Eigenvector

An eigenvector is a nonzero vector that, when multiplied by a matrix, results in a scaled version of itself. The scaling factor is called the eigenvalue.

Given a square matrix $A$ , a nonzero vector $\mathbf{v}$ is called an eigenvector of $A$ if there exists a scalar $\lambda$ such that $A\mathbf{v} = \lambda\mathbf{v}$ . The scalar $\lambda$ is the eigenvalue associated with that eigenvector. Geometrically, multiplying by $A$ does not change the direction of $\mathbf{v}$ — it only stretches, compresses, or reverses it along the same line.

Key Formula

A\mathbf{v} = \lambda\mathbf{v}

Where:

$A$ = a square matrix
$v$ = the eigenvector (must be nonzero)
$λ$ = the eigenvalue, a scalar that indicates how the eigenvector is scaled

Worked Example

Problem: Verify that

\mathbf{v} = \begin{pmatrix} 1 \\ 2 \end{pmatrix}

is an eigenvector of

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

, and find its eigenvalue.

Step 1: Compute the product

A\mathbf{v}

A\mathbf{v} = \begin{pmatrix} 4 & 1 \\ 2 & 3 \end{pmatrix}\begin{pmatrix} 1 \\ 2 \end{pmatrix} = \begin{pmatrix} 4(1)+1(2) \\ 2(1)+3(2) \end{pmatrix} = \begin{pmatrix} 6 \\ 8 \end{pmatrix}

Step 2: Check whether the result is a scalar multiple of

\mathbf{v}

\begin{pmatrix} 6 \\ 8 \end{pmatrix} = 2\begin{pmatrix} 3 \\ 4 \end{pmatrix}?

Step 3: That doesn't look right — try the scalar directly. Divide each component of

A\mathbf{v}

by the corresponding component of

\mathbf{v}

\frac{6}{1} = 6, \quad \frac{8}{2} = 4

Step 4: The ratios are not equal (6 ≠ 4), so

A\mathbf{v}

is not a scalar multiple of

\mathbf{v}

Answer: The vector

\mathbf{v} = \begin{pmatrix} 1 \\ 2 \end{pmatrix}

is **not** an eigenvector of

A

. For it to be an eigenvector, every component ratio would need to yield the same scalar

\lambda

. You can verify that

\mathbf{v} = \begin{pmatrix} 1 \\ 1 \end{pmatrix}

is an eigenvector with

\lambda = 5

, since

A\begin{pmatrix} 1 \\ 1 \end{pmatrix} = \begin{pmatrix} 5 \\ 5 \end{pmatrix} = 5\begin{pmatrix} 1 \\ 1 \end{pmatrix}

Why It Matters

Eigenvectors reveal the fundamental directions along which a linear transformation acts by pure scaling. They appear throughout science and engineering — in principal component analysis for data science, in vibration modes of mechanical structures, and in quantum mechanics where measurable states are eigenvectors of operators. Understanding eigenvectors is essential for diagonalizing matrices, solving systems of differential equations, and analyzing stability in dynamical systems.

Common Mistakes

Mistake: Claiming the zero vector is an eigenvector.

Correction: By definition, eigenvectors must be nonzero. The equation

A\mathbf{0} = \lambda\mathbf{0}

is satisfied for any

\lambda

and any matrix, so it gives no useful information.

Mistake: Forgetting that any nonzero scalar multiple of an eigenvector is also an eigenvector.

Correction: If

\mathbf{v}

is an eigenvector with eigenvalue

\lambda

, then

c\mathbf{v}

(for any

c \neq 0

) satisfies

A(c\mathbf{v}) = cA\mathbf{v} = c\lambda\mathbf{v} = \lambda(c\mathbf{v})

. Eigenvectors are not unique — they define a direction (or subspace), not a single vector.

Related Terms

Eigenvalue — The scalar paired with each eigenvector
Matrix — Eigenvectors are defined for square matrices
Vector — An eigenvector is a special type of vector
Scalar — The eigenvalue is a scalar multiplier