How this instrument works
A matrix A has directions that survive it unrotated — nonzero vectors v where the only effect is a change in length, captured by Av = λv. Each such v is an eigenvector, and the scale factor λ is its eigenvalue. Every other direction gets bent as well as stretched; eigenvectors are the axes a transformation leaves alone. For a 2×2 matrix, both eigenvalues fall out of one quadratic, λ² − (trace)λ + det = 0, built from nothing more than the four entries a, b, c, and d, where trace = a + d and det = ad − bc, then solved the ordinary way with the quadratic formula.
That quadratic is exactly where the characteristic-polynomial sheet elsewhere on this site stops, reporting trace and determinant and leaving the factoring to you. This instrument finishes the job, substituting into λ = (trace ± √(trace² − 4·det)) ⁄ 2 and returning both roots as actual numbers. A shortcut worth knowing by heart: whenever b equals c — a symmetric matrix — the discriminant trace² − 4·det works out to (a − d)² + 4b², an expression that can never go negative, guaranteeing real eigenvalues, and whenever the two come out distinct, their eigenvectors meet at a right angle.
Solving for λ is only half the story, and this sheet deliberately leaves the other half to hand-work: an eigenvector is not one vector but an entire line through the origin, since Av = λv holds for any scalar multiple of a solution just as well as the solution itself. Substitute a found λ back into (A − λI)v = 0 and solve. Two distinct eigenvalues usually hand you two independent directions, but a repeated root is not a guarantee of that — a shear like [[1, 1], [0, 1]] has one eigenvalue counted twice yet only a single eigenvector direction, a genuinely 'defective' matrix — a fate a scaled identity, where every direction qualifies, never suffers.
- Enter the four matrix entries in fields a (row 1, col 1), b (row 1, col 2), c (row 2, col 1), and d (row 2, col 2), read as [[a, b], [c, d]].
- Read Eigenvalue λ₁ (larger) — the bigger root of the characteristic equation, the larger of the two scaling factors along the matrix's own directions.
- Read Eigenvalue λ₂ (smaller) — the other root; the two multiply to the determinant and add to the trace, a quick sanity check on both.
- If a warning about no real eigenvalues appears, the matrix rotates rather than stretches — trace² is smaller than 4·det, and the roots form a complex-conjugate pair.
- To find the actual eigenvector for either root, substitute it back into (A − λI)v = 0 by hand — this sheet locates the scaling factors, not the vectors themselves.
Worked example — eigenvalues and eigenvectors of [[2, 1], [1, 2]]
This example plugs in a = 2 and b = 1 for the top row, then c = 1 and d = 2 for the bottom, giving the matrix [[2, 1], [1, 2]]. Trace comes to 2 + 2 = 4 and determinant to 2×2 − 1×1 = 3, so the quadratic reads λ² − 4λ + 3 = 0; substituting directly, λ = (4 ± √(16 − 12)) ⁄ 2 = (4 ± 2) ⁄ 2. Eigenvalue λ₁ (larger) reads 3, and Eigenvalue λ₂ (smaller) reads 1 — the two roots of (λ − 1)(λ − 3) = 0.
Neither root is the end of the story. For λ₁ = 3, solving (A − 3I)v = 0 gives [[−1, 1], [1, −1]]v = 0, satisfied by any multiple of (1, 1); check it by hand, A(1, 1) = (2+1, 1+2) = (3, 3), exactly three times the original vector. For λ₂ = 1, (A − I)v = 0 gives [[1, 1], [1, 1]]v = 0, satisfied by (1, −1); A(1, −1) = (2−1, 1−2) = (1, −1), a direction the matrix leaves entirely untouched. Because a = d and b = c here, the matrix is symmetric, and its two eigenvector directions meet at a right angle: (1, 1)·(1, −1) = 1×1 + 1×(−1) = 0, the orthogonality holding exactly as promised.
Questions
What exactly is an eigenvector?
A nonzero vector v that a matrix A sends to a scalar multiple of itself: Av = λv. The direction of v does not change under the transformation — only its length, scaled by the eigenvalue λ. Every other direction gets rotated as well as stretched; eigenvectors are the ones a matrix leaves alone.
How do you find the eigenvector once you already have an eigenvalue?
Substitute the eigenvalue back into (A − λI)v = 0 and solve the resulting linear system by hand. For [[2, 1], [1, 2]] with λ = 3, that system reduces to −v₁ + v₂ = 0, solved by any multiple of (1, 1); with λ = 1 it becomes v₁ + v₂ = 0, solved by any multiple of (1, −1).
Why does this calculator report two numbers, λ₁ and λ₂, rather than vector components?
Because an eigenvector is not a single point — any nonzero scalar multiple of a solution to (A − λI)v = 0 satisfies the equation equally well, so 'the' eigenvector is really a whole line through the origin. A numeric field can hold one number cleanly; it cannot hold a line. This sheet locates the two scaling factors precisely and leaves the direction, and how you normalize it, to you.
How does this relate to the characteristic polynomial?
λ² − (trace)λ + det = 0 is exactly the characteristic polynomial belonging to this matrix, and its two roots are λ₁ and λ₂. A separate characteristic-polynomial sheet on this site reports only trace and determinant, the polynomial's coefficients; this one carries that same equation one step further, through the quadratic formula, to the actual numbers.
Does a repeated eigenvalue always come with two independent eigenvector directions?
No — that failure is common enough to have its own name, 'defective'. Take [[7, 0], [0, 7]], seven times the identity: its eigenvalue 7 repeats, and because the matrix is a uniform scaling, every direction still qualifies as an eigenvector. A shear such as [[1, 1], [0, 1]] also repeats an eigenvalue, at 1, yet solving (A − I)v = 0 leaves only the single direction (1, 0) — one dimension short of a full set.
What is the most common hand-calculation mistake with this formula?
Losing a sign in the determinant term, ad − bc, especially when b or c is itself negative — a stray sign flips the discriminant and can turn two real eigenvalues into a complex pair or the reverse. Recompute trace and det separately before touching the square root, and check that λ₁ + λ₂ equals the trace as a fast sanity check.