Eigenvalue Calculator

About the Eigenvalue Calculator

Eigenvalues and eigenvectors describe the directions a matrix leaves unchanged and how much it stretches them. They are behind Google's original PageRank algorithm, the vibration modes of bridges and molecules, principal component analysis in statistics, the stability of differential equations and the energy levels in quantum mechanics. In a linear algebra course they are usually the topic where everything learned so far — determinants, row reduction and null spaces — comes together.

This calculator follows the textbook method. It forms the characteristic polynomial $\det(\lambda I - A)$ , finds its roots exactly using the rational root theorem, the quadratic formula and polynomial division, and then solves $(A - \lambda I)\mathbf v = \mathbf 0$ for each eigenvalue to produce eigenvectors with simple integer entries wherever possible.

Irrational eigenvalues are kept in exact radical form, complex eigenvalues are reported as conjugate pairs, and repeated eigenvalues come with both their algebraic and geometric multiplicities, so you can tell at a glance whether the matrix is diagonalisable. Matrices from 2×2 to 4×4 are supported, which covers essentially every homework and exam problem.

What are eigenvalues and eigenvectors?

A nonzero vector $\mathbf v$ is an eigenvector of a square matrix $A$ if multiplying by $A$ only scales it: $A\mathbf v = \lambda \mathbf v$ for some number $\lambda$ . That number is the eigenvalue belonging to $\mathbf v$ . Most vectors change direction when multiplied by a matrix; eigenvectors are the special ones that stay on their own line through the origin.

Rewriting the definition as $(A - \lambda I)\mathbf v = \mathbf 0$ shows how to find them. A nonzero solution exists only when $A - \lambda I$ is singular, which means its determinant is zero. The equation $\det(A - \lambda I) = 0$ is called the characteristic equation; for an $n \times n$ matrix it is a polynomial of degree $n$ , so there are exactly $n$ eigenvalues counted with multiplicity, some of which may be complex.

Every eigenvalue has infinitely many eigenvectors, because any nonzero multiple of an eigenvector is another one. Answers are therefore given as a representative vector, or a basis when there are several independent directions, and your textbook may show a different multiple of the same vector.

How to use the Eigenvalue Calculator

Type the matrix one row per line, separating entries with spaces, for example 4 1 on the first line and 2 3 on the second.
You can also put a whole matrix on one line with semicolons between rows, such as 4 1; 2 3.
Entries can be integers, decimals, fractions or negative numbers. The matrix must be square, from 2×2 to 4×4.
Press Find eigenvalues to see the characteristic polynomial, its roots and the eigenvector calculation for each eigenvalue.
Check the multiplicities reported for repeated eigenvalues if you need to know whether the matrix is diagonalisable.

Formulas

Eigenvalue equation

A\mathbf v = \lambda \mathbf v,\quad \mathbf v \ne \mathbf 0

Characteristic equation

\det(\lambda I - A) = 0

2×2 shortcut

\lambda^2 - (\operatorname{tr} A)\,\lambda + \det A = 0

Trace and determinant

\lambda_1 + \cdots + \lambda_n = \operatorname{tr} A,\quad \lambda_1 \cdots \lambda_n = \det A

Diagonalisation

A = PDP^{-1},\quad D = \operatorname{diag}(\lambda_1, \ldots, \lambda_n)

Worked example: 2×2, rational

The worked example below is computed live for $A = \begin{bmatrix}4 & 1 \\ 2 & 3\end{bmatrix}$ . The trace is 7 and the determinant is $4\cdot3 - 1\cdot2 = 10$ , so the characteristic polynomial is $\lambda^2 - 7\lambda + 10 = (\lambda - 5)(\lambda - 2)$ and the eigenvalues are 5 and 2.

For $\lambda = 5$ , the matrix $A - 5I$ has rows $(-1, 1)$ and $(2, -2)$ , whose null space is spanned by $\langle 1, 1 \rangle$ . For $\lambda = 2$ , $A - 2I$ has rows $(2, 1)$ and $(2, 1)$ , giving $\langle -1, 2 \rangle$ . Checking one is quick: $A\langle 1, 1\rangle = \langle 5, 5\rangle$ , exactly five times the vector.

The solution below is generated by the same engine that powers the calculator above.

Final answer

\lambda = 5,\ 2

How we read your problem

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

\begin{bmatrix}4 & 1 \\ 2 & 3\end{bmatrix}

Step-by-step solution

Set up the characteristic equation
$\det(\lambda I - A) = 0,\quad A = \begin{bmatrix}4 & 1 \\ 2 & 3\end{bmatrix}$
For a 2×2 matrix it is λ² − (trace)λ + det
$\lambda^2 - (7)\lambda + (10) = 0$
Characteristic polynomial
$p(\lambda) = \lambda^{2} - 7\lambda + 10$
Rational roots found with the rational root theorem
$\lambda = 2,\ \lambda = 5$
Eigenvectors for λ = 5: solve (A − λI)v = 0
$\begin{bmatrix}-1 & 1 \\ 2 & -2\end{bmatrix}\,v = 0 \Rightarrow v = \langle 1, 1 \rangle$
Eigenvectors for λ = 2: solve (A − λI)v = 0
$\begin{bmatrix}2 & 1 \\ 2 & 1\end{bmatrix}\,v = 0 \Rightarrow v = \langle -1, 2 \rangle$

Formulas used

Characteristic equation
$\det(\lambda I - A) = 0$
Eigenvector equation
$(A - \lambda I)\,v = 0,\ v \neq 0$

The eigenvalues are the roots of the characteristic polynomial. Eigenvectors: λ = 5: ⟨1, 1⟩; λ = 2: ⟨-1, 2⟩.

Verification

Sum of eigenvalues ≈ 7 equals the trace 7.
Product of eigenvalues equals the determinant 10.

Detailed explanation

Step 1: the characteristic polynomial

For a 2×2 matrix the polynomial can be written down immediately as $\lambda^2 - (\text{trace})\lambda + \det$ . For larger matrices the calculator expands $\det(\lambda I - A)$ by cofactors. The 3×3 symmetric matrix with rows $(2, 0, 0)$ , $(0, 3, 4)$ and $(0, 4, 9)$ gives $\lambda^3 - 14\lambda^2 + 35\lambda - 22$ .

Two quick checks catch most arithmetic errors. The coefficient of $\lambda^{n-1}$ is minus the trace, here $-(2 + 3 + 9) = -14$ , and the constant term is $(-1)^n \det A$ , here $-22$ because the determinant is 22.

Step 2: finding the roots

With integer entries, any rational eigenvalue must divide the constant term, so the calculator tests the divisors of 22 and finds $\lambda = 1$ , then divides the polynomial by $(\lambda - 1)$ and factors the quadratic that remains. The eigenvalues are 11, 2 and 1, and their sum, 14, matches the trace as it must.

When the remaining quadratic does not factor over the rationals, the quadratic formula gives exact radicals. For $\begin{bmatrix}1 & 2 \\ 3 & 4\end{bmatrix}$ , the polynomial $\lambda^2 - 5\lambda - 2$ has roots $\frac52 \pm \frac{\sqrt{33}}{2}$ , approximately 5.372 and $-0.372$ . If the discriminant is negative the roots are complex conjugates; the rotation matrix $\begin{bmatrix}0 & -1 \\ 1 & 0\end{bmatrix}$ has $\lambda^2 + 1 = 0$ , so $\lambda = \pm i$ , reflecting the fact that a 90° rotation leaves no real direction unchanged.

Step 3: eigenvectors from the null space

For each eigenvalue, substitute it into $A - \lambda I$ and solve the homogeneous system by row reduction. Because the matrix is singular, at least one row reduces to zeros, leaving a free variable. Setting the free variable to a convenient value and back-substituting gives the eigenvector, which the calculator scales to small integers.

For $\lambda = 11$ in the 3×3 example, $A - 11I$ has rows $(-9, 0, 0)$ , $(0, -8, 4)$ and $(0, 4, -2)$ . The first row forces $v_1 = 0$ and the others both say $v_3 = 2v_2$ , so $\mathbf v = \langle 0, 1, 2 \rangle$ . Symmetric matrices like this one always have real eigenvalues and mutually perpendicular eigenvectors, which you can confirm with the dot product.

Repeated eigenvalues and diagonalisation

An eigenvalue that appears $k$ times as a root has algebraic multiplicity $k$ . The number of independent eigenvectors it has is its geometric multiplicity, which can be smaller. Both $\begin{bmatrix}2 & 1 \\ 0 & 2\end{bmatrix}$ and $2I$ have the double eigenvalue 2, but the first has only one eigenvector direction, $\langle 1, 0 \rangle$ , while the second has two, $\langle 1, 0 \rangle$ and $\langle 0, 1 \rangle$ .

A matrix can be written as $PDP^{-1}$ , with the eigenvalues on the diagonal of $D$ and the eigenvectors as the columns of $P$ , exactly when every eigenvalue's geometric multiplicity equals its algebraic multiplicity. Diagonalisation makes powers easy, since $A^k = PD^kP^{-1}$ , and that is how eigenvalues predict long-term behaviour in population models and Markov chains.

Shortcuts worth knowing

The eigenvalues of a triangular matrix are its diagonal entries, so the matrix with rows $(1, 2, 3)$ , $(0, 4, 5)$ and $(0, 0, 6)$ has eigenvalues 6, 4 and 1 without any calculation. A matrix is invertible exactly when 0 is not an eigenvalue, and the eigenvalues of $A^{-1}$ are the reciprocals of those of $A$ , with the same eigenvectors.

Simple changes to a matrix move its eigenvalues in predictable ways. Adding $c$ times the identity adds $c$ to every eigenvalue, multiplying the matrix by a scalar multiplies every eigenvalue by that scalar, and raising it to the power $k$ raises every eigenvalue to the power $k$ , all without changing the eigenvectors. A matrix and its transpose share the same eigenvalues, though usually not the same eigenvectors. These facts let you check answers and predict results before computing anything.

Common mistakes

Computing $\det(A) - \lambda$ instead of $\det(A - \lambda I)$ ; $\lambda$ must be subtracted from every diagonal entry.
Accepting the zero vector as an eigenvector; eigenvectors must be nonzero by definition.
Making an arithmetic slip in the polynomial and then finding a null space containing only the zero vector.
Assuming every matrix has real eigenvalues; rotations and many non-symmetric matrices have complex ones.
Thinking an eigenvector is wrong because it differs from the textbook by a constant multiple.
Concluding a matrix is diagonalisable without checking the geometric multiplicity of repeated eigenvalues.

Tips

Check that the eigenvalues add up to the trace and multiply to the determinant.
Verify each eigenvector by computing $A\mathbf v$ and comparing it with $\lambda\mathbf v$ .
Read eigenvalues straight off the diagonal of triangular matrices.
For 3×3 matrices, expand the determinant along a row or column with the most zeros.
Use the determinant calculator to confirm that $A - \lambda I$ is singular for each eigenvalue.
Expect real eigenvalues and perpendicular eigenvectors whenever the matrix is symmetric.
Factor out common numbers from a matrix first; the eigenvalues simply scale by that factor.

Frequently asked questions

How do I enter a matrix into the eigenvalue calculator?

Type one row per line with spaces between entries, or put all rows on one line separated by semicolons, such as 4 1; 2 3. Fractions, decimals and negative numbers are accepted.

What size matrices are supported?

Square matrices from 2×2 up to 4×4. Eigenvalues are only defined for square matrices, so a 3×2 matrix, for example, is rejected with an explanation.

Can the calculator find complex eigenvalues?

Yes. When the characteristic polynomial has a negative discriminant, the eigenvalues are reported as a complex-conjugate pair, such as λ = i and −i for a 90° rotation matrix.

Why is my eigenvector different from the one in my textbook?

Any nonzero multiple of an eigenvector is also an eigenvector. ⟨1, 1⟩ and ⟨3, 3⟩ describe the same direction. Check that one vector is a multiple of the other.

What do algebraic and geometric multiplicity mean?

Algebraic multiplicity is how many times an eigenvalue repeats as a root of the characteristic polynomial. Geometric multiplicity is how many independent eigenvectors it has. The calculator reports both for repeated eigenvalues.

Does the calculator give exact or decimal eigenvalues?

Exact whenever possible: integers, fractions and radicals such as 5/2 ± √33/2. Only when the roots of a cubic or quartic cannot be found exactly does it fall back to accurate decimal values.

How are eigenvalues related to the determinant and trace?

The product of all eigenvalues equals the determinant and their sum equals the trace. For the matrix with rows (4, 1) and (2, 3), 5 × 2 = 10 is the determinant and 5 + 2 = 7 is the trace.

What does an eigenvalue of zero mean?

The matrix is singular: it has no inverse and its determinant is zero. The eigenvectors for λ = 0 span the null space of the matrix.

Where are eigenvalues used outside the classroom?

In vibration analysis, stability of control systems, principal component analysis, quantum mechanics, Markov chains and search-engine ranking, wherever repeated application of a linear transformation needs to be understood.

Find your matrix's eigenvalues

Enter a square matrix above to get the characteristic polynomial, exact eigenvalues and an eigenvector for each one, with every step shown.

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