Linear Algebra - Eigenvalues and Eigenvectors

Eigenvalues and Eigenvectors

Definition

For $n \times n$matrix $A$, eigenvector $v eq 0$and eigenvalue $\lambda$satisfy: $Av = \lambda v$

Geometric meaning: $v$scaled by $\lambda$, direction unchanged (or reversed).

Characteristic Polynomial

Characteristic equation: $\det(A - \lambda I) = 0$

Characteristic polynomial: $p(\lambda) = \det(A - \lambda I)$

Eigenvalues are roots of $p(\lambda)$.

Eigenvalue Properties

Sum of eigenvalues: $\text{tr}(A) = \text{trace} = \text{sum of diagonal entries}$

Product of eigenvalues: $\det(A)$

Eigenspaces

Eigenspace $E_\lambda$: Null space of $(A - \lambda I)$

$E_\lambda = \{v : (A - \lambda I)v = 0\}$

Diagonalization

$A$is diagonalizable if $A = PDP^{-1}$where $D$is diagonal (eigenvalues) and columns of $P$are eigenvectors.

Necessary and sufficient: $A$has $n$linearly independent eigenvectors (not all matrices diagonalizable).

Process:

1. Find eigenvalues $\lambda_i$
2. Find linearly independent eigenvectors $v_i$
3. $P$has eigenvectors as columns
4. $D$has eigenvalues on diagonal
5. $A = PDP^{-1}$

Power of matrix: $A^n = PD^nP^{-1}$(easy: just raise eigenvalues to power)

Matrix exponential: $e^A = P e^D P^{-1}$

Similar Matrices

$A$and $B$are similar if $B = P^{-1}AP$for invertible $P$.

Similar matrices have same eigenvalues, trace, determinant.

Spectral Theorem (Real)

If $A$is symmetric ($A = A^T$), then:

• All eigenvalues are real
• $A$is diagonalizable by orthogonal matrix ($A = PDP^T$)
• Eigenvectors corresponding to different eigenvalues are orthogonal