Understanding Eigenvectors of a Matrix

Let $A$ be an $n \times n$ matrix over the field of real or complex numbers. For many matrices, there exist nonzero vectors $X$ and scalars $\lambda$ such that $AX = \lambda X$. The vectors $X$ satisfying this equation are called the eigenvectors of $A$, and the corresponding scalars $\lambda$ are called eigenvalues.

Mathematical Formulation of the Eigenvector Equation for a Matrix

Given an $n \times n$ matrix $A$, a nonzero vector $X \in \mathbb{F}^n$ (where $\mathbb{F}$ is $\mathbb{R}$ or $\mathbb{C}$) is called an eigenvector of $A$ with corresponding eigenvalue $\lambda \in \mathbb{F}$ if and only if

$AX = \lambda X$

where $X \neq 0$.

The eigenvector equation can be rewritten in the following form by bringing all terms to one side:

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

where $I$ is the $n \times n$ identity matrix of the same order as $A$.

Explicit Condition for Existence of Nontrivial Eigenvectors

The linear equation $(A - \lambda I)X = 0$ has a nonzero solution if and only if the matrix $A - \lambda I$ is singular, which is mathematically equivalent to demanding

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

The above equation is called the characteristic equation of $A$. The roots $\lambda_1, \lambda_2, \ldots, \lambda_n$ (counted with algebraic multiplicity) are the eigenvalues of $A$. For each eigenvalue $\lambda$, the associated eigenvectors are the nonzero solutions to $(A - \lambda I)X = 0$.

For detailed information about solving the characteristic equation, refer to How to Determine Eigenvalues.

Explicit Description of the Eigenvector Solution Space

For a given eigenvalue $\lambda$ of $A$, the set of all solutions of $(A - \lambda I)X = 0$ forms a vector space called the eigenspace of $A$ corresponding to $\lambda$. This eigenspace is a subspace of $\mathbb{F}^n$ and contains all eigenvectors corresponding to $\lambda$, along with the zero vector (which is not regarded as an eigenvector in most definitions).

For each eigenvalue, the dimension of its eigenspace is called its geometric multiplicity, and it satisfies $1 \leq \text{geometric multiplicity} \leq \text{algebraic multiplicity}$.

Procedure to Find Eigenvectors of a Matrix for a Given Eigenvalue

To determine the eigenvectors of a matrix $A$ associated with a particular eigenvalue $\lambda$:

1. Compute the matrix $A - \lambda I$. 2. Form the homogeneous system $(A - \lambda I)X = 0$. 3. Solve for all nonzero vectors $X$ that satisfy the system. This involves expressing the system as a set of linear equations and reducing it to find the general solution. 4. Every nonzero scalar multiple $kX$, where $k \in \mathbb{F} \setminus \{0\}$, is also an eigenvector associated with $\lambda$.

For foundational concepts in matrices and determinants, consult Properties of Determinants and Matrices Overview.

Detailed Stepwise Computation: Eigenvectors of a $2 \times 2$ Matrix

Let $A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$ be a $2 \times 2$ matrix. Its characteristic equation is obtained by solving:

$\det(A - \lambda I) = \det\begin{bmatrix} a - \lambda & b \\ c & d - \lambda \end{bmatrix} = (a - \lambda)(d - \lambda) - bc = 0$

Let $\lambda_1$ be one root of this equation. The associated eigenvectors $X = \begin{bmatrix} x \\ y \end{bmatrix} \neq 0$ satisfy

$\begin{bmatrix} a - \lambda_1 & b \\ c & d - \lambda_1 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix}$

This yields two homogeneous equations in $x$ and $y$:

$(a - \lambda_1)x + b y = 0$

$c x + (d - \lambda_1)y = 0$

Since the two equations are linearly dependent for eigenvalues (as the determinant vanishes), a nontrivial solution exists. Express $y$ in terms of $x$ (or vice versa):

$y = -\dfrac{a - \lambda_1}{b}x$   (assuming $b \neq 0$)

Thus, all eigenvectors are of the form $k\begin{bmatrix} 1 \\ -\frac{a-\lambda_1}{b} \end{bmatrix}$ for $k \neq 0$.

Worked Example: Calculation of Eigenvectors for a Specific Matrix

Given: $A = \begin{bmatrix} 1 & 4 \\ -4 & -7 \end{bmatrix}$

Step 1: Characteristic Equation

$\det(A - \lambda I) = \left| \begin{matrix} 1-\lambda & 4 \\ -4 & -7-\lambda \end{matrix} \right| = (1 - \lambda)(-7 - \lambda) - (4)(-4)$

$= (1 - \lambda)(-7 - \lambda) + 16$

$= [-(7 + \lambda) + \lambda(7+\lambda)] + 16$

$= [-(7 + \lambda) + (1-\lambda)(-7 - \lambda)] + 16$ (To maintain all steps, expand using distributive law)

$(1-\lambda)(-7-\lambda) = 1 \cdot (-7) + 1 \cdot (-\lambda) - \lambda \cdot (-7) - \lambda \cdot \lambda$

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

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

$= -7 + 6\lambda - \lambda^2$

Adding $16$, we get:

$\det(A-\lambda I) = -7 + 6\lambda - \lambda^2 + 16$

$= 6\lambda - \lambda^2 + 9$

$= -\lambda^2 + 6\lambda + 9$

$= -(\lambda^2 - 6\lambda - 9)$

The sign does not affect roots; equate to zero and solve:

$\lambda^2 + 6\lambda + 9 = 0$

$(\lambda + 3)^2 = 0$

Result: The eigenvalue is $\lambda = -3$ (algebraic multiplicity $2$).

Step 2: Find Eigenvectors

Substitute $\lambda = -3$ in $A - \lambda I$:

$A - (-3)I = A + 3I = \begin{bmatrix} 1+3 & 4 \\ -4 & -7+3 \end{bmatrix} = \begin{bmatrix} 4 & 4 \\ -4 & -4 \end{bmatrix}$

Let $X = \begin{bmatrix} x \\ y \end{bmatrix}$, then

$\begin{bmatrix} 4 & 4 \\ -4 & -4 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix}$

This gives two equations:

$4x + 4y = 0$

$-4x - 4y = 0$

Both equations are dependent; use the first:

$x + y = 0$

Set $x = k$, then $y = -k$ for arbitrary nonzero $k \in \mathbb{R}$.

Final result: All eigenvectors for $\lambda = -3$ are $k\begin{bmatrix} 1 \\ -1 \end{bmatrix}$ with $k \neq 0$.

For additional worked examples and practice, see Eigenvectors of a Matrix.

Properties of Eigenvectors and Further Concepts

Eigenvectors corresponding to distinct eigenvalues of a real symmetric matrix are linearly independent and orthogonal. For arbitrary matrices, eigenvectors associated with different eigenvalues are always linearly independent, but not necessarily orthogonal. 

For each eigenvalue, every nonzero scalar multiple of an eigenvector is also an eigenvector for the same eigenvalue.

For $n \times n$ matrices, if there are $n$ linearly independent eigenvectors, $A$ is said to be diagonalizable. The collection of all eigenvectors for $A$ (including all eigenspaces) is always linearly independent if all eigenvalues are distinct.

Computation of eigenvectors is crucial in the diagonalization of matrices, the determination of matrix powers, and in applications including vibration analysis and quantum mechanics.