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