How do I find eigenvectors once I have the eigenvalues?

For each eigenvalue \lambda, solve (A - \lambda I)\mathbf{v} = \mathbf{0} by row reduction or inspection. The solution set (the null space) gives you all eigenvectors for that eigenvalue. Pick any nonzero vector from that null space.

Why do I need both eigenvalues and eigenvectors?

Eigenvalues tell you the strength of transformation (the scaling factor); eigenvectors tell you the direction in which that scaling happens. Together they reveal how the matrix acts on preferred directions, essential for diagonalization, stability analysis, and solving systems of differential equations.

Can a matrix have no real eigenvalues?

Yes. A rotation matrix like \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix} (a 90° rotation) has only complex eigenvalues, because rotating a vector always changes its direction for every direction—no vector is merely scaled.

What does it mean if I get a repeated eigenvalue?

A repeated eigenvalue (where the same value appears more than once in the characteristic polynomial) can mean the eigenvector space has fewer dimensions than expected. You may find fewer linearly independent eigenvectors than the multiplicity of that eigenvalue, a situation called defect or deficiency.

Eigenvalues and Eigenvectors

Eigenvalues and eigenvectors show which vectors a matrix scales without rotating, revealing fundamental properties of linear transformations and systems.

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

Solve a problem with Eigenvalues and Eigenvectors

Type the problem. The solver will use Eigenvalues and Eigenvectors where Eigenvalues and Eigenvectors is the right tool, and tell you when it is not.

What each symbol means

What Eigenvalues and Eigenvectors takes

Symbol	Meaning
$A$	The matrix whose behavior you are analyzing—the linear transformation itself. If you confuse this with a vector, you lose the entire framework for understanding the transformation.
$\mathbf{v}$	The eigenvector—a nonzero vector that the matrix transforms without changing its direction, only its length. If you allow this to be zero, the eigenvalue becomes meaningless since every scalar works.
$\lambda$	The eigenvalue—the scalar (real or complex) that measures how much the matrix stretches or compresses the vector in the direction of $\mathbf{v}$ . Negative values mean the direction reverses as well as scales.

When to use it

Use eigenvalues and eigenvectors when you need to understand how a linear transformation behaves on certain special directions, or solve systems of differential equations.

Level

Usually taught in: Linear Algebra

Worked examples

1. Find eigenvalues and eigenvectors of a 2×2 matrix

Problem

Find all eigenvalues and eigenvectors of

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

$\det(A - \lambda I) = \det\begin{pmatrix} 4-\lambda & 1 \\ 2 & 3-\lambda \end{pmatrix}$
Write the characteristic equation by setting up the determinant of $A - \lambda I$ equal to zero; this finds values of $\lambda$ where the matrix is singular.
$(4-\lambda)(3-\lambda) - 1 \cdot 2 = 0$
Expand the $2 \times 2$ determinant using the formula $ad - bc$ .
$12 - 7\lambda + \lambda^2 - 2 = 0$
Distribute and simplify the left side: $(4-\lambda)(3-\lambda) = 12 - 7\lambda + \lambda^2$ .
$\lambda^2 - 7\lambda + 10 = 0$
Rearrange into standard quadratic form.
$(\lambda - 5)(\lambda - 2) = 0$
Factor the quadratic by finding factors of 10 that sum to $-7$ .
$\lambda_1 = 5, \quad \lambda_2 = 2$
Solve each factor; the eigenvalues are 5 and 2.
$(A - 5I)\mathbf{v} = \begin{pmatrix} -1 & 1 \\ 2 & -2 \end{pmatrix}\begin{pmatrix} v_1 \\ v_2 \end{pmatrix} = \mathbf{0}$
For $\lambda_1 = 5$ , substitute back into $(A - \lambda I)\mathbf{v} = \mathbf{0}$ and solve.
$-v_1 + v_2 = 0 \Rightarrow v_1 = v_2$
From the first row, we see that $v_1$ and $v_2$ must be equal.
$\mathbf{v}_1 = \begin{pmatrix} 1 \\ 1 \end{pmatrix}$
Choose $v_1 = 1$ ; any nonzero scalar multiple is also an eigenvector.
$(A - 2I)\mathbf{v} = \begin{pmatrix} 2 & 1 \\ 2 & 1 \end{pmatrix}\begin{pmatrix} v_1 \\ v_2 \end{pmatrix} = \mathbf{0}$
For $\lambda_2 = 2$ , substitute and solve the system.
$2v_1 + v_2 = 0 \Rightarrow v_2 = -2v_1$
From the first row, $v_2$ is $-2$ times $v_1$ .
$\mathbf{v}_2 = \begin{pmatrix} 1 \\ -2 \end{pmatrix}$
Choose $v_1 = 1$ ; then $v_2 = -2$ .

Answer: $\lambda_1 = 5 \text{ with } \mathbf{v}_1 = \begin{pmatrix} 1 \\ 1 \end{pmatrix}; \quad \lambda_2 = 2 \text{ with } \mathbf{v}_2 = \begin{pmatrix} 1 \\ -2 \end{pmatrix}$

This shows the two directions in which the matrix acts purely as a scalar multiplier: stretching by a factor of 5 in the direction $(1, 1)$ and by a factor of 2 in the direction $(1, -2)$ . Every other vector is both scaled and rotated.

2. Find eigenvalues with negative eigenvalue and harder factoring

Problem

Find all eigenvalues and eigenvectors of

B = \begin{pmatrix} 2 & 4 \\ 1 & -1 \end{pmatrix}

$\det(B - \lambda I) = \det\begin{pmatrix} 2-\lambda & 4 \\ 1 & -1-\lambda \end{pmatrix}$
Set up the characteristic equation; note that this matrix will produce a negative eigenvalue, indicating reflection in addition to scaling.
$(2-\lambda)(-1-\lambda) - 4 \cdot 1 = 0$
Apply the $2 \times 2$ determinant formula $ad - bc$ .
$-2 - 2\lambda + \lambda + \lambda^2 - 4 = 0$
Expand the product $(2-\lambda)(-1-\lambda) = -2 - 2\lambda + \lambda + \lambda^2 = \lambda^2 - \lambda - 2$ .
$\lambda^2 - \lambda - 6 = 0$
Combine like terms: $-2 - 4 = -6$ and $-2\lambda + \lambda = -\lambda$ .
$(\lambda - 3)(\lambda + 2) = 0$
Factor by finding factors of $-6$ that sum to $-1$ : we need $-3$ and $+2$ .
$\lambda_1 = 3, \quad \lambda_2 = -2$
The eigenvalues are 3 and $-2$ ; the negative value indicates reflection.
$(B - 3I)\mathbf{v} = \begin{pmatrix} -1 & 4 \\ 1 & -4 \end{pmatrix}\begin{pmatrix} v_1 \\ v_2 \end{pmatrix} = \mathbf{0}$
For $\lambda_1 = 3$ , form the system $(B - 3I)\mathbf{v} = \mathbf{0}$ .
$-v_1 + 4v_2 = 0 \Rightarrow v_1 = 4v_2$
From the first row, $v_1$ must be 4 times $v_2$ .
$\mathbf{v}_1 = \begin{pmatrix} 4 \\ 1 \end{pmatrix}$
Choose $v_2 = 1$ so that $v_1 = 4$ .
$(B - (-2)I)\mathbf{v} = \begin{pmatrix} 4 & 4 \\ 1 & 1 \end{pmatrix}\begin{pmatrix} v_1 \\ v_2 \end{pmatrix} = \mathbf{0}$
For $\lambda_2 = -2$ , form the system; note that $B - (-2I) = B + 2I$ .
$4v_1 + 4v_2 = 0 \Rightarrow v_1 = -v_2$
From the first row, $v_1$ and $v_2$ must have opposite signs.
$\mathbf{v}_2 = \begin{pmatrix} 1 \\ -1 \end{pmatrix}$
Choose $v_1 = 1$ ; then $v_2 = -1$ .

Answer: $\lambda_1 = 3 \text{ with } \mathbf{v}_1 = \begin{pmatrix} 4 \\ 1 \end{pmatrix}; \quad \lambda_2 = -2 \text{ with } \mathbf{v}_2 = \begin{pmatrix} 1 \\ -1 \end{pmatrix}$

The negative eigenvalue $\lambda_2 = -2$ means that in the direction $(1, -1)$ , the matrix scales by a factor of 2 AND reverses direction. This is why negative eigenvalues appear in systems with reflection or oscillation. The eigenvalue of 3 represents pure scaling without reversal.

3. Find eigenvalues for a word problem on cyclist power

Problem

A cyclist riding on a mountain route experiences power transfer between uphill and downhill segments modeled by

R = \begin{pmatrix} 3 & 1 \\ 1 & 3 \end{pmatrix}

, where entries represent effort coupling. Find the eigenvalues to determine the principal directions of sustained power output.

$\det(R - \lambda I) = \det\begin{pmatrix} 3-\lambda & 1 \\ 1 & 3-\lambda \end{pmatrix}$
Set up the characteristic equation; this matrix is symmetric, so we expect real eigenvalues.
$(3-\lambda)(3-\lambda) - 1 \cdot 1 = 0$
Apply the determinant formula for a $2 \times 2$ matrix.
$(3-\lambda)^2 - 1 = 0$
Notice that the diagonal entries are equal; this simplifies to $(3-\lambda)^2 - 1^2$ , a difference of squares.
$9 - 6\lambda + \lambda^2 - 1 = 0$
Expand $(3-\lambda)^2 = 9 - 6\lambda + \lambda^2$ .
$\lambda^2 - 6\lambda + 8 = 0$
Simplify by collecting constants: $9 - 1 = 8$ .
$(\lambda - 4)(\lambda - 2) = 0$
Factor by finding factors of 8 that sum to $-6$ : we need $-4$ and $-2$ .
$\lambda_1 = 4, \quad \lambda_2 = 2$
The eigenvalues are 4 and 2, both positive since $R$ is symmetric positive-definite.
$(R - 4I)\mathbf{v} = \begin{pmatrix} -1 & 1 \\ 1 & -1 \end{pmatrix}\begin{pmatrix} v_1 \\ v_2 \end{pmatrix} = \mathbf{0}$
For $\lambda_1 = 4$ , set up the eigenvector equation.
$-v_1 + v_2 = 0 \Rightarrow v_1 = v_2$
From the first row, the components must be equal.
$\mathbf{v}_1 = \begin{pmatrix} 1 \\ 1 \end{pmatrix}$
Choose $v_1 = 1$ and $v_2 = 1$ ; this represents equal uphill-downhill effort.
$(R - 2I)\mathbf{v} = \begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix}\begin{pmatrix} v_1 \\ v_2 \end{pmatrix} = \mathbf{0}$
For $\lambda_2 = 2$ , set up the eigenvector equation.
$v_1 + v_2 = 0 \Rightarrow v_1 = -v_2$
From the first row, the components must be opposite in sign.
$\mathbf{v}_2 = \begin{pmatrix} 1 \\ -1 \end{pmatrix}$
Choose $v_1 = 1$ and $v_2 = -1$ ; this represents opposite efforts on uphill versus downhill.

Answer: $\lambda_1 = 4 \text{ with } \mathbf{v}_1 = \begin{pmatrix} 1 \\ 1 \end{pmatrix}; \quad \lambda_2 = 2 \text{ with } \mathbf{v}_2 = \begin{pmatrix} 1 \\ -1 \end{pmatrix}$

The larger eigenvalue $\lambda_1 = 4$ represents the most amplified direction: when the cyclist applies equal force to uphill and downhill segments in ratio $1:1$ , the total effort multiplies by 4 over the cycle. The smaller eigenvalue $\lambda_2 = 2$ represents a less efficient pattern. This helps identify the most sustainable power distribution on repeated climbs.

Common mistakes

Where Eigenvalues and Eigenvectors usually goes wrong

The mistake	Why it is wrong	The fix
Computing $\det\begin{pmatrix} a-\lambda & b \\ c & d-\lambda \end{pmatrix}$ by adding the diagonal terms instead: $(a-\lambda) + (d-\lambda)$ .	Students sometimes confuse the trace (sum of diagonal) with the determinant and forget that the determinant formula requires cross-multiplication: $(a-\lambda)(d-\lambda) - bc$ , not addition.	Always use the determinant formula $ps - qr$ for a $2 \times 2$ matrix: $(a-\lambda)(d-\lambda) - bc$ . The trace comes later when you verify your eigenvalues, not when computing the characteristic polynomial.
Treating the zero vector $\mathbf{0}$ as an eigenvector for an eigenvalue.	While the zero vector technically satisfies $A\mathbf{0} = \lambda\mathbf{0}$ for any $\lambda$ , eigenvectors must be nonzero by definition—otherwise the eigenvalue is not uniquely determined and the whole concept collapses.	Always check that your eigenvector is nonzero. If you get the zero vector, your calculation went wrong; go back and check whether you correctly solved $(A - \lambda I)\mathbf{v} = \mathbf{0}$ .
Setting up $\det(A) = 0$ or $\det\begin{pmatrix} a & b \\ c & d \end{pmatrix} = 0$ instead of $\det\begin{pmatrix} a-\lambda & b \\ c & d-\lambda \end{pmatrix} = 0$ .	The characteristic equation must come from $(A - \lambda I)\mathbf{v} = \mathbf{0}$ ; without subtracting $\lambda I$ , you compute the determinant of the original matrix, which does not give eigenvalues.	Always form $A - \lambda I$ first, then set its determinant equal to zero. The $\lambda$ terms in the resulting polynomial are what you solve for.

Tips and when to use something else

Symmetric matrices (where $A = A^T$ ) always have real eigenvalues and orthogonal eigenvectors—useful for quick error checks.
The sum of all eigenvalues equals the trace of $A$ (sum of diagonal entries), and their product equals the determinant; use this to verify that you found all eigenvalues.
For just solving $Ax = b$ , don't use eigenvalues—instead apply Gaussian Elimination, which is faster and simpler for linear systems.
Eigenvector scaling does not matter: if $\mathbf{v}$ is an eigenvector for eigenvalue $\lambda$ , then so is $5\mathbf{v}$ or any nonzero scalar multiple.

Related formulas

Characteristic PolynomialSolve the characteristic polynomial equation to find a matrix's eigenvalues—the scaling factors that reveal how the matrix transforms vectors.Matrix MultiplicationMatrix multiplication combines rows from the first matrix with columns of the second to produce a new matrix, useful for solving systems and transformations.Determinant of a 2x2 MatrixA single number computed from a 2x2 matrix that tells whether the matrix is invertible and helps solve systems of linear equations.Linear IndependenceLinear Independence tells you when a set of vectors has no redundancy: none can be written as a combination of the others.Magnitude of a VectorMagnitude of a vector is its length or size, measured as the distance from the origin; use it to find how far something has traveled overall.

Frequently asked questions

How do I find eigenvectors once I have the eigenvalues?: For each eigenvalue $\lambda$ , solve $(A - \lambda I)\mathbf{v} = \mathbf{0}$ by row reduction or inspection. The solution set (the null space) gives you all eigenvectors for that eigenvalue. Pick any nonzero vector from that null space.
Why do I need both eigenvalues and eigenvectors?: Eigenvalues tell you the strength of transformation (the scaling factor); eigenvectors tell you the direction in which that scaling happens. Together they reveal how the matrix acts on preferred directions, essential for diagonalization, stability analysis, and solving systems of differential equations.
Can a matrix have no real eigenvalues?: Yes. A rotation matrix like $\begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}$ (a 90° rotation) has only complex eigenvalues, because rotating a vector always changes its direction for every direction—no vector is merely scaled.
What does it mean if I get a repeated eigenvalue?: A repeated eigenvalue (where the same value appears more than once in the characteristic polynomial) can mean the eigenvector space has fewer dimensions than expected. You may find fewer linearly independent eigenvectors than the multiplicity of that eigenvalue, a situation called defect or deficiency.

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Reviewed 2026-09-18