Finding Eigenvalues of a Matrix Explained Simply

The calculation of eigenvalues of a matrix is a fundamental operation in linear algebra. Eigenvalues are defined for a square matrix, and their determination relies on the characteristic polynomial associated with that matrix.

Definition of Eigenvalues Using the Characteristic Equation

Let $A$ be an $n \times n$ square matrix over a field (typically $\mathbb{R}$ or $\mathbb{C}$). A scalar $\lambda \in \mathbb{R}$ (or $\mathbb{C}$) is called an eigenvalue of $A$ if there exists a non-zero vector $X$ such that

$A X = \lambda X$

where $X$ is a non-zero $n \times 1$ column vector (the corresponding eigenvector). This equation can be rewritten by bringing all terms to one side:

$A X - \lambda X = 0$

Factoring $X$ out gives

$(A - \lambda I) X = 0$

where $I$ denotes the $n \times n$ identity matrix. This forms a homogeneous system of equations.

Criterion for Non-Trivial Solutions and the Characteristic Polynomial

The existence of a non-trivial solution $X \neq 0$ is possible if and only if the matrix $(A - \lambda I)$ is singular. The necessary and sufficient condition for singularity is

$\det(A - \lambda I) = 0$

The left-hand side, $\det(A - \lambda I)$, is a polynomial in $\lambda$ of degree $n$, known as the characteristic polynomial of $A$. Every root of this polynomial is an eigenvalue of $A$.

Stepwise Procedure to Determine the Eigenvalues of a Matrix

Given an $n \times n$ matrix $A$, the following systematic approach is used:

Step 1: Write the matrix $(A - \lambda I)$ explicitly by subtracting $\lambda$ from each diagonal entry of $A$ while keeping other entries unchanged.

Step 2: Calculate the determinant $\det(A - \lambda I)$. This step expands to a polynomial equation in terms of $\lambda$.

Step 3: Set $\det(A - \lambda I) = 0$ and solve for $\lambda$. The solutions are all the eigenvalues of $A$.

Explicit Computation: Eigenvalues for a $2 \times 2$ Matrix

Let $A$ be a $2\times 2$ matrix given by $A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$. To determine its eigenvalues, proceed as follows:

Step 1: Consider $A - \lambda I = \begin{pmatrix} a - \lambda & b \\ c & d - \lambda \end{pmatrix}$.

Step 2: Compute the determinant:

$\det(A - \lambda I) = (a - \lambda)(d - \lambda) - b c$

Step 3: Set the determinant equal to zero:

$(a - \lambda)(d - \lambda) - b c = 0$

Expanding the left-hand side yields:

$a d - a \lambda - d \lambda + \lambda^2 - b c = 0$

Rearrange terms to obtain:

$\lambda^2 - (a + d)\lambda + (a d - b c) = 0$

Thus, the characteristic equation is a quadratic in $\lambda$. The two eigenvalues are the solutions of this quadratic equation:

$\lambda = \frac{(a + d) \pm \sqrt{(a + d)^2 - 4(a d - b c)}}{2}$

Explicit Computation: Eigenvalues for a $3 \times 3$ Matrix

Let $A = \begin{pmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{pmatrix}$ be a $3 \times 3$ matrix. The characteristic polynomial is constructed as follows:

Step 1: Write $A - \lambda I$:

$A - \lambda I = \begin{pmatrix} a_{11} - \lambda & a_{12} & a_{13} \\ a_{21} & a_{22} - \lambda & a_{23} \\ a_{31} & a_{32} & a_{33} - \lambda \end{pmatrix}$

Step 2: Compute $\det(A - \lambda I)$, expanding along any row or column, with each minor explicitly written. The computation yields a cubic polynomial in $\lambda$:

$\det(A - \lambda I) = (a_{11} - \lambda)\left[(a_{22} - \lambda)(a_{33} - \lambda) - a_{23} a_{32}\right]$

$ - a_{12}\left[a_{21}(a_{33} - \lambda) - a_{23} a_{31}\right]$

$ + a_{13}\left[a_{21} a_{32} - (a_{22} - \lambda)a_{31}\right]$

Setting this cubic expression equal to zero:

$\det(A - \lambda I) = 0$

The roots of this cubic polynomial are the eigenvalues of $A$.

Illustrative Example: Finding Eigenvalues of a Specific $2 \times 2$ Matrix

Given: $A = \begin{pmatrix} 4 & 2 \\ 1 & 3 \end{pmatrix}$

Substitution: $A - \lambda I = \begin{pmatrix} 4 - \lambda & 2 \\ 1 & 3 - \lambda \end{pmatrix}$

Simplification:

$\det(A - \lambda I) = (4 - \lambda)(3 - \lambda) - 2 \cdot 1$

$= (4 \times 3) - 4\lambda - 3\lambda + \lambda^2 - 2$

$= 12 - 4\lambda - 3\lambda + \lambda^2 - 2$

$= \lambda^2 - 7\lambda + 10$

Final result: Solve $\lambda^2 - 7\lambda + 10 = 0$

$\lambda = \frac{7 \pm \sqrt{49 - 40}}{2}$

$\lambda = \frac{7 \pm 3}{2}$

$\lambda_1 = \frac{7 + 3}{2} = 5$, $\lambda_2 = \frac{7 - 3}{2} = 2$

Role of Characteristic Polynomial Degree and Algebraic Multiplicity

For an $n \times n$ matrix $A$, the characteristic polynomial is always of degree $n$. The total number of (complex) eigenvalues, counting multiplicity, is always $n$. The algebraic multiplicity of an eigenvalue is its number of occurrences as a root of the characteristic polynomial.

For further procedures concerning eigenvectors, readers can refer to Eigenvectors Of A Matrix, which details their explicit calculation for any given eigenvalue.

Summary: Essential Steps in Calculating Eigenvalues

Calculating the eigenvalues of a square matrix involves constructing the characteristic equation, $\det(A - \lambda I) = 0$, and solving it for $\lambda$. For each solution $\lambda$, a corresponding eigenvector satisfies $(A - \lambda I) X = 0$. These steps are foundational in advanced study, forms the basis for matrix diagonalization, and underpin concepts in Matrices And Determinants.