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TopicLinear Algebra used inMSIAI-written · not yet proofread

Determinants and Eigenvalues

Also Eigenvalues and Eigenvectors

How the determinant measures volume change and singularity, and how it leads through the characteristic polynomial to eigenvalues and eigenvectors, the directions a matrix only scales.

Concepts#

  1. Determinantsigned volume factor det⁡(A)\det(A), test for invertibility start here
  2. Eigenvaluethe scaling factors λ\lambda, roots of pA(λ)=det⁡(A−λI)p_A(\lambda) = \det(A - \lambda I)
  3. Eigenvectorthe directions x≠0x \neq 0 with Ax=λxAx = \lambda x, eigenspaces, eigendecomposition

How it fits together#

The chain. Each concept is built on the previous one:

det⁡(A−λI)=0⏟Determinant  ⟺  λ is an eigenvalue⏟root of pA  ⟺  ker⁡(A−λI)≠{0}⏟Eλ holds eigenvectors.\underbrace{\det(A - \lambda I) = 0}_{\text{Determinant}} \iff \underbrace{\lambda \text{ is an eigenvalue}}_{\text{root of } p_A} \iff \underbrace{\ker(A - \lambda I) \neq \{0\}}_{E_\lambda \text{ holds eigenvectors}}.

In practice: compute pA(λ)p_A(\lambda) with the determinant, find its roots (the eigenvalues), then solve the homogeneous system (A−λI)x=0(A - \lambda I)x = 0 for each root (the eigenspace). That is MML Example 4.5.

One picture behind all three. A linear map in R2\mathbb{R}^2 turns the unit square into a parallelogram:

  • ∣det⁡(A)∣|\det(A)| is its area, and sign⁡det⁡(A)\operatorname{sign}\det(A) says whether it is mirrored.
  • Eigenvectors are the directions that stay on their own line, and eigenvalues the stretch factors along them.
  • If the eigenvectors form a basis, the parallelogram spanned by them is mapped to one with sides scaled by λi\lambda_i. So det⁡(A)=∏iλi\det(A) = \prod_i \lambda_i.

Singularity tests. For A∈Rn×nA \in \mathbb{R}^{n \times n} these are all the same statement (Invertible Matrix > Properties, Rank and Invertibility):

TestGeometric reading
det⁡(A)=0\det(A) = 0volume collapses to 00
00 is an eigenvaluesome direction is mapped to 00
ker⁡(A)=E0≠{0}\ker(A) = E_0 \neq \{0\}there is an eigenvector for λ=0\lambda = 0
rank⁡(A)<n\operatorname{rank}(A) < nthe image is a proper subspace

Invariants of a linear map. det⁡\det, tr⁡\operatorname{tr} and the eigenvalues don’t depend on the basis: similar matrices S−1ASS^{-1}AS share them. The eigenvectors do change with the basis, as vectors x↦S−1xx \mapsto S^{-1}x, but they represent the same directions of the map.

How many eigenvectors? The algebraic multiplicity (from pAp_A) is an upper bound for the geometric multiplicity (dim⁡Eλ\dim E_\lambda).

CaseEigenvectorsDecomposition
nn distinct real eigenvaluesbasis (Thm. 4.12)A=PDP−1A = PDP^{-1}
symmetric AAorthonormal basis, real λ\lambda (spectral theorem)A=PDP⊤A = PDP^\top
repeated eigenvalue, geometric < algebraic multiplicitytoo few (defective)not diagonalizable
complex eigenvalues (e.g. rotation)no real eigenvectors for themnot over R\mathbb{R}

Symmetric matrices in MSI. The symmetric case is the one that matters for least squares and covariances. A symmetric matrix is positive (semi)definite iff all its eigenvalues are >0> 0 (≥0\ge 0). By det⁡=∏λi\det = \prod\lambda_i, a positive definite matrix has det⁡>0\det > 0 and is invertible. The converse fails: diag⁡(−1,−1)\operatorname{diag}(-1, -1) has det⁡=1\det = 1.

Key Formulas#

  • det⁡(abcd)=ad−bc\det\begin{pmatrix} a & b \\ c & d \end{pmatrix} = ad - bc (Determinant)
  • det⁡(AB)=det⁡(A)det⁡(B)\det(AB) = \det(A)\det(B), det⁡(A⊤)=det⁡(A)\det(A^\top) = \det(A), det⁡(A−1)=1/det⁡(A)\det(A^{-1}) = 1/\det(A), det⁡(λA)=λndet⁡(A)\det(\lambda A) = \lambda^n\det(A) (Determinant)
  • Ax=λxAx = \lambda x, x≠0x \neq 0 (Eigenvalue, Eigenvector)
  • pA(λ)=det⁡(A−λI)p_A(\lambda) = \det(A - \lambda I), eigenvalues = roots of pAp_A (Eigenvalue > Characteristic polynomial)
  • det⁡(A)=∏iλi\det(A) = \prod_i \lambda_i, tr⁡(A)=∑iλi\operatorname{tr}(A) = \sum_i \lambda_i (Eigenvalue > Determinant, trace and invariance)
  • Eλ=ker⁡(A−λI)E_\lambda = \ker(A - \lambda I) (Eigenvector)
  • A=PDP−1A = PDP^{-1} iff the eigenvectors form a basis; A=PDP⊤A = PDP^\top for symmetric AA (Eigenvector > Properties)

Lectures#

  • Rank and Invertibility: rank as the cause of invertibility; det⁡\det and eigenvalues give equivalent tests for square matrices

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