​)v=0, giving v=(1,1)\mathbf v=(1,1)v=(1,1). For λ=2\lambda=2λ=2 solve (2121)v=0\begin{pmatrix}2&1\\2&1\end{pmatrix}\mathbf v=\mathbf 0(22​11​)v=0, giving v=(1,−2)\mathbf v=(1,-2)v=(1,−2). Check: A(1,1)=(5,5)=5 (1,1)A(1,1)=(5,5)=5\,(1,1)A(1,1)=(5,5)=5(1,1). The eigenvalues of A2A^2A2 are 252525 and 444.
  • Eigenpair: Av=λvA\mathbf v=\lambda\mathbf vAv=λv with v≠0\mathbf v\neq\mathbf 0v=0; an eigenvector's direction is unchanged, its length is scaled by λ\lambdaλ.
  • Characteristic equation: det⁡(A−λI)=0\det(A-\lambda I)=0det(A−λI)=0; for 2×22\times22×2, λ2−tr⁡A λ+det⁡A=0\lambda^2-\operatorname{tr}A\,\lambda+\det A=0λ2−trAλ+detA=0.
  • Checks: eigenvalues sum to the trace and multiply to the determinant; AkA^kAk has eigenvalues λk\lambda^kλk.
Why a quant caresThe eigenvectors of a covariance matrix are the independent directions of risk in a portfolio of assets, and the eigenvalues are the variances along them. A large eigenvalue is a bet that many assets move together, a tiny one is a combination of assets that is almost riskless on paper. The rest of this skill builds on that picture: how to find the eigenpairs of a big matrix, why they are real for covariance matrices, and what goes wrong when they are estimated from too little data.

Lab · what's graded

  • What an eigenpair is15 pts
  • The larger eigenvalue20 pts
  • Determinant from eigenvalues10 pts
  • The characteristic equation15 pts
  • Squaring a matrix15 pts
  • The method in order10 pts
  • Why finance cares15 pts
0/ 7 answered
Tasks
1Choice15 ptsNot answered

Which statement defines an eigenvector v\mathbf vv of a square matrix AAA with eigenvalue λ\lambdaλ?

2Numeric answer20 ptsNot answered

Find the larger eigenvalue of A=(4123)A=\begin{pmatrix}4&1\\2&3\end{pmatrix}A=(42​13​).

lambda=lambda =lambda=accepted: ± 0.01
3Numeric answer10 ptsNot answered

A 3×33\times33×3 matrix has eigenvalues 2, 3 and 5. What is its determinant?

detA=det A =detA=accepted: ± 0.01
4Fill the blanks15 ptsNot answered

For A=(4123)A=\begin{pmatrix}4&1\\2&3\end{pmatrix}A=(42​13​) fill in the characteristic equation.

λ2− λ+=0\displaystyle \lambda^2-\htmlData{blank=a}{\phantom{\rule[-0.35em]{2.4em}{1.35em}}}\,\lambda+\htmlData{blank=b}{\phantom{\rule[-0.35em]{2.4em}{1.35em}}}=0λ2−λ+=0
5Matrix · vector15 ptsNot answered

Compute A2A^2A2 for A=(4123)A=\begin{pmatrix}4&1\\2&3\end{pmatrix}A=(42​13​). Its eigenvalues are the squares 25 and 4.

A2=A^2 =A2=
6Order the steps10 ptsNot answered

Put the steps for finding eigenvalues and eigenvectors in order.

Drag to reorder, or focus an item, press Space to lift it, the arrow keys to move it and Space to drop it.

  1. 1For each eigenvalue solve (A−λI)v=0(A-\lambda I)\mathbf v=\mathbf 0(A−λI)v=0
  2. 2Form A−λIA-\lambda IA−λI and set det⁡(A−λI)=0\det(A-\lambda I)=0det(A−λI)=0
  3. 3Normalise the vectors if unit length is needed
  4. 4Solve the polynomial in λ\lambdaλ for its roots
7Explain in words15 ptsNot answered

In two sentences, say what the eigenvectors and eigenvalues of a covariance matrix of asset returns tell you about risk.

Math modeKeyboard or LaTeX inputTask 1 of 7 · Ctrl ⏎ to check

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