Lecture 007

Eigenvalue and Eigenvector

Eigenvalue and (non-trivial) Eigenvector: \lambda \in \mathbb{R} is a eigenvalue of the matrix A^{n \times n} if

(\exists v \in \mathbb{R}^n)(v \neq 0 \land Av = \lambda v)

where v is a eigenvector

\begin{align*} Av &= \lambda v\\ Av - \lambda v &= 0\\ Av - \lambda I v &= 0\\ (A - \lambda I) v &= 0\\ \end{align*}
\begin{align*} &\lambda \text{ is eigenvalue of } A\\ \iff &(A - \lambda I)x = 0 \text{ has non-trivial solution} \\ \iff &(A - \lambda I) \text{ is singular}\\ \iff &\det(A - \lambda I) = 0\\ \end{align*}

Eigenspace (E_\lambda correspond to A): nullspace of (A - \lambda I), all the eigenvector correspond to A, \lambda

Eigenfunction: eigenfunction of a linear operator (linear transform) D is any non-zero function f such that for scalar eigenvalue \lambda: Here we treat eigenfunction f as a vector.

Df = \lambda f


// TODO: understand below Properties of Eigenvector and Eigenvalue

  1. 0 is an eigenvalue of A iff A is singular (determinant zero, squashing space)
  2. If \lambda_1, \lambda_2, ..., \lambda_n are different eigenvalues of A \in \mathbb{R}^{n \times n} and v_1, v_2, ..., v_n are non-trivial eigenvectors corresponding to them, then v_1, v_2, ..., v_n are linearly independent // TODO: proof
  3. By above, A^{n \times n} can have at most n different eigenvalues (by algebratic multiplicities)
  4. If \lambda is an eigenvalue of A, then \lambda^k is an eigenvalue of A^k (by induction)

Trace: trace of A^{n \times n} is the sum of its entries on the diagonal (\text{tr}(A) = \sum_i a_{ii} given A = (a_{ij})_{1 \leq i, j, \leq n})

  1. Assume that together with multiplicities A^{n \times n} has n eigenvalues, then \text{tr}(A) = \lambda_1 + \lambda_2 + ... + \lambda_n and \det A = \lambda_1 \cdot \lambda_2 \cdot ... \cdot \lambda_n // TODO: 3b1b proof

  2. A \simeq B \implies \text{tr}(A) = \text{tr}(B) (but not vise-versa)


Similar: A^{n \times n} \simeq B^{n \times n} \iff (\exists S^{n \times n})(S^{-1}AS = B)

Theorem: A^{n \times n} \simeq B^{n \times n} \implies

  1. \det A = \det B
  2. A \text{ is invertible} \iff B \text{ is invertible}
  3. \text{rank}(A) = \text{rank}B
  4. A, B have the same characteristic polynomial
  5. A, B have the same eigenvalues
  6. (\forall m \geq 0)(A^m \simeq B^m) // TODO: proof above

Diagonalizable Matrix: A^{n \times n} is diagonalizable if (\exists \text{diagonal matrix} D)(A \simeq D)

Application: given diagonalizable A^{n \times n}, calculate A^k

  1. find \lambda_1, \lambda_2, ..., \lambda_n
  2. find v_1, v_2, ..., v_n
  3. Put v_1, v_2, ..., v_n into columns of S
  4. find S^{-1}
  5. calculate D = S^{-1}AS
  6. then A = SDS^{-1}
  7. then A^k = SD^kS^{-1}

Table of Content