Properties of Eigenvalues

Eigenvalues possess algebraic and geometric properties that connect them to matrix invariants like trace and determinant. Understanding these relationships provides computational shortcuts and theoretical insights.

TheoremTrace and Determinant

Let $A$ be an $n \times n$ matrix with eigenvalues $\lambda_1, \lambda_2, \ldots, \lambda_n$ (counted with multiplicity). Then:

Sum of eigenvalues: $\sum_{i=1}^n \lambda_i = \text{tr}(A)$ (trace of $A$ )
Product of eigenvalues: $\prod_{i=1}^n \lambda_i = \det(A)$

The trace $\text{tr}(A) = \sum_{i=1}^n a_{ii}$ is the sum of diagonal entries.

These formulas reveal that trace and determinant are eigenvalue invariants—they can be computed from eigenvalues without finding eigenvectors.

ExampleQuick Eigenvalue Check

For $A = \begin{bmatrix} 4 & 2 \\ 1 & 3 \end{bmatrix}$ , we have $\text{tr}(A) = 7$ and $\det(A) = 10$ .

If the eigenvalues are $\lambda_1, \lambda_2$ , then $\lambda_1 + \lambda_2 = 7$ and $\lambda_1\lambda_2 = 10$ .

Solving: $\lambda_1, \lambda_2$ satisfy $\lambda^2 - 7\lambda + 10 = 0$ , giving $\lambda = 5, 2$ .

TheoremProperties Under Matrix Operations

Let $\lambda$ be an eigenvalue of $A$ with eigenvector $\mathbf{v}$ . Then:

$c\lambda$ is an eigenvalue of $cA$ (same eigenvector)
$\lambda^k$ is an eigenvalue of $A^k$ (same eigenvector)
If $A$ is invertible, $\lambda^{-1}$ is an eigenvalue of $A^{-1}$ (same eigenvector)
$\lambda + c$ is an eigenvalue of $A + cI$ (same eigenvector)
If $p(\lambda)$ is a polynomial, $p(\lambda)$ is an eigenvalue of $p(A)$ (same eigenvector)

Proof sketch: If $A\mathbf{v} = \lambda\mathbf{v}$ , then $A^2\mathbf{v} = A(A\mathbf{v}) = A(\lambda\mathbf{v}) = \lambda A\mathbf{v} = \lambda^2\mathbf{v}$ , and so on.

DefinitionAlgebraic vs Geometric Multiplicity

For eigenvalue $\lambda$ of matrix $A$ :

Algebraic multiplicity: The multiplicity of $\lambda$ as a root of the characteristic polynomial $p_A(\lambda)$
Geometric multiplicity: The dimension of the eigenspace $E_\lambda = \dim(\ker(A - \lambda I))$

Always: $1 \leq \text{geometric multiplicity} \leq \text{algebraic multiplicity}$

TheoremLinear Independence of Eigenvectors

Eigenvectors corresponding to distinct eigenvalues are linearly independent.

More generally, if $\mathbf{v}_1, \ldots, \mathbf{v}_k$ are eigenvectors for distinct eigenvalues $\lambda_1, \ldots, \lambda_k$ , then $\{\mathbf{v}_1, \ldots, \mathbf{v}_k\}$ is linearly independent.

Remark

The independence of eigenvectors for distinct eigenvalues is fundamental to diagonalization. If an $n \times n$ matrix has $n$ distinct eigenvalues, it automatically has $n$ linearly independent eigenvectors, guaranteeing diagonalizability. This geometric independence reflects the algebraic fact that different eigenvalues represent fundamentally different "modes" of the transformation.