# Eigenvalues: Definition, How to Find Them & Examples

### TL;DR

An eigenvalue is the scalar $\lambda$ by which a matrix stretches or shrinks a special vector — its eigenvector — without turning it. This article covers the defining equation $A\mathbf{v} = \lambda\mathbf{v}$, how to find eigenvalues from the characteristic equation $\det(A - \lambda I) = 0$, the trace-and-determinant properties, and the sign and factoring mistakes that derail the calculation.

## What Eigenvalues Are

An eigenvalue of a square matrix $A$ is a scalar $\lambda$ for which there is a non-zero vector $\mathbf{v}$ satisfying

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

The vector $\mathbf{v}$ is the corresponding eigenvector. In words: most vectors get knocked off their line when a matrix multiplies them, but an eigenvector keeps its direction — the matrix only stretches it (or shrinks or flips it) by the factor $\lambda$. That factor is the eigenvalue. Eigenvalues are also called characteristic roots, characteristic values, or latent roots.

The eigenvalue is the _how much_ and the eigenvector is the _which direction_. They always come in pairs, and this article focuses on the eigenvalue — the scaling number itself. For the partner concept, the direction that stays fixed, see the companion article on [eigenvectors](/content/math/algebra/eigenvectors/index.html); this page is the degree-first, "find the number" view of the same equation.

## How Do You Find Eigenvalues?

Eigenvalues come from the **characteristic equation**. Rewrite $A\mathbf{v} = \lambda\mathbf{v}$ as

$$A\mathbf{v} - \lambda\mathbf{v} = \mathbf{0}$$

$$(A - \lambda I)\mathbf{v} = \mathbf{0}$$

where $I$ is the identity matrix of the same order as $A$. For a non-zero $\mathbf{v}$ to exist, the matrix $A - \lambda I$ must be singular, which means its [determinant](/content/math/algebra/determinant-of-matrix/index.html) is zero:

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

This is the characteristic equation. Expanding the determinant gives a polynomial in $\lambda$ — the characteristic polynomial — whose roots are the eigenvalues. The steps:

1. Form $A - \lambda I$ by subtracting $\lambda$ from each diagonal entry of $A$.
2. Compute $\det(A - \lambda I)$.
3. Set that determinant to 0 and solve for $\lambda$.

An $n \times n$ matrix gives a degree-$n$ polynomial, so it has at most $n$ eigenvalues.

**Variable glossary.** $A$ is the square matrix; $\lambda$ (lambda) is an eigenvalue; $\mathbf{v}$ is the eigenvector; $I$ is the identity matrix; $\det(\cdot)$ is the determinant; the _characteristic polynomial_ is $\det(A - \lambda I)$ written as a polynomial in $\lambda$.

## What Are the Properties of Eigenvalues?

A few properties let you check your work and shortcut the algebra.

- **Sum equals the trace.** The eigenvalues add up to the trace of $A$ (the sum of its diagonal entries). If the diagonal sums to 7, so do the eigenvalues.
- **Product equals the determinant.** Multiply the eigenvalues together and you get $\det A$. A zero among the eigenvalues means $\det A = 0$, so the matrix is not [invertible](/content/math/algebra/invertible-matrix/index.html).
- **Triangular matrices give their diagonal free.** For an upper- or lower-triangular matrix, the eigenvalues are simply the diagonal entries — no characteristic equation needed.
- **Symmetric matrices have real eigenvalues.** A [symmetric matrix](/content/math/algebra/symmetric-matrix/index.html) always has real eigenvalues, which is why they appear so often in physics and statistics.
- **Transpose shares eigenvalues.** $A$ and $A^{\mathsf{T}}$ have the same eigenvalues.

The trace and determinant checks are the fastest sanity tests you have: after solving, add your eigenvalues and multiply them, then compare with the trace and determinant of the original matrix.

## Examples of Eigenvalues

### Example 1

**Find the eigenvalues of $A = \begin{bmatrix} 2 & 0 \\ 0 & 3 \end{bmatrix}$.**

The matrix is diagonal (a special triangular case), so the eigenvalues are the diagonal entries. **Final answer:** $\lambda = 2$ and $\lambda = 3$. (Check: trace $= 5 = 2 + 3$ ✓; determinant $= 6 = 2 \times 3$ ✓.)

### Example 2

**Find the eigenvalues of $A = \begin{bmatrix} 5 & 4 \\ 1 & 2 \end{bmatrix}$.**

Form $A - \lambda I$:

$$A - \lambda I = \begin{bmatrix} 5 - \lambda & 4 \\ 1 & 2 - \lambda \end{bmatrix}$$

Set the determinant to zero:

$$(5 - \lambda)(2 - \lambda) - (4)(1) = 0$$

$$\lambda^2 - 7\lambda + 10 - 4 = 0$$

$$\lambda^2 - 7\lambda + 6 = 0$$

$$(\lambda - 6)(\lambda - 1) = 0$$

**Final answer:** $\lambda = 6$ and $\lambda = 1$. (Check: trace $= 7 = 6 + 1$ ✓; determinant $= 10 - 4 = 6 = 6 \times 1$ ✓.)

### Example 3 (the mistake worth making once)

**Find the eigenvalues of $A = \begin{bmatrix} 4 & 1 \\ 2 & 3 \end{bmatrix}$.**

_Wrong attempt._ A student forms $A - \lambda I$ but forgets to subtract $\lambda$ from _both_ diagonal entries, writing the determinant as $(4 - \lambda)(3) - (1)(2)$ — leaving the bottom-right entry as plain 3.

_The break._ That gives $12 - 3\lambda - 2 = 0$, a linear equation with a single root $\lambda = \tfrac{10}{3}$. But a $2 \times 2$ matrix must produce a _quadratic_ with up to two eigenvalues. A linear characteristic equation is the tell that a $\lambda$ was dropped.

_Correct._ Subtract $\lambda$ from both diagonal entries:

$$\det\begin{bmatrix} 4 - \lambda & 1 \\ 2 & 3 - \lambda \end{bmatrix} = (4 - \lambda)(3 - \lambda) - (1)(2) = 0$$

$$\lambda^2 - 7\lambda + 12 - 2 = 0$$

$$\lambda^2 - 7\lambda + 10 = 0$$

$$(\lambda - 5)(\lambda - 2) = 0$$

**Final answer:** $\lambda = 5$ and $\lambda = 2$. (Check: trace $= 7$ ✓; determinant $= 12 - 2 = 10 = 5 \times 2$ ✓.)

### Example 4

**Find the eigenvalues of the triangular matrix $A = \begin{bmatrix} 3 & 7 & -1 \\ 0 & 2 & 4 \\ 0 & 0 & -3 \end{bmatrix}$.**

The matrix is upper-triangular, so the eigenvalues are the diagonal entries directly, no expansion needed. **Final answer:** $\lambda = 3$, $\lambda = 2$, $\lambda = -3$.

### Example 5

**Find the eigenvalues of $A = \begin{bmatrix} 2 & -1 \\ 1 & 2 \end{bmatrix}$.**

$$\det\begin{bmatrix} 2 - \lambda & -1 \\ 1 & 2 - \lambda \end{bmatrix} = (2 - \lambda)^2 + 1 = 0$$

$$(2 - \lambda)^2 = -1$$

$$2 - \lambda = \pm i$$

$$\lambda = 2 \mp i$$

**Final answer:** $\lambda = 2 + i$ and $\lambda = 2 - i$ — a complex conjugate pair. (This is exactly what a rotation-and-scaling matrix produces; a real matrix can have complex eigenvalues.)

### Example 6

**Find the eigenvalues of $A = \begin{bmatrix} 3 & 1 & 1 \\ 2 & 4 & 2 \\ 1 & 1 & 3 \end{bmatrix}$, given that one eigenvalue is $\lambda = 2$.**

The characteristic polynomial of this $3 \times 3$ matrix factors as $(\lambda - 2)^2(\lambda - 6) = 0$. Using the property checks: trace $= 3 + 4 + 3 = 10$, and $2 + 2 + 6 = 10$ ✓; determinant $= 2 \times 2 \times 6 = 24$. **Final answer:** $\lambda = 2$ (repeated) and $\lambda = 6$.

## Why Eigenvalues Matter — "the few numbers that define a matrix"

Eigenvalues were not invented as abstract algebra. They surfaced in the 1700s and 1800s in the study of _rotating bodies_ and _vibrating systems_ — the principal axes of a spinning top, the natural frequencies of an oscillating string. The question was always the same: as a system transforms, what stays aligned, and by how much does it grow? That is the eigenvalue question.

- **PageRank.** Google's original ranking was the dominant eigenvalue problem for the web's link matrix — the eigenvector of that matrix is the ranking.
- **Vibration and resonance.** The eigenvalues of a structure's stiffness matrix are its natural frequencies. The [Tacoma Narrows Bridge](https://en.wikipedia.org/wiki/Tacoma_Narrows_Bridge_(1940)) collapse in 1940 is the textbook warning about what happens when a driving frequency meets a structure's natural one.
- **Quantum mechanics.** The allowed energy levels of a system are the eigenvalues of its energy operator — discrete, fixed numbers, exactly like the eigenvalues of a matrix.
- **Data science.** Principal component analysis ranks the directions of greatest variance in data by their eigenvalues, the basis of dimensionality reduction.
- **Stability analysis.** Whether a system settles or blows up is read from the signs of its eigenvalues.

A connection worth carrying: a [rotation matrix](/content/math/algebra/rotation-matrix/index.html) has complex eigenvalues (as in Example 5), because a pure rotation leaves _no_ real direction unchanged — there is nothing for a real eigenvector to point along. That single fact ties the geometry of rotation to the algebra of eigenvalues.

## The Mathematicians Behind Eigenvalues

**Augustin-Louis Cauchy** (1789–1857, France) gave the eigenvalue problem its modern footing, proving that symmetric matrices have real eigenvalues and connecting the idea to the principal axes of quadratic forms.

**David Hilbert** (1862–1943, Germany) extended eigenvalues from finite matrices to infinite-dimensional spaces and coined the term _Eigenwert_ ("proper value"), the German root of the word we still use.

## Common Confusions Cleared Up

### Mistake 1: Subtracting λ from only one diagonal entry

**Where it slips in:** Forming $A - \lambda I$ for the characteristic equation.

**Don't do this:** Subtracting $\lambda$ from just the top-left entry. That changes the polynomial's degree and loses an eigenvalue.

**The correct way:** Subtract $\lambda$ from _every_ diagonal entry — that is what $\lambda I$ does. A quick check: an $n \times n$ matrix should give a degree-$n$ characteristic polynomial, so a $2 \times 2$ that produces a linear equation means a $\lambda$ was missed. The first instinct is to treat $\lambda I$ as a single number subtracted once, when it is subtracted along the whole diagonal.

### Mistake 2: Multiplying out a polynomial that is already factored

**Where it slips in:** After reaching something like $(\lambda - 3)(\lambda - 5) = 0$.

**Don't do this:** Expanding it back to $\lambda^2 - 8\lambda + 15 = 0$ and then re-factoring. That is wasted work and a fresh chance for an arithmetic slip.

**The correct way:** If the characteristic equation arrives partly factored, read the roots straight off. The habit that fixes this is pausing to ask "is this already factored?" before expanding anything.

### Mistake 3: Forgetting that eigenvalues can be complex

**Where it slips in:** Matrices that rotate, like Example 5.

**Don't do this:** Assuming a real matrix must have real eigenvalues and treating $(2 - \lambda)^2 = -1$ as having no answer.

**The correct way:** A real matrix can have complex eigenvalues, appearing in conjugate pairs. Solve over the complex numbers rather than declaring "no solution."

## The Short Version

- An **eigenvalue** is the scalar $\lambda$ by which a matrix scales its eigenvector, defined by $A\mathbf{v} = \lambda\mathbf{v}$.  
- Find eigenvalues by solving the characteristic equation $\det(A - \lambda I) = 0$.
- The eigenvalues sum to the trace and multiply to the determinant — two quick checks.
- The most common mistake is subtracting $\lambda$ from only one diagonal entry, which drops an eigenvalue.
- Real matrices can have complex eigenvalues, and a zero eigenvalue signals a singular matrix.

## Practice Questions on Eigenvalues

Work through these, then check your answers below.

1. Find the eigenvalues of $\begin{bmatrix} 4 & 0 \\ 0 & 7 \end{bmatrix}$.
2. Solve $\det(A - \lambda I) = 0$ for $\begin{bmatrix} 3 & 2 \\ 1 & 4 \end{bmatrix}$, then confirm with the trace-and-determinant check.
3. Find the eigenvalues of the $3 \times 3$ matrix $\begin{bmatrix} 2 & 0 & 0 \\ 1 & 3 & 0 \\ 4 & 5 & 1 \end{bmatrix}$ from scratch.
4. The eigenvalues of a $2 \times 2$ matrix are $3$ and $-2$. What are its trace and determinant?

**Answer to Question 1:** The matrix is diagonal, so the eigenvalues are the diagonal entries: $\lambda = 4$ and $\lambda = 7$.

**Answer to Question 2:** $\det\begin{bmatrix} 3 - \lambda & 2 \\ 1 & 4 - \lambda \end{bmatrix} = (3 - \lambda)(4 - \lambda) - (2)(1) = 0$. $$\lambda^2 - 7\lambda + 12 - 2 = 0$$ $$\lambda^2 - 7\lambda + 10 = 0$$ $$(\lambda - 5)(\lambda - 2) = 0$$ So $\lambda = 5$ and $\lambda = 2$. Check: trace $= 3 + 4 = 7 = 5 + 2$ ✓; determinant $= 12 - 2 = 10 = 5 \times 2$ ✓.

**Answer to Question 3:** The matrix is lower-triangular, so the eigenvalues are the diagonal entries directly: $\lambda = 2$, $\lambda = 3$, $\lambda = 1$. (Verifying from scratch, the characteristic polynomial expands along the first row to $(2 - \lambda)(3 - \lambda)(1 - \lambda) = 0$, giving the same three roots.)

**Answer to Question 4:** The eigenvalues sum to the trace and multiply to the determinant. So trace $= 3 + (-2) = 1$ and determinant $= (3)(-2) = -6$.