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Characteristic Polynomial Calculator

Solve characteristic polynomial problems with clear steps, notation, and a final check.

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Evaluate the governing formula locally in your browser. Define each known quantity once, then change values to test another case instantly.

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Algebra

Characteristic Polynomial Calculator explained

The short version

  • The characteristic polynomial is what you get when you compute det(A − λI) and leave λ as a letter.
  • Setting it to zero gives the eigenvalues: the stretch factors along the directions a matrix does not turn.
  • For any 2×2 matrix there is a shortcut: λ² − (trace)λ + (determinant).

The formula this page uses

p(λ) = det(A − λI) for 2×2: p(λ) = λ² − (trace A)·λ + det A

What each part means

SymbolWhat it means
A — The square matrixMust have the same number of rows as columns, or det(A − λI) is not defined.
λ — EigenvalueThe unknown stretch factor. A pure number with no unit.
I — Identity matrix1s down the diagonal, 0s elsewhere, so λI subtracts λ from each diagonal entry only.
trace A — TraceThe sum of the diagonal entries. It equals the sum of the eigenvalues.

Show your work: a full example

  1. The matrixA = [[4, 1], [2, 3]]
  2. Subtract λ down the diagonalA − λI = [[4 − λ, 1], [2, 3 − λ]]
  3. Take the determinant(4 − λ)(3 − λ) − (1)(2)
  4. Expand12 − 7λ + λ² − 2 = λ² − 7λ + 10
  5. Cross-check with the shortcuttrace = 4 + 3 = 7 and det = 12 − 2 = 10, giving λ² − 7λ + 10
  6. Set it to zero and factor(λ − 2)(λ − 5) = 0, so λ = 2 and λ = 5
  7. Check2 + 5 = 7 matches the trace and 2 × 5 = 10 matches the determinant

A second, different case

  1. A different case: a quarter-turn matrixB = [[0, −1], [1, 0]]
  2. Trace0 + 0 = 0
  3. Determinant(0)(0) − (−1)(1) = 1
  4. Apply the shortcutp(λ) = λ² − 0·λ + 1 = λ² + 1
  5. Solveλ² = −1, so λ = i and λ = −i
  6. Check the discriminant0² − 4(1)(1) = −4, negative, so no real eigenvalue exists
  7. Read it geometricallyB rotates every vector by 90°, so no direction is left pointing where it started
Copy-ready example

A = [[4, 1], [2, 3]]

The matrix

2×2 matrices, their traces and determinants, and the polynomial that follows

Matrix Atracedetp(λ)Eigenvalues
[[4, 1], [2, 3]]710λ² − 7λ + 102 and 5
[[3, 0], [0, −1]]2−3λ² − 2λ − 33 and −1
[[2, 0], [0, 2]]44λ² − 4λ + 42, repeated twice
[[1, 1], [0, 1]]21λ² − 2λ + 11, repeated twice
[[0, −1], [1, 0]]01λ² + 1i and −i

Three mistakes to check for

What students writeWhy it's wrongDo this instead
p(λ) = det(A) − λλ must be subtracted from every diagonal entry before the determinant is taken, not from the finished number.det([[4 − λ, 1], [2, 3 − λ]]) = λ² − 7λ + 10.
With trace 7, writing λ² + 7λ + 10The trace enters the shortcut with a minus sign, so a positive trace gives a negative middle coefficient.λ² − 7λ + 10, whose roots 2 and 5 add back to 7.
A negative discriminant means the matrix was entered wrongReal matrices routinely have complex eigenvalues, and every rotation does.λ² + 1 gives λ = ±i, which is the correct answer for a 90° turn.

Questions about the Characteristic Polynomial Calculator

Why is it called the characteristic polynomial?

Because it is a fingerprint that survives a change of basis. Two matrices that describe the same transformation in different coordinate systems produce the identical polynomial, so its coefficients describe the transformation rather than the numbers you happened to write down.

Does row reducing A first make the calculation easier?

No, it breaks it. Row operations change eigenvalues, so the polynomial you get afterwards belongs to a different matrix. Expand det(A − λI) directly, or use the trace-and-determinant shortcut for a 2×2.

Is the 2×2 shortcut worth memorising?

Yes, it turns a determinant expansion into two glances. Sum the diagonal for the trace, cross-multiply for the determinant, and write λ² − (trace)λ + det. The 3×3 version exists too but involves the sum of the three 2×2 minors.

Can two different matrices have the same characteristic polynomial?

Yes. [[2, 0], [0, 2]] and [[1, 1], [0, 1]] are genuinely different transformations, yet the table above shows both give a polynomial with a single repeated root. The polynomial pins down eigenvalues but not the full behaviour.

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