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Eigenvectors &
Eigenspaces

Learn how to find the special directions that a matrix preserves, compute eigenvectors from eigenvalues, and describe eigenspaces as solution sets.

Linear Algebra

What Is an Eigenvector?

An eigenvector of a square matrix \(A\) is a nonzero vector \(\mathbf{v}\) such that multiplying by \(A\) only scales the vector, without changing its direction:

The word "eigen" is German for "own" or "characteristic": these are the directions that the matrix owns.

Visual Intuition

\(\mathbf{v}\) \(A\mathbf{v} = \lambda\mathbf{v}\) \(\mathbf{u}\) \(A\mathbf{u}\) Eigenvector: stays on same line Non-eigenvector: direction changes

Why Do Eigenvectors Matter?

Diagonalisation
If \(A\) has enough eigenvectors, we can write \(A = PDP^{-1}\), making powers of \(A\) trivial to compute.
Differential Equations
Solutions to \(\mathbf{x}' = A\mathbf{x}\) are built from eigenvectors, giving exponential modes \(e^{\lambda t}\mathbf{v}\).
Data Science
PCA projects data onto eigenvectors of the covariance matrix, revealing the directions of greatest variance.

The Core Equation

Starting from \(A\mathbf{v} = \lambda\mathbf{v}\), rearrange:

This is a homogeneous system. For nonzero solutions to exist, the coefficient matrix must be singular:

What Is an Eigenspace?

Once you have an eigenvalue \(\lambda\), the eigenspace \(E_\lambda\) is the set of all vectors satisfying \((A - \lambda I)\mathbf{v} = \mathbf{0}\):

The dimension of \(E_\lambda\) is called the geometric multiplicity of \(\lambda\). It tells you how many linearly independent eigenvectors belong to that eigenvalue.

Key Distinctions

Eigenvalue \(\lambda\) Eigenvector \(\mathbf{v}\) Eigenspace \(E_\lambda\)
What is it? A scalar A nonzero vector A subspace
Found by \(\det(A-\lambda I)=0\) Solving \((A-\lambda I)\mathbf{v}=\mathbf{0}\) \(\text{Nul}(A-\lambda I)\)
Can be zero? Yes, \(\lambda=0\) is valid No, \(\mathbf{v}\neq\mathbf{0}\) Contains \(\mathbf{0}\), but \(\dim \geq 1\)
Unique? Finitely many Not unique (any scalar multiple works) Unique for each \(\lambda\)

How to Compute Eigenvectors

1
Find eigenvalues. Compute \(\det(A - \lambda I) = 0\) and solve for \(\lambda\).
2
Form \(A - \lambda I\). Substitute each eigenvalue into \(A - \lambda I\).
3
Row reduce. Bring \(A - \lambda I\) to row echelon form.
4
Solve the homogeneous system. Express the general solution in parametric vector form.
5
State the eigenspace. The span of the basis vectors is \(E_\lambda\).

Example 1: Finding Eigenvectors

Find the eigenvalues, eigenvectors, and eigenspaces of:

Step 1: Characteristic equation:

Example 1: Eigenspaces

\(\lambda_1 = 5\)
\(A - 5I = \begin{pmatrix}-1 & 1\\2 & -2\end{pmatrix}\)
Row reduce: \(\begin{pmatrix}1 & -1\\0 & 0\end{pmatrix}\)
Free variable: \(x_2 = t\)
\(E_5 = \text{span}\left\{\begin{pmatrix}1\\1\end{pmatrix}\right\}\)
\(\lambda_2 = 2\)
\(A - 2I = \begin{pmatrix}2 & 1\\2 & 1\end{pmatrix}\)
Row reduce: \(\begin{pmatrix}1 & \frac{1}{2}\\0 & 0\end{pmatrix}\)
Free variable: \(x_2 = t\)
\(E_2 = \text{span}\left\{\begin{pmatrix}-1\\2\end{pmatrix}\right\}\)

Example 1: Geometric View

\(\begin{pmatrix}1\\1\end{pmatrix}\) \(\begin{pmatrix}-1\\2\end{pmatrix}\) \(E_5\) \(E_2\) \(x_1\) \(x_2\)

Example 2: Repeated Eigenvalue

Consider:

Characteristic equation: \((3-\lambda)^2 = 0\), so \(\lambda = 3\) with algebraic multiplicity 2.

Example 3: A \(3 \times 3\) Matrix

Eigenvalues: \(\det(C-\lambda I) = (2-\lambda)^2(5-\lambda) = 0\), giving \(\lambda_1=2\) (alg. mult. 2) and \(\lambda_2=5\) (alg. mult. 1).

\(\lambda_1 = 2\)
\(C-2I = \begin{pmatrix}0&1&0\\0&0&0\\0&0&3\end{pmatrix}\)
One free variable: \(x_1\)
\(E_2 = \text{span}\left\{\begin{pmatrix}1\\0\\0\end{pmatrix}\right\}\)
Geometric mult. = 1 < 2 (deficient)
\(\lambda_2 = 5\)
\(C-5I = \begin{pmatrix}-3&1&0\\0&-3&0\\0&0&0\end{pmatrix}\)
One free variable: \(x_3\)
\(E_5 = \text{span}\left\{\begin{pmatrix}0\\0\\1\end{pmatrix}\right\}\)
Geometric mult. = 1

Algebraic vs. Geometric Multiplicity

Algebraic Multiplicity
The number of times \(\lambda\) appears as a root of the characteristic polynomial. Denoted \(\text{AM}(\lambda)\).
Geometric Multiplicity
The dimension of the eigenspace \(E_\lambda\): the number of linearly independent eigenvectors. Denoted \(\text{GM}(\lambda)\).

Common Mistakes

✕ Forgetting \(I\)
Writing \(A - \lambda\) instead of \(A - \lambda I\). You cannot subtract a scalar from a matrix without the identity.
✕ Including \(\mathbf{0}\)
Listing the zero vector as an eigenvector. By definition, eigenvectors must be nonzero.
✕ Stopping at Eigenvalues
Finding \(\lambda\) but not solving \((A-\lambda I)\mathbf{v}=\mathbf{0}\). The eigenvector requires row reduction.

Summary Checklist

Step Action Result
1 Compute \(\det(A-\lambda I)=0\) Eigenvalues \(\lambda_1, \lambda_2, \ldots\)
2 Form \(A - \lambda_i I\) Coefficient matrix
3 Row reduce to echelon form Identify free variables
4 Write parametric solution Eigenvectors (basis for \(E_\lambda\))
5 State \(E_\lambda = \text{span}\{\ldots\}\) Eigenspace as subspace

Question 1 of 5

If \(A\mathbf{v} = 3\mathbf{v}\) and \(\mathbf{v} \neq \mathbf{0}\), which statement is true?

Question 2 of 5

Given \(A = \begin{pmatrix}5 & 0\\0 & 2\end{pmatrix}\), what is the eigenspace \(E_5\)?

Question 3 of 5

The eigenvalues of \(\begin{pmatrix}1&2\\0&3\end{pmatrix}\) are \(\lambda = 1\) and \(\lambda = 3\). What is a basis for \(E_3\)?

Question 4 of 5

If \(\lambda = 4\) has algebraic multiplicity 3 but \(\dim(E_4) = 2\), the matrix is:

Question 5 of 5

Which of the following is always true about an eigenspace \(E_\lambda\)?