Search the complete library

What do you want to learn or calculate?

Quick linksAll calculatorsMath subjectsPractice questionsFormula library
← Linear Algebra formulas

Characteristic equation

Find eigenvalues as roots of a polynomial.

Linear Algebra · Eigenvalues
$$\det(A-\lambda I)=0$$

Characteristic equation is one of 2 eigenvalues formulas in the linear algebra section of this library, and it is used at university level.

Why characteristic equation works

Move everything to one side of Av = λv and it reads (A − λI)v = 0. A nonzero v solving that means the matrix A − λI crushes something to the origin, which happens exactly when it is not invertible, which happens exactly when its determinant is 0. So the determinant condition is a test for the eigenvalues, not a coincidence.

What each symbol means

$I$ is identity matrix and $\lambda$ is the unknown eigenvalue.

Characteristic equation: when it holds

$A$ must be square.

When it stops applying

The determinant is only defined for square matrices, so a rectangular A has no characteristic equation at all. Beyond about size 4 the polynomial route also stops being practical, since no general formula solves a degree-5 polynomial and small coefficient changes can move the roots a lot.

Characteristic equation: a worked example

For $A=\operatorname{diag}(2,3)$, $(2-\lambda)(3-\lambda)=0$.

The mistake to avoid

What people do: Students subtract λ from every entry of A instead of only from the diagonal.

Why it goes wrong: The I in the expression is the identity matrix, which holds ones only along its diagonal and zeros everywhere else, so λI has nothing to subtract off the other entries.

Do this instead: Subtract λ down the main diagonal alone. For A = [[2, 1], [1, 2]] you get (2 − λ)² − 1 = 0, giving λ = 1 and λ = 3.

Characteristic equation: step by step

  1. Name the unknown, and the unit the answer has to come out in.
  2. Match the symbols to your values. $I$ is identity matrix and $\lambda$ is the unknown eigenvalue.
  3. Check the conditions before substituting. $A$ must be square.
  4. Substitute, keep exact values to the last line, then test the sign, size, and unit against a rough estimate — the check that catches most linear algebra slips.

Where this formula fits

Subject
Linear Algebra formulas — 17 entries in this library
Topic
Eigenvalues
Level
University

Formulas are easiest to keep when they sit inside a method rather than on a list. Use the links below to see where characteristic equation comes from, to check a calculation against a tool, and to practise it until you can recall it without looking.

Questions about characteristic equation

How many eigenvalues does an n by n matrix have?

Exactly n, once you count complex values and repeats, because the determinant expands into a degree-n polynomial in λ. A 3 by 3 might show only two distinct values if one of them is a double root.

Does solving this give me the eigenvectors too?

No, it gives only the eigenvalues. Substitute each λ back into (A − λI)v = 0 and solve that system for v, which typically leaves one free variable and therefore a whole direction.

What if the roots turn out complex?

That is normal for matrices with rotation in them, and for real matrices the complex roots always come in conjugate pairs. The eigenvectors are complex too, and together they describe a rotating-and-scaling action.

Is there a shortcut for a 2 by 2?

Yes: the polynomial is always λ² − (trace)λ + determinant = 0. For [[2, 1], [1, 2]] that reads λ² − 4λ + 3 = 0, which factors to give 1 and 3 without expanding anything.

Stuck on a problem?

Work a characteristic equation problem step by step

Type your own problem, or upload a photo of it. You get the method, the answer, and a check you can repeat yourself.