9.3 Eigenvalues, Eigenvectors, and Diagonalization

9.3 Eigenvalues, Eigenvectors, and Diagonalization

Spectral theory investigates how linear operators act invariantly on specific one-dimensional subspaces. On the GRE Mathematics Subject Test, questions test eigenvalue calculations via trace and determinant tricks, algebraic versus geometric multiplicity inequalities, diagonalizability criteria, matrix powers, and the Cayley-Hamilton Theorem.

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Eigenvalue Equation and the Characteristic Polynomial

Let $A \in M_{n \times n}(\mathbb{F})$ where $\mathbb{F}$ is $\mathbb{R}$ or $\mathbb{C}$. A scalar $\lambda \in \mathbb{F}$ is an eigenvalue of $A$ if there exists a non-zero vector $\mathbf{v} \in \mathbb{F}^n$ such that: $A\mathbf{v} = \lambda \mathbf{v} \iff (A - \lambda I_n)\mathbf{v} = \mathbf{0}.$ Because non-zero solutions exist if and only if the operator $(A - \lambda I_n)$ has a non-trivial null space, eigenvalues are precisely the roots of the characteristic equation: $p_A(\lambda) = \det(\lambda I_n - A) = 0.$

Characteristic Polynomial Expansion

The characteristic polynomial of an $n \times n$ matrix is a monic polynomial of degree $n$: $p_A(\lambda) = \lambda^n - \operatorname{tr}(A) \lambda^{n-1} + \dots + (-1)^n \det(A).$ From Vieta's formulas, the roots $\lambda_1, \dots, \lambda_n$ (counted with multiplicity) satisfy:

• Trace Formula: $\operatorname{tr}(A) = \sum_{i=1}^n a_{ii} = \sum_{i=1}^n \lambda_i$.
• Determinant Formula: $\det(A) = \prod_{i=1}^n \lambda_i$.

These two scalar identities allow candidates to find unknown eigenvalues without expanding determinants.

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Eigenspaces, Algebraic Multiplicity, and Geometric Multiplicity

For each eigenvalue $\lambda$:

• The eigenspace $E_\lambda = \operatorname{null}(A - \lambda I_n)$ is the subspace of all eigenvectors corresponding to $\lambda$, together with the zero vector.
• The algebraic multiplicity, $m_{\text{alg}}(\lambda)$, is the multiplicity of $\lambda$ as a root of $p_A(\lambda)$.
• The geometric multiplicity, $m_{\text{geo}}(\lambda)$, is the dimension of the eigenspace: $m_{\text{geo}}(\lambda) = \operatorname{dim}(E_\lambda) = n - \operatorname{rank}(A - \lambda I_n).$

The Fundamental Multiplicity Inequality

For every eigenvalue $\lambda$ of any square matrix $A$: $1 \le m_{\text{geo}}(\lambda) \le m_{\text{alg}}(\lambda).$ An eigenvalue is termed defective if $m_{\text{geo}}(\lambda) < m_{\text{alg}}(\lambda)$.

Linear Independence of Eigenvectors

Eigenvectors corresponding to mutually distinct eigenvalues are linearly independent. If $A$ has $n$ distinct eigenvalues, the corresponding eigenvectors form a complete basis for $\mathbb{F}^n$.

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Criterion for Matrix Diagonalizability

A matrix $A \in M_{n \times n}(\mathbb{F})$ is diagonalizable over $\mathbb{F}$ if there exists an invertible matrix $P$ and a diagonal matrix $D$ such that: $A = P D P^{-1}.$ The columns of $P = [\mathbf{v}_1 \dots \mathbf{v}_n]$ are linearly independent eigenvectors of $A$, and the diagonal entries of $D = \operatorname{diag}(\lambda_1, \dots, \lambda_n)$ are the corresponding eigenvalues.

Necessary and Sufficient Conditions for Diagonalizability

A matrix $A$ is diagonalizable over $\mathbb{F}$ if and only if:

1. The characteristic polynomial $p_A(\lambda)$ splits completely into linear factors over $\mathbb{F}$.
2. For every distinct eigenvalue $\lambda_i$, the geometric multiplicity equals the algebraic multiplicity: $m_{\text{geo}}(\lambda_i) = m_{\text{alg}}(\lambda_i).$ Equivalently, the sum of geometric multiplicities equals the matrix dimension: $\sum m_{\text{geo}}(\lambda_i) = n$.

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Matrix Powers and Matrix Exponentials

Diagonalization drastically simplifies computing matrix powers and functions: $A^k = (P D P^{-1})^k = P D^k P^{-1} = P \begin{pmatrix} \lambda_1^k & & 0 \\ & \ddots & \\ 0 & & \lambda_n^k \end{pmatrix} P^{-1}.$ Similarly, the matrix exponential $e^{At}$ is given by: $e^{At} = P e^{Dt} P^{-1} = P \operatorname{diag}(e^{\lambda_1 t}, \dots, e^{\lambda_n t}) P^{-1}.$

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The Cayley-Hamilton Theorem and Minimal Polynomials

The Cayley-Hamilton Theorem

Every square matrix satisfies its own characteristic equation: $p_A(A) = O_n.$ If $p_A(\lambda) = \lambda^n + c_{n-1}\lambda^{n-1} + \dots + c_1 \lambda + c_0$, then: $A^n + c_{n-1} A^{n-1} + \dots + c_1 A + c_0 I_n = O_n.$

Application to Matrix Inverses and Powers

When $\det(A) = (-1)^n c_0 \neq 0$, multiplying by $A^{-1}$ yields: $A^{-1} = -\frac{1}{c_0} (A^{n-1} + c_{n-1}A^{n-2} + \dots + c_1 I_n).$ Furthermore, any high power $A^m$ ($m \ge n$) can be reduced to a polynomial in $A$ of degree at most $n - 1$ by polynomial long division: $x^m = q(x) p_A(x) + r(x) \implies A^m = r(A)$.

The Minimal Polynomial

The minimal polynomial $m_A(\lambda)$ is the unique monic polynomial of lowest degree annihilating $A$ ($m_A(A) = O_n$).

• $m_A(\lambda)$ divides $p_A(\lambda)$.
• $m_A(\lambda)$ and $p_A(\lambda)$ have identical roots (ignoring multiplicities).
• Diagonalizability Criterion: $A$ is diagonalizable over $\mathbb{F}$ if and only if $m_A(\lambda)$ factors into distinct linear factors over $\mathbb{F}$ (contains no repeated roots).

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Step-by-Step Worked Problems

Problem 1: Matrix Powers via Diagonalization

Compute $A^8$ for $A = \begin{pmatrix} 1 & 4 \ 2 & 3 \end{pmatrix}$.

Solution:

1. Find eigenvalues of $A$: $\operatorname{tr}(A) = 1 + 3 = 4, \quad \det(A) = 1(3) - 4(2) = 3 - 8 = -5.$ Roots of $\lambda^2 - 4\lambda - 5 = 0 \implies (\lambda - 5)(\lambda + 1) = 0$. Thus $\lambda_1 = 5, \lambda_2 = -1$.
2. Find eigenvectors:
   • For $\lambda_1 = 5$: $A - 5I = \begin{pmatrix} -4 & 4 \ 2 & -2 \end{pmatrix} \implies -x_1 + x_2 = 0 \implies \mathbf{v}_1 = \begin{pmatrix} 1 \ 1 \end{pmatrix}$.
   • For $\lambda_2 = -1$: $A - (-1)I = \begin{pmatrix} 2 & 4 \ 2 & 4 \end{pmatrix} \implies x_1 + 2x_2 = 0 \implies \mathbf{v}_2 = \begin{pmatrix} -2 \ 1 \end{pmatrix}$.
3. Construct $P$ and $P^{-1}$: $P = \begin{pmatrix} 1 & -2 \\ 1 & 1 \end{pmatrix}, \quad \det(P) = 1(1) - (-2)(1) = 3 \implies P^{-1} = \frac{1}{3} \begin{pmatrix} 1 & 2 \\ -1 & 1 \end{pmatrix}.$
4. Compute $A^8 = P D^8 P^{-1}$ with $D^8 = \operatorname{diag}(5^8, (-1)^8) = \operatorname{diag}(5^8, 1)$: $A^8 = \begin{pmatrix} 1 & -2 \\ 1 & 1 \end{pmatrix} \begin{pmatrix} 5^8 & 0 \\ 0 & 1 \end{pmatrix} \frac{1}{3} \begin{pmatrix} 1 & 2 \\ -1 & 1 \end{pmatrix} = \frac{1}{3} \begin{pmatrix} 5^8 + 2 & 2 \cdot 5^8 - 2 \\ 5^8 - 1 & 2 \cdot 5^8 + 1 \end{pmatrix}.$

Problem 2: Minimal Polynomial and Matrix Exponents via Cayley-Hamilton

Let $A$ be a $2 \times 2$ matrix satisfying $\operatorname{tr}(A) = 3$ and $\det(A) = 2$. Express $A^4$ in the form $c_1 A + c_0 I$.

Solution:

1. The characteristic polynomial is $p_A(\lambda) = \lambda^2 - 3\lambda + 2 = (\lambda - 1)(\lambda - 2)$.
2. By the Cayley-Hamilton Theorem, $A^2 - 3A + 2I = 0$, so $A^2 = 3A - 2I$.
3. Divide $\lambda^4$ by $(\lambda - 1)(\lambda - 2)$: $\lambda^4 = q(\lambda)(\lambda^2 - 3\lambda + 2) + (c_1 \lambda + c_0).$
4. Evaluate at the roots $\lambda = 1$ and $\lambda = 2$:
   • $\lambda = 1 \implies 1^4 = c_1(1) + c_0 \implies c_1 + c_0 = 1$.
   • $\lambda = 2 \implies 2^4 = c_1(2) + c_0 \implies 2c_1 + c_0 = 16$.
5. Subtract the equations: $c_1 = 15$, and $c_0 = 1 - 15 = -14$.
6. Therefore, $A^4 = 15A - 14I$.

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GRE Exam Traps & Pitfalls

Trap 1: Confusing Multiplicity Types An algebraic multiplicity $m_{\text{alg}}(\lambda) > 1$ does not prevent diagonalizability! The identity matrix $I_n$ has $\lambda = 1$ with $m_{\text{alg}} = n$, yet it is already diagonal because $m_{\text{geo}}(1) = n$.

Trap 2: Ignoring the Underlying Field The rotation matrix $R_{\pi/2} = \begin{pmatrix} 0 & -1 \ 1 & 0 \end{pmatrix}$ has eigenvalues $\pm i$. It is diagonalizable over $\mathbb{C}$, but non-diagonalizable over $\mathbb{R}$ because its characteristic polynomial $\lambda^2 + 1$ does not split over $\mathbb{R}$.

Trap 3: Sign Errors in Cayley-Hamilton for 3x3 Matrices For $3 \times 3$ matrices, the characteristic polynomial is $\lambda^3 - \operatorname{tr}(A)\lambda^2 + c_1 \lambda - \det(A) = 0$. Forgetting the negative sign on $\det(A)$ disrupts polynomial division and inverse calculations.