The eigenstructure of 𝐴 is what “unpacks” the matrix exponential and the LTI response.

Definition

  • Eigenvalues/eigenvectors: a pair 𝜆∈ℂ, 𝐯∈ℂ𝑛, 𝐯≠𝟎 such that
    • 𝐴𝐯=𝜆𝐯
  • Equivalently:
    • (𝜆𝐼−𝐴)𝐯=𝟎
  • Hence 𝜆 is such that (𝜆𝐼−𝐴) is singular, and 𝐯∈𝒩︀(𝜆𝐼−𝐴)

Characteristic Polynomial

  • (𝜆𝐼−𝐴) singular ⇔det(𝜆𝐼−𝐴)=0
  • det(𝜆𝐼−𝐴)=𝑝𝐴(𝜆)=𝜆𝑛+𝛼𝑛−1𝜆𝑛−1+…+𝛼1𝜆+𝛼0
    • the “characteristic polynomial” of 𝐴
  • Eigenvalues are the roots of this polynomial:
    • 𝑝𝐴(𝜆)=0(char. eq'n)
    • ⇒𝑛 values satisfy this: 𝜆1,…,𝜆𝑛
  • For each eigenvalue 𝜆𝑘, the corresponding eigenvector 𝐯𝑘 satisfies
    • (𝜆𝑘𝐼−𝐴)𝐯𝑘=𝟎𝑘=1,…,𝑛
  • The char. polynomial also gives the Cayley-Hamilton Theorem

Zero eigenvalues

  • If 𝜆=0 is an eigenvalue of 𝐴 then
    • det(𝜆𝐼−𝐴)=det(𝐴)=0
  • hence 𝐴 is singular and its columns do not define a basis for ℝ𝑛

Eigenvector Matrix and Diagonalization

  • Is there significance also of the eigenvectors of 𝐴 in “unpacking” 𝑒𝐴𝑡? Recall, for each 𝜆𝑘, the eigenvector 𝐯𝑘 satisfies
    • 𝐴𝐯𝑘=𝜆𝑘𝐯𝑘,𝐯𝑘≠𝟎⇒𝐯𝑘∈𝒩︀(𝜆𝑘𝐼−𝐴)
  • Let 𝑉=[𝐯1,𝐯2,…,𝐯𝑛] (eigenvector matrix). Then
    • 𝐴𝑉=[(𝐴𝐯1),(𝐴𝐯2),…,(𝐴𝐯𝑛)]=[(𝜆1𝐯1),(𝜆2𝐯2),…,(𝜆𝑛𝐯𝑛)]=𝑉Λ
    • where Λ=diag{𝜆𝑘}𝑛𝑘=1 (diagonal matrix of eigenvalues)

Diagonalization

  • 𝐴𝑉=𝑉Λ
  • 𝐴=𝑉Λ𝑉−1Λ=𝑉−1𝐴𝑉
  • (assuming 𝑉−1 exists!)
  • Fact: the eigenvectors for distinct eigenvalues are linearly independent
    • ⇒ cols of 𝑉 linearly independent
    • ⇒det𝑉≠0
    • ⇒𝑉−1 exists
  • How this is exploited: Dyadic Expansion; when eigenvalues repeat, see Jordan Normal Form

Left and Right Eigenvectors

  • Consider
    • 𝐴𝑉=𝑉Λ⇒𝑉−1𝐴=Λ𝑉−1
  • Row-partition 𝑉−1 (see Dyadic Expansion):
    • [𝐰𝖳1𝐰𝖳2⋮𝐰𝖳𝑛]𝐴=(𝜆1𝜆2⋱𝜆𝑛)[𝐰𝖳1𝐰𝖳2⋮𝐰𝖳𝑛]
    • [𝐰𝖳1𝐴𝐰𝖳2𝐴⋮𝐰𝖳𝑛𝐴]=[𝜆1𝐰𝖳1𝜆2𝐰𝖳2⋮𝜆𝑛𝐰𝖳𝑛]
  • ⇒𝐰𝖳𝑘𝐴=𝜆𝑘𝐰𝖳𝑘
  • Compare with 𝐴𝐯𝑘=𝜆𝑘𝐯𝑘:
    • 𝐯𝑘: “right” eigenvectors (columns of 𝑉)
    • 𝐰𝖳𝑘: “left” eigenvectors (rows of 𝑉−1)
  • Equivalently:
    • (𝐰𝖳𝑘𝐴)𝖳=(𝜆𝑘𝐰𝖳𝑘)𝖳⇒𝐴𝖳𝐰𝑘=𝜆𝑘𝐰𝑘
    • ⇒𝐰𝑘 are the ordinary (right) eigenvectors of 𝐴𝖳

NOTE

  • 𝐴 and 𝐴𝖳 have the same char. poly, hence the same eigenvalues 𝜆𝑘:
    • det𝐴𝖳=det𝐴, so
    • det(𝑠𝐼−𝐴)𝖳=det(𝑠𝐼−𝐴𝖳)=det(𝑠𝐼−𝐴)

Bi-Orthonormality

  • Now 𝑉−1𝑉=𝐼, or
    • [𝐰𝖳1⋮𝐰𝖳𝑛][𝐯1,…,𝐯𝑛]=𝐼
  • so that
    • [𝐰𝖳1𝐯1𝐰𝖳2𝐯1⋮𝐰𝖳𝑛𝐯1𝐰𝖳1𝐯2………𝐰𝖳1𝐯𝑛𝐰𝖳2𝐯𝑛⋮𝐰𝖳𝑛𝐯𝑛]=[10⋮0010…0……00⋮1]

Bi-orthonormality

  • Hence
  • 𝐰𝖳𝑖𝐯𝑗={1if𝑖=𝑗0otherwise
  • Thus {𝐯𝑘}𝑛𝑘=1 and {𝐰𝑘}𝑛𝑘=1 are “bi-orthonormal” sets