Linear Algebra

Eigenvalues and Eigenvectors Explained with Examples

Q: What is the characteristic polynomial?

It is det(A - lambda I), a polynomial in lambda whose roots are the eigenvalues. For a 2 × 2 matrix it equals lambda^2 - tr(A)lambda + det(A).

By MathSolver Pro Editorial Team · Published October 25, 2026

An eigenvector of a matrix is a direction that the matrix only stretches or flips, never turns, and the eigenvalue is the stretch factor. They reveal the essential behaviour of a matrix and underpin diagonalization, stability analysis, Markov chains and data compression.

The defining equation

Most vectors change direction when you multiply them by a matrix. A few special vectors do not: the matrix simply stretches, shrinks or reverses them along the same line. These are the eigenvectors, and the factor by which each one is scaled is its eigenvalue. In symbols, a nonzero vector $\mathbf{v}$ is an eigenvector of the square matrix $A$ with eigenvalue $\lambda$ when:

A\mathbf{v} = \lambda\mathbf{v}, \qquad \mathbf{v} \ne \mathbf{0}

The zero vector is excluded because it would satisfy the equation for every $\lambda$ . Eigenvalues, however, can be zero, negative or even complex. An eigenvalue of 2 means vectors in that direction double in length, $-1$ means they flip, and 0 means they are flattened to nothing, which happens exactly when the matrix is singular.

Finding eigenvalues with the characteristic equation

Rewrite the defining equation as $(A - \lambda I)\mathbf{v} = \mathbf{0}$ . For a nonzero solution to exist, the matrix $A - \lambda I$ must be singular, so its determinant must be zero. That gives the characteristic equation:

\det(A - \lambda I) = 0

For an $n \times n$ matrix, this is a polynomial equation of degree $n$ in $\lambda$ , so there are $n$ eigenvalues when they are counted with multiplicity and complex values are allowed. For a $2 \times 2$ matrix the polynomial is a quadratic, which you can solve with the methods in how to solve quadratic equations.

A complete 2 by 2 example

Example 1: eigenvalues and eigenvectors of a 2 by 2 matrix

Let $A = \begin{pmatrix} 4 & 1 \\ 2 & 3 \end{pmatrix}$ . Then $\det(A - \lambda I) = (4 - \lambda)(3 - \lambda) - 2 = \lambda^2 - 7\lambda + 10$ .
Factor: $(\lambda - 5)(\lambda - 2) = 0$ , so the eigenvalues are $\lambda = 5$ and $\lambda = 2$ .
For $\lambda = 5$ : $A - 5I = \begin{pmatrix} -1 & 1 \\ 2 & -2 \end{pmatrix}$ . The first row says $-x + y = 0$ , so $y = x$ and an eigenvector is $(1, 1)$ .
For $\lambda = 2$ : $A - 2I = \begin{pmatrix} 2 & 1 \\ 2 & 1 \end{pmatrix}$ , so $2x + y = 0$ and an eigenvector is $(1, -2)$ .
Check: $A(1, -2) = (4 - 2, 2 - 6) = (2, -4) = 2(1, -2)$ .

Any nonzero multiple of an eigenvector is also an eigenvector, so $(3, 3)$ or $(-2, 4)$ would be equally correct answers. What matters is the direction, not the length. Notice too that the two rows of $A - \lambda I$ were multiples of each other each time. That is guaranteed when $\lambda$ is a true eigenvalue, and it is a handy check on your arithmetic.

Two shortcuts: trace and determinant

The eigenvalues of any square matrix are linked to two easy numbers. Their sum equals the trace, the sum of the diagonal entries, and their product equals the determinant:

\lambda_1 + \lambda_2 + \cdots + \lambda_n = \operatorname{tr}(A), \qquad \lambda_1\lambda_2\cdots\lambda_n = \det(A)

In Example 1, the trace is $4 + 3 = 7 = 5 + 2$ and the determinant is $12 - 2 = 10 = 5 \cdot 2$ . For a $2 \times 2$ matrix, this means the characteristic equation is always $\lambda^2 - \operatorname{tr}(A)\lambda + \det(A) = 0$ , which saves expanding the determinant by hand. Triangular matrices are even easier: their eigenvalues are simply the diagonal entries.

Complex and repeated eigenvalues

The rotation matrix $\begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}$ turns every vector by 90 degrees, so no real direction stays on its own line. Its characteristic equation is $\lambda^2 + 1 = 0$ , giving the complex eigenvalues $\lambda = \pm i$ . Complex eigenvalues of a real matrix always come in conjugate pairs and signal rotation.

Eigenvalues can also repeat. The matrix $\begin{pmatrix} 2 & 1 \\ 0 & 2 \end{pmatrix}$ has the double eigenvalue 2, yet $A - 2I = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}$ forces $y = 0$ , so the only eigenvectors are multiples of $(1, 0)$ . A matrix like this, with fewer independent eigenvectors than its size, is called defective. The identity matrix, in contrast, also has the repeated eigenvalue 1, but every nonzero vector is an eigenvector.

Symmetric matrices avoid both complications. Their eigenvalues are always real, and eigenvectors for different eigenvalues are perpendicular. For $\begin{pmatrix} 2 & 1 \\ 1 & 2 \end{pmatrix}$ , the eigenvalues are 3 and 1 with eigenvectors $(1, 1)$ and $(1, -1)$ , whose dot product is zero.

Diagonalization and matrix powers

When an $n \times n$ matrix has $n$ independent eigenvectors, it can be diagonalized. Put the eigenvectors in the columns of $P$ and the eigenvalues on the diagonal of $D$ , and then:

A = PDP^{-1} \quad \Rightarrow \quad A^k = PD^kP^{-1}

Raising a diagonal matrix to a power just raises each diagonal entry to that power, so computing $A^{100}$ becomes easy. The eigenvalues also predict long-term behaviour: when one eigenvalue is larger in absolute value than all the others, repeated multiplication by $A$ pushes almost every starting vector towards that eigenvalue's eigenvector.

Example 2: the long-run state of a Markov chain

Each year 90 percent of customers of company X stay and 10 percent switch to Y, while Y keeps 50 percent and loses 50 percent to X. The transition matrix is $\begin{pmatrix} 0.9 & 0.5 \\ 0.1 & 0.5 \end{pmatrix}$ .
The trace is 1.4 and the determinant is $0.45 - 0.05 = 0.4$ , so the eigenvalues are 1 and 0.4.
For $\lambda = 1$ : $-0.1x + 0.5y = 0$ , giving the eigenvector $(5, 1)$ .
Scaled to total 1, the long-run market shares are five sixths for X and one sixth for Y, whatever the starting shares, because the 0.4 component shrinks away each year.

Where eigenvalues matter

Search engines: the original PageRank algorithm ranks web pages using the dominant eigenvector of a link matrix.
Engineering: the natural vibration frequencies of bridges and buildings come from eigenvalues, which designers keep away from expected forcing frequencies.
Data science: principal component analysis uses the eigenvectors of a covariance matrix to find the directions of greatest variation.
Differential equations: the signs of eigenvalues decide whether an equilibrium is stable or unstable.
Quantum mechanics: measurable quantities such as energy levels are eigenvalues of operators.

The eigenvalue calculator finds eigenvalues and eigenvectors, including complex ones, with the characteristic polynomial shown step by step. The determinant calculator is useful for checking $\det(A - \lambda I)$ at a candidate value.

A 3 by 3 example

Larger matrices follow the same recipe, and zeros help just as they do for determinants. Take the matrix with rows $(2, 0, 0)$ , $(0, 3, 4)$ and $(0, 4, 9)$ . Expanding $\det(A - \lambda I)$ along the first row gives $(2 - \lambda)$ times the characteristic polynomial of the smaller block with rows $(3, 4)$ and $(4, 9)$ . That block has trace 12 and determinant $27 - 16 = 11$ , so its polynomial is $\lambda^2 - 12\lambda + 11 = (\lambda - 1)(\lambda - 11)$ .

The eigenvalues are therefore 2, 1 and 11. As a check, they add up to 14, the trace of the full matrix, and multiply to 22, its determinant. The eigenvectors are $(1, 0, 0)$ for 2, $(0, -2, 1)$ for 1 and $(0, 1, 2)$ for 11. The matrix is symmetric, and as promised these three vectors are mutually perpendicular.

Common mistakes

Subtracting $\lambda$ from every entry instead of only the diagonal entries.
Accepting the zero vector as an eigenvector after solving $(A - \lambda I)\mathbf{v} = \mathbf{0}$ .
Assuming a repeated eigenvalue always comes with two independent eigenvectors.
Forgetting to check the result by computing $A\mathbf{v}$ and comparing it with $\lambda\mathbf{v}$ .

The final check is the most valuable habit of all. It takes one matrix multiplication, and if $A\mathbf{v}$ is not a multiple of $\mathbf{v}$ , either the eigenvalue or the eigenvector is wrong.

A geometric picture

Imagine applying the matrix from Example 1 to every point of a sheet of graph paper. Most arrows drawn from the origin are both stretched and turned. Arrows along the line through $(1, 1)$ , however, simply become five times longer, and arrows along the line through $(1, -2)$ become twice as long, staying on their lines. The eigenvectors are the skeleton of the transformation: once you know what happens along these two lines, you know what happens everywhere, because every vector can be split into pieces along them.

Frequently asked questions

Can an eigenvalue be zero?

Yes. An eigenvalue of zero means some nonzero vector is sent to the zero vector, so the matrix is singular and has no inverse. Eigenvectors, however, can never be the zero vector.

Are eigenvectors unique?

No. Any nonzero multiple of an eigenvector is also an eigenvector for the same eigenvalue, so answers are often scaled to have simple whole-number entries or length 1.

Do all matrices have real eigenvalues?

No. Rotation matrices, for instance, have complex eigenvalues. Real symmetric matrices, however, always have real eigenvalues.

What is the characteristic polynomial?

It is $\det(A - \lambda I)$ , a polynomial in $\lambda$ whose roots are the eigenvalues. For a $2 \times 2$ matrix it equals $\lambda^2 - \operatorname{tr}(A)\lambda + \det(A)$ .

Where does the word eigen come from?

It is German for own or characteristic. An eigenvector is a matrix's own direction, one it maps onto itself, and an eigenvalue is its characteristic scaling factor.

MathSolver Pro Editorial Team

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