Eigenvalues and Eigenvectors
Before An eigenvector is a direction the rule x goes to Ax only stretches, and its eigenvalue makes A - λI not invertible, which the determinant detects. Builds on: Determinants · Matrix Operations and Invertibility · The Invertible Matrix Theorem · Linear Transformations and Their Matrices
Next Eigenvectors lead to diagonalization, which turns powers of a matrix into simple arithmetic. Leads to: Diagonalization
Before You Start: Prerequisite Check
Can you do these? (Click to reveal self-test)
Test yourself on these prerequisite skills:
Determinant: Compute $\det\begin{pmatrix}a-\lambda&b\\c&d-\lambda\end{pmatrix}$.
Check
$(a-\lambda)(d-\lambda) - bc = \lambda^2 - (a+d)\lambda + (ad - bc)$.
Null space: How do you find the null space of a matrix?
Check
Row-reduce $[A \mid \mathbf{0}]$ to RREF; write the parametric solution; the direction vectors form a basis for $\text{Nul}(A)$.
Invertibility: How is the determinant connected to invertibility?
Check
$A$ is invertible iff $\det(A) \neq 0$. Equivalently, $A$ is singular (not invertible) iff $\det(A) = 0$.
If you struggled: Review Determinants and The Invertible Matrix Theorem.
Try This First
Let $A = \begin{pmatrix}2&1\\0&3\end{pmatrix}$.
For each vector, compute $A\mathbf{v}$ and check whether the result is a scalar multiple of $\mathbf{v}$. If so, what is the scalar?
(a) $\mathbf{v}_1 = \begin{pmatrix}1\\1\end{pmatrix}$
(b) $\mathbf{v}_2 = \begin{pmatrix}1\\0\end{pmatrix}$
(c) $\mathbf{v}_3 = \begin{pmatrix}1\\2\end{pmatrix}$
Show Solution
(a) $A\mathbf{v}_1 = \begin{pmatrix}3\\3\end{pmatrix} = 3\begin{pmatrix}1\\1\end{pmatrix}$. Yes: scalar $= 3$. $\mathbf{v}_1$ is an eigenvector with eigenvalue $\lambda = 3$.
(b) $A\mathbf{v}_2 = \begin{pmatrix}2\\0\end{pmatrix} = 2\begin{pmatrix}1\\0\end{pmatrix}$. Yes: scalar $= 2$. $\mathbf{v}_2$ is an eigenvector with eigenvalue $\lambda = 2$.
(c) $A\mathbf{v}_3 = \begin{pmatrix}4\\6\end{pmatrix}$. Is this a multiple of $\begin{pmatrix}1\\2\end{pmatrix}$? If so, both entries must have the same scaling: $4/1 = 4$ but $6/2 = 3$. The scalings differ. $\mathbf{v}_3$ is NOT an eigenvector.
The definition: a nonzero vector $\mathbf{v}$ is an eigenvector of $A$ if $A\mathbf{v} = \lambda\mathbf{v}$ for some scalar $\lambda$. That scalar is the corresponding eigenvalue. The matrix “stretches” or “shrinks” $\mathbf{v}$ without rotating it.
The Geometric Picture
Multiplying by a matrix usually rotates a vector and changes its length. Eigenvectors are the special directions that the matrix leaves unchanged in orientation: $A\mathbf{v}$ points in the same direction as $\mathbf{v}$ (or the opposite direction if $\lambda < 0$). The eigenvalue tells you how much the length is scaled.
Eigenvalues and eigenvectors matter because they reveal the “natural axes” of a transformation. In those directions, the transformation is pure scaling.
Prerequisite Map
Quick Reference
| Property | Value |
|---|---|
| Course | MATH 231 |
| Difficulty | Advanced |
| Time | ~30 minutes |
| Source | Lay, Linear Algebra and Its Applications, Sections 5.1 and 5.2 |
Algorithm Summary
| Step | Action |
|---|---|
| 1 | Form the characteristic polynomial $\det(A - \lambda I)$ |
| 2 | Solve $\det(A - \lambda I) = 0$ to find all eigenvalues $\lambda$ |
| 3 | For each $\lambda$, row-reduce $(A - \lambda I)$ and find $\text{Nul}(A - \lambda I)$ |
| 4 | A basis for $\text{Nul}(A - \lambda I)$ gives the corresponding eigenvectors |
Key Concepts
Definition
A scalar $\lambda$ is an eigenvalue of a square matrix $A$ if there exists a nonzero vector $\mathbf{v}$ such that:
\[ A\mathbf{v} = \lambda\mathbf{v} \]
The vector $\mathbf{v}$ is an eigenvector corresponding to $\lambda$.
The zero vector is never an eigenvector, even though $A\mathbf{0} = 0 = \lambda\mathbf{0}$ for any $\lambda$. The definition explicitly requires $\mathbf{v} \neq \mathbf{0}$.
Rewriting the condition: $A\mathbf{v} = \lambda\mathbf{v}$ is equivalent to $(A - \lambda I)\mathbf{v} = \mathbf{0}$. This means $\mathbf{v}$ is in the null space of $(A - \lambda I)$.
For a nonzero $\mathbf{v}$ to exist in $\text{Nul}(A - \lambda I)$, the matrix $(A - \lambda I)$ must be singular (not invertible). By the IMT, this means $\det(A - \lambda I) = 0$.
Source: Lay, Linear Algebra and Its Applications, Section 5.1: Eigenvectors and Eigenvalues.
The Characteristic Equation
The characteristic polynomial of an $n \times n$ matrix $A$ is: \[ p(\lambda) = \det(A - \lambda I) \]
The characteristic equation is $\det(A - \lambda I) = 0$.
The eigenvalues of $A$ are exactly the roots of the characteristic equation.
For a $2 \times 2$ matrix $A = \begin{pmatrix}a&b\\c&d\end{pmatrix}$: \[ \det(A - \lambda I) = (a-\lambda)(d-\lambda) - bc = \lambda^2 - (a+d)\lambda + (ad-bc) \]
This is a degree-2 polynomial in $\lambda$. For an $n \times n$ matrix, the characteristic polynomial has degree $n$.
Source: Lay, Linear Algebra and Its Applications, Section 5.2: The Characteristic Equation.
Finding Eigenvectors: The Eigenspace
For each eigenvalue $\lambda$, the corresponding eigenspace is: \[ E_\lambda = \text{Nul}(A - \lambda I) \]
The eigenspace always contains $\mathbf{0}$ (the trivial solution), but the eigenvectors are the nonzero elements: $E_\lambda \setminus \{\mathbf{0}\}$.
The eigenspace can be multi-dimensional. If $(A - \lambda I)$ has 2 free variables, the eigenspace is 2-dimensional and there are two linearly independent eigenvectors for that eigenvalue. The geometric multiplicity of $\lambda$ is $\dim(E_\lambda)$.
Worked Example
Find all eigenvalues and eigenvectors of $A = \begin{pmatrix}2&1\\1&2\end{pmatrix}$.
Step 1. Characteristic polynomial: \[ \det(A - \lambda I) = \det\begin{pmatrix}2-\lambda&1\\1&2-\lambda\end{pmatrix} = (2-\lambda)^2 - 1 = \lambda^2 - 4\lambda + 3 \]
Step 2. Solve $\lambda^2 - 4\lambda + 3 = 0$: $(\lambda-1)(\lambda-3) = 0$. Eigenvalues: $\lambda = 1$ and $\lambda = 3$.
Step 3a. Eigenspace for $\lambda = 1$: row-reduce $A - I = \begin{pmatrix}1&1\\1&1\end{pmatrix}$.
$R_2 - R_1$: $\begin{pmatrix}1&1\\0&0\end{pmatrix}$. Free variable $x_2 = t$, $x_1 = -t$.
Basis for $E_1$: $\left\{\begin{pmatrix}-1\\1\end{pmatrix}\right\}$.
Step 3b. Eigenspace for $\lambda = 3$: row-reduce $A - 3I = \begin{pmatrix}-1&1\\1&-1\end{pmatrix}$.
$R_2 + R_1$: $\begin{pmatrix}-1&1\\0&0\end{pmatrix}$. Free variable $x_2 = t$, $x_1 = t$.
Basis for $E_3$: $\left\{\begin{pmatrix}1\\1\end{pmatrix}\right\}$.
Verify $\lambda = 1$: $A\begin{pmatrix}-1\\1\end{pmatrix} = \begin{pmatrix}-2+1\\-1+2\end{pmatrix} = \begin{pmatrix}-1\\1\end{pmatrix} = 1\begin{pmatrix}-1\\1\end{pmatrix}$ (correct).
Verify $\lambda = 3$: $A\begin{pmatrix}1\\1\end{pmatrix} = \begin{pmatrix}3\\3\end{pmatrix} = 3\begin{pmatrix}1\\1\end{pmatrix}$ (correct).
Common Pitfalls
| Mistake | Why It Is Wrong | Correct Approach |
|---|---|---|
| Including $\mathbf{0}$ as an eigenvector | The definition requires the eigenvector to be nonzero | State eigenvectors explicitly as nonzero; the eigenspace includes $\mathbf{0}$ but eigenvectors do not |
| Assuming the eigenspace is always 1-dimensional | The eigenspace is $\text{Nul}(A-\lambda I)$, which can have any dimension from 1 to $n$ | Find the full null space; report all free variable directions |
| Computing $\det(\lambda I - A)$ instead of $\det(A - \lambda I)$ | Both are valid up to a sign change; $\det(\lambda I - A) = (-1)^n\det(A - \lambda I)$. Either gives the correct eigenvalues since both have the same roots | Be consistent with one convention; Lay uses $\det(A - \lambda I)$ |
| Finding the null space of $A$, not $(A - \lambda I)$ | Eigenvalues require solving $(A - \lambda I)\mathbf{v} = \mathbf{0}$, not $A\mathbf{v} = \mathbf{0}$ | Subtract $\lambda$ from each diagonal entry, then find the null space |
Practice Problems
For $A = \begin{pmatrix}3&2\\3&-2\end{pmatrix}$, determine whether each vector is an eigenvector of $A$. If yes, state the eigenvalue.
(a) $\mathbf{v}_1 = \begin{pmatrix}2\\1\end{pmatrix}$
(b) $\mathbf{v}_2 = \begin{pmatrix}-1\\3\end{pmatrix}$
(c) $\mathbf{v}_3 = \begin{pmatrix}1\\1\end{pmatrix}$
Find all eigenvalues of $A = \begin{pmatrix}3&2\\3&-2\end{pmatrix}$ by solving the characteristic equation $\det(A - \lambda I) = 0$.
Find all eigenvalues and a basis for each eigenspace of $A = \begin{pmatrix}4&-1\\2&1\end{pmatrix}$.
Find all eigenvalues and a basis for each eigenspace of $A = \begin{pmatrix}2&1&3\\0&1&-1\\0&0&4\end{pmatrix}$.
Before computing, predict the eigenvalues from the structure of $A$.
Suppose $A\mathbf{v} = \lambda\mathbf{v}$ with $\mathbf{v} \neq \mathbf{0}$.
(a) Show that $A^2\mathbf{v} = \lambda^2\mathbf{v}$.
(b) Show that if $A$ is invertible, then $A^{-1}\mathbf{v} = \frac{1}{\lambda}\mathbf{v}$.
(c) Verify using $A = \begin{pmatrix}2&1\\1&2\end{pmatrix}$ with eigenvalue $\lambda = 3$ and eigenvector $\mathbf{v} = \begin{pmatrix}1\\1\end{pmatrix}$: compute $A^2\mathbf{v}$ and $A^{-1}\mathbf{v}$ directly and confirm they match $\lambda^2\mathbf{v}$ and $\frac{1}{\lambda}\mathbf{v}$.
Common Misconceptions
the zero vector is an eigenvector. The definition requires $\mathbf{v} \neq \mathbf{0}$. The equation $A\mathbf{0} = \lambda\mathbf{0}$ is satisfied trivially for any $\lambda$, which is exactly why the zero vector is excluded: allowing it would make every $\lambda$ an eigenvalue of every matrix. Eigenvectors must be nonzero.
each eigenvalue has exactly one eigenvector. The eigenspace $\text{Nul}(A - \lambda I)$ can have any dimension from 1 to $n$. If the eigenspace is 2-dimensional, there are infinitely many eigenvectors for $\lambda$, and any two independent ones form a basis for the eigenspace. A basis for the eigenspace contains all the essential eigenvector information.
real matrices cannot have complex eigenvalues. A real matrix can have complex eigenvalues when the characteristic polynomial has no real roots. For example, $A = \begin{pmatrix}0&-1\\1&0\end{pmatrix}$ (rotation by 90°) has characteristic polynomial $\lambda^2 + 1$, with complex roots $\pm i$. Complex eigenvalues come in conjugate pairs for real matrices. This topic is developed further in advanced linear algebra courses.
Mastery Checklist
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Mental Model
The Balanced Directions
Imagine applying a transformation to a room full of arrows pointing in all directions. Most arrows will tilt when the transformation is applied: they rotate away from their original direction. Eigenvectors are the rare arrows that keep their direction: the transformation may stretch or shrink them, but they stay aligned with their original orientation.
The eigenvalue tells you the stretch factor: $\lambda > 1$ means the arrow gets longer, $0 < \lambda < 1$ means it gets shorter, $\lambda < 0$ means the arrow flips to the opposite direction (and possibly changes length), and $\lambda = 0$ means the arrow collapses to zero (the corresponding eigenvector is in the null space of $A$, which is why $\lambda = 0$ is an eigenvalue iff $A$ is singular).
Eigenvalues and eigenvectors reveal the hidden skeleton of a transformation: the axes along which it acts purely as scaling.
Connections
Looking ahead in MATH 231:
- Diagonalization: if $A$ has enough eigenvectors to form a basis, then $A = PDP^{-1}$ where $P$ holds the eigenvectors and $D$ is diagonal. Diagonalization makes matrix powers easy.
Real-world connections:
- In structural engineering, the eigenvectors of a stiffness matrix are the “natural modes” of vibration of a structure; the eigenvalues are related to the natural frequencies.
- In Google’s PageRank algorithm, the ranking of web pages is the eigenvector corresponding to the largest eigenvalue of the link matrix.
- Principal Component Analysis (PCA) in data science finds the eigenvectors of the data covariance matrix, revealing the directions of greatest variance.
Source: Lay, Linear Algebra and Its Applications, Section 5.1: Eigenvectors and Eigenvalues; Section 5.2: The Characteristic Equation.
| Previous | Up | Next |
|---|---|---|
| Dimension, Rank, and Coordinates | MATH231 Skills Index | Diagonalization |
Last updated: 2026-06-16
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