Eigenvalues and Eigenvectors Calculator
This eigenvalues and eigenvectors calculator gives exact answers for a 2×2 or 3×3 matrix — fractions and surds, never a rounded decimal.
Type 1 2; 0 3 and you get λ = 1 with eigenvector (1, 0), and λ = 3 with (1, 1).
What an Eigenvector Actually Is
Multiply a vector by a matrix and it usually comes out pointing somewhere else. An eigenvector is a direction that survives — it may be stretched, squashed or flipped, but it still lies along the same line it started on.
The eigenvalue is how much it was stretched by. Two, and the vector doubles in length. A half, and it shrinks. Minus one, and it flips end for end without changing size.
Av = λv
That is the whole definition, and everything below is the business of finding which v and which λ satisfy it.
Getting From That to Something You Can Solve
Av = λv looks unhelpful because there are two unknowns tangled together. Move everything to one side and it improves:
(A − λI)v = 0
Now the question is: for which λ does the matrix A − λI send some non-zero vector to zero? A matrix does that only when its determinant vanishes. So:
det(A − λI) = 0
Expand that and you have a polynomial in λ — the characteristic equation. Its roots are the eigenvalues, and there is nothing else to find them from. The determinant calculator handles that determinant on its own if you want to see it built by hand.
Exact, Not 1.618
Here is the thing this page is really for.
Take the matrix 1 1; 1 0. Its characteristic equation is λ² − λ − 1 = 0, so its eigenvalues are (1 ± √5)/2 — the golden ratio and its partner, exactly. Every calculator I could find returns 1.618 and −0.618.
Those decimals are the same numbers with the interesting part filed off. They cannot be checked by hand, they will not match a mark scheme, and they hide the fact that the answer was something worth recognising. Pearson's own page states that for 3×3 matrices it solves the characteristic polynomial numerically; another well-known one offers a menu of significant digits. This page keeps the surd.
| Matrix | Eigenvalues |
|---|---|
| 1 2; 0 3 | 1 and 3 |
| 1 1; 1 0 | (1 ± √5)/2 |
| 1 1; 0 1 | 1, twice — and defective |
| 0 −1; 1 0 | ±i — no real ones |
Two Checks That Cost Nothing
Before you go anywhere near the eigenvectors, check the eigenvalues against two numbers you already have.
They must add up to the trace — the sum down the main diagonal. And they must multiply to the determinant. Both follow from the characteristic polynomial's coefficients, and both take a few seconds.
For 1 2; 0 3: the eigenvalues 1 and 3 add to 4, which is 1 + 3 down the diagonal. They multiply to 3, which is the determinant. Two ticks, and an error would have shown up before it spread into four eigenvector calculations.
Finding the Directions
Once you have an eigenvalue, put it back in. A − λI becomes an ordinary matrix of numbers, and (A − λI)v = 0 becomes an ordinary system to solve by row reduction.
The answer that comes out is a direction, not a single vector. If (1, 2) is an eigenvector then so is (2, 4), and so is (−3, −6). This page gives the smallest whole numbers that work, with the first non-zero entry positive, because a convention is needed and that one is the least surprising. If your textbook scaled it differently, both of you are right.
The row reduction itself is the same operation as anywhere else in linear algebra, and the reduced row echelon form calculator is the page for watching it happen step by step.
When a Repeat Is Not a Problem, and When It Is
An eigenvalue can appear twice. Whether that matters depends on something the repeat itself does not tell you.
The matrix 3 0; 0 3 has λ = 3 twice, and two independent eigenvectors to go with it — every direction is an eigenvector, since the matrix just triples everything. Perfectly well behaved.
The matrix 1 1; 0 1 also has a repeated eigenvalue, λ = 1 twice. But it has only one eigenvector. There are not enough directions to build a basis from, and a matrix in that position is called defective: it cannot be diagonalised, no matter what you do.
Counting roots does not settle it. You have to count eigenvectors, and this page does.
Why Anyone Cares
Because a diagonalizable matrix is one you can change your point of view on and make simple, and having enough eigenvectors is exactly what makes it one.
Put the eigenvectors in the columns of a matrix P and the eigenvalues down the diagonal of D, and A = PDP⁻¹. In that basis the matrix does nothing but stretch along its own axes.
Raising it to a power becomes raising a diagonal matrix to a power, which is raising a handful of numbers to a power — the whole reason eigenvalues turn up in everything from population models to vibration analysis to the algorithm that ranked this page. The one piece of work that formula still asks for is P⁻¹, and the inverse matrix calculator is the page for that half.
Complex Eigenvalues Mean Rotation
The matrix 0 −1; 1 0 rotates the plane a quarter turn. Its eigenvalues are ±i, and it has no real eigenvectors at all.
That is not a failure of the method — it is the method telling you the truth. If every direction gets turned, then no direction survives, and there is nothing real for an eigenvector to be. Complex eigenvalues always arrive in conjugate pairs, which is also why a 3×3 matrix always has at least one real eigenvalue: three roots cannot pair up completely.
How to Use This Eigenvalues and Eigenvectors Calculator
Type the rows separated by semicolons: 1 2; 0 3. Numbers within a row can be separated by spaces or commas, and a pasted [[1,2],[0,3]] works too.
Entries may be negative, decimal or fractions. Sizes 2×2 and 3×3 are covered, which is where exact answers exist — past that the characteristic polynomial reaches degree four and exact roots stop being available in general.
Eigenvalues and Eigenvectors Calculator FAQ
What is an eigenvector?
How do you find eigenvalues?
How many eigenvalues does a matrix have?
Can I check my eigenvalues quickly?
Why is my eigenvector different from the book's?
What does defective mean?
What do complex eigenvalues mean?
Why bother with surds instead of decimals?
Expand det(A − λI), solve it, then put each root back and row reduce. Check against the trace and the determinant before you trust anything, and count eigenvectors rather than roots when the question is whether it diagonalises.
