Characteristic Polynomial

p_A(λ) = (λ I - A) — its roots are the eigenvalues.

The idea

The characteristic polynomial of a square matrix is the polynomial whose roots are the matrix's eigenvalues; it turns the search for eigenvalues into the familiar problem of finding the roots of a polynomial.

A number $\lambda$ is an eigenvalue of $A$ exactly when $(A - \lambda I)v = 0$ for some nonzero vector $v$, and a square matrix sends a nonzero vector to zero exactly when its determinant is $0$. Expanding $\det(\lambda I - A)$ produces a polynomial in $\lambda$, the characteristic polynomial $p_{A}(\lambda) = \det(\lambda I - A),$ and its roots are precisely the eigenvalues of $A$. For an $n \times n$ matrix it has degree $n$ and leading coefficient $1$.

Listing the eigenvalues with multiplicity factors it as $p_{A}(\lambda) = (\lambda - \lambda_{1}) \cdots (\lambda - \lambda_{n})$. Expanding this product and comparing coefficients shows that the eigenvalues sum to $\operatorname{tr} A$ — the trace, the sum of the diagonal entries — and multiply to $\det A$. For a $2 \times 2$ matrix these two numbers determine the whole polynomial: $p_{A}(\lambda) = \lambda^{2} - (\operatorname{tr} A)\,\lambda + \det A.$

Ways to work on it

• Walkthrough. Build p_A(λ) for a 2 × 2 matrix; read off sum and product of eigenvalues.
• Practice. Random 2 × 2: compute trace and determinant; identify what the constant term measures.
• Hardest. Read coefficients of p_A(λ) off a known eigenvalue list (size 3).

Not sure where to start? Take the ten-question placement test.