Characteristic Polynomial

The characteristic polynomial of a matrix $A$ encodes all eigenvalue information in a single polynomial. Its roots are the eigenvalues, its coefficients are symmetric functions of the eigenvalues, and it is invariant under similarity -- making it one of the most important invariants in linear algebra.

Definition

Definition5.5Characteristic polynomial

Let $A \in M_{n \times n}(F)$ . The characteristic polynomial of $A$ is:

$p_A(\lambda) = \det(A - \lambda I).$

This is a polynomial of degree $n$ in $\lambda$ with leading coefficient $(-1)^n$ :

$p_A(\lambda) = (-1)^n \lambda^n + (-1)^{n-1} \operatorname{tr}(A) \lambda^{n-1} + \cdots + \det(A).$

The eigenvalues of $A$ are exactly the roots of $p_A(\lambda) = 0$ .

RemarkSign convention

Some authors define $p_A(\lambda) = \det(\lambda I - A)$ instead, which gives a monic polynomial (leading coefficient $1$ ). In this convention, $p_A(\lambda) = \lambda^n - \operatorname{tr}(A)\lambda^{n-1} + \cdots + (-1)^n \det(A)$ . We will follow the convention $\det(A - \lambda I)$ but note when the monic form is used.

Computing characteristic polynomials

ExampleGeneral 2x2 matrix

For $A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$ :

$p_A(\lambda) = \det \begin{pmatrix} a - \lambda & b \\ c & d - \lambda \end{pmatrix} = (a - \lambda)(d - \lambda) - bc = \lambda^2 - (a+d)\lambda + (ad - bc).$

So $p_A(\lambda) = \lambda^2 - \operatorname{tr}(A)\lambda + \det(A)$ , and the eigenvalues are:

$\lambda = \frac{\operatorname{tr}(A) \pm \sqrt{\operatorname{tr}(A)^2 - 4\det(A)}}{2}.$

ExampleA specific 3x3 matrix

Let $A = \begin{pmatrix} 2 & 1 & 0 \\ 0 & 3 & 1 \\ 0 & 0 & 2 \end{pmatrix}$ . Since $A$ is upper triangular:

$p_A(\lambda) = (2 - \lambda)(3 - \lambda)(2 - \lambda) = -({\lambda - 2})^2(\lambda - 3).$

Eigenvalues: $\lambda = 2$ (algebraic multiplicity $2$ ) and $\lambda = 3$ (algebraic multiplicity $1$ ).

ExampleCompanion matrix

The companion matrix of the polynomial $p(x) = x^n + a_{n-1}x^{n-1} + \cdots + a_1 x + a_0$ is:

$C = \begin{pmatrix} 0 & 0 & \cdots & 0 & -a_0 \\ 1 & 0 & \cdots & 0 & -a_1 \\ 0 & 1 & \cdots & 0 & -a_2 \\ \vdots & & \ddots & & \vdots \\ 0 & 0 & \cdots & 1 & -a_{n-1} \end{pmatrix}.$

The characteristic polynomial of $C$ (in the monic convention) is exactly $p(x)$ . This proves that every monic polynomial is the characteristic polynomial of some matrix.

For $p(x) = x^3 - 6x^2 + 11x - 6 = (x-1)(x-2)(x-3)$ : the companion matrix $C = \begin{pmatrix} 0 & 0 & 6 \\ 1 & 0 & -11 \\ 0 & 1 & 6 \end{pmatrix}$ has eigenvalues $1, 2, 3$ .

Coefficients of the characteristic polynomial

Theorem5.1Coefficients and symmetric functions

Let $A \in M_{n \times n}(F)$ with eigenvalues $\lambda_1, \ldots, \lambda_n$ (over the algebraic closure). In the monic convention $\det(\lambda I - A) = \prod_{i=1}^n (\lambda - \lambda_i)$ , the coefficients are the elementary symmetric polynomials:

$\det(\lambda I - A) = \lambda^n - e_1 \lambda^{n-1} + e_2 \lambda^{n-2} - \cdots + (-1)^n e_n,$

where:

$e_1 = \sum_i \lambda_i = \operatorname{tr}(A)$ ,
$e_2 = \sum_{i < j} \lambda_i \lambda_j$ ,
$e_k = \sum_{i_1 < \cdots < i_k} \lambda_{i_1} \cdots \lambda_{i_k}$ ,
$e_n = \lambda_1 \cdots \lambda_n = \det(A)$ .

ExampleSymmetric functions for a 3x3 matrix

Let $A$ have eigenvalues $1, 2, 4$ . Then:

$e_1 = 1 + 2 + 4 = 7 = \operatorname{tr}(A)$ .
$e_2 = 1 \cdot 2 + 1 \cdot 4 + 2 \cdot 4 = 14$ .
$e_3 = 1 \cdot 2 \cdot 4 = 8 = \det(A)$ .

Characteristic polynomial (monic): $\lambda^3 - 7\lambda^2 + 14\lambda - 8 = (\lambda - 1)(\lambda - 2)(\lambda - 4)$ .

ExampleThe coefficient e_2 as sum of 2x2 minors

The coefficient $e_2$ can be computed directly from $A$ as the sum of all $2 \times 2$ principal minors:

$e_2 = \sum_{i < j} \det \begin{pmatrix} a_{ii} & a_{ij} \\ a_{ji} & a_{jj} \end{pmatrix}.$

For $A = \begin{pmatrix} 1 & 2 & 3 \\ 0 & 4 & 5 \\ 0 & 0 & 6 \end{pmatrix}$ : $e_2 = \det\begin{pmatrix}1&2\\0&4\end{pmatrix} + \det\begin{pmatrix}1&3\\0&6\end{pmatrix} + \det\begin{pmatrix}4&5\\0&6\end{pmatrix} = 4 + 6 + 24 = 34$ .

Check: eigenvalues $1, 4, 6$ , so $e_2 = 1 \cdot 4 + 1 \cdot 6 + 4 \cdot 6 = 4 + 6 + 24 = 34$ .

Similarity invariance

Theorem5.2Characteristic polynomial is a similarity invariant

If $B = P^{-1}AP$ for some invertible $P$ , then $p_B(\lambda) = p_A(\lambda)$ .

In particular, similar matrices have the same eigenvalues (with the same algebraic multiplicities), the same trace, and the same determinant.

ProofProof of similarity invariance

$\det(B - \lambda I) = \det(P^{-1}AP - \lambda I) = \det(P^{-1}(A - \lambda I)P) = \det(P^{-1}) \det(A - \lambda I) \det(P) = \det(A - \lambda I)$ .

■

ExampleSimilar matrices share characteristic polynomial

Let $A = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}$ and $P = \begin{pmatrix} 1 & 1 \\ i & -i \end{pmatrix}$ (over $\mathbb{C}$ ). Then $P^{-1}AP = \begin{pmatrix} i & 0 \\ 0 & -i \end{pmatrix} = B$ .

Both have $p(\lambda) = \lambda^2 + 1$ , eigenvalues $\pm i$ .

ExampleSame eigenvalues but not similar

$A = \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}$ and $B = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = I$ both have characteristic polynomial $(\lambda - 1)^2$ and eigenvalues $\{1, 1\}$ .

But $A \neq I$ while $B = I$ , and since $P^{-1}IP = I$ for all $P$ , the matrices $A$ and $B$ are not similar. The characteristic polynomial alone does not determine the similarity class.

Characteristic polynomial of a linear transformation

Definition5.6Characteristic polynomial of a linear map

For a linear transformation $T: V \to V$ on a finite-dimensional vector space, the characteristic polynomial is defined as $p_T(\lambda) = \det([T]_\mathcal{B} - \lambda I)$ for any basis $\mathcal{B}$ of $V$ .

This is well-defined (independent of the basis choice) by the similarity invariance theorem.

ExampleDifferentiation operator on polynomials

Let $V = \{p(x) : \deg p \leq 2\}$ with basis $\{1, x, x^2\}$ , and $T = d/dx$ .

The matrix of $T$ : $T(1) = 0$ , $T(x) = 1$ , $T(x^2) = 2x$ , so $[T] = \begin{pmatrix} 0 & 1 & 0 \\ 0 & 0 & 2 \\ 0 & 0 & 0 \end{pmatrix}$ .

Characteristic polynomial: $p_T(\lambda) = (-\lambda)^3 = -\lambda^3$ . Only eigenvalue: $\lambda = 0$ with $m_a = 3$ . The eigenspace is $\ker T = \operatorname{span}\{1\}$ , so $m_g = 1$ . This confirms $T$ is nilpotent ( $T^3 = 0$ ).

ExampleShift operator

On $V = \mathbb{R}^3$ , the right shift operator $S(a, b, c) = (0, a, b)$ has matrix $[S] = \begin{pmatrix} 0 & 0 & 0 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \end{pmatrix}$ .

Characteristic polynomial: $-\lambda^3$ . Eigenvalue $\lambda = 0$ with $m_a = 3$ , $m_g = \dim\ker S = 1$ (kernel is $\operatorname{span}\{(0, 0, 1)\}$ ).

Factoring characteristic polynomials

ExampleFactoring over Q using rational root theorem

$A = \begin{pmatrix} 1 & 2 & 0 \\ 0 & 1 & 2 \\ 2 & 0 & 1 \end{pmatrix}$ .

Characteristic polynomial (monic): $\lambda^3 - 3\lambda^2 + 3\lambda + 7$ ... let us compute carefully.

$\det(\lambda I - A) = \det \begin{pmatrix} \lambda - 1 & -2 & 0 \\ 0 & \lambda - 1 & -2 \\ -2 & 0 & \lambda - 1 \end{pmatrix}$ .

Expanding along the first row: $(\lambda - 1)[(\lambda-1)^2 - 0] + 2[0 - 4] = (\lambda-1)^3 - 8$ .

So $p(\lambda) = (\lambda - 1)^3 - 8$ . Setting $u = \lambda - 1$ : $u^3 = 8$ , so $u = 2$ , giving $\lambda = 3$ . The other roots (over $\mathbb{C}$ ) are $u = 2\omega, 2\omega^2$ where $\omega = e^{2\pi i/3}$ .

Over $\mathbb{R}$ : only $\lambda = 3$ is a real eigenvalue.

ExampleIrreducible quadratic factor

$A = \begin{pmatrix} 0 & 1 & 0 \\ -1 & 0 & 0 \\ 0 & 0 & 2 \end{pmatrix}$ .

$p_A(\lambda) = \det \begin{pmatrix} -\lambda & 1 & 0 \\ -1 & -\lambda & 0 \\ 0 & 0 & 2 - \lambda \end{pmatrix} = (2 - \lambda)(\lambda^2 + 1)$ .

Over $\mathbb{R}$ : eigenvalue $\lambda = 2$ only. Over $\mathbb{C}$ : eigenvalues $2, i, -i$ .

The factor $\lambda^2 + 1$ is irreducible over $\mathbb{R}$ but splits as $(\lambda - i)(\lambda + i)$ over $\mathbb{C}$ .

ExampleRepeated roots and their meaning

$A = \begin{pmatrix} 5 & 4 & 2 \\ 4 & 5 & 2 \\ 2 & 2 & 2 \end{pmatrix}$ .

$\operatorname{tr}(A) = 12$ , and one can compute $p_A(\lambda) = -\lambda^3 + 12\lambda^2 - 21\lambda + 10$ , or in monic form: $\lambda^3 - 12\lambda^2 + 21\lambda - 10 = (\lambda - 1)^2(\lambda - 10)$ .

Eigenvalues: $\lambda = 1$ ( $m_a = 2$ ) and $\lambda = 10$ ( $m_a = 1$ ).

The discriminant

Definition5.7Discriminant of the characteristic polynomial

For a $2 \times 2$ matrix $A$ with $p_A(\lambda) = \lambda^2 - \operatorname{tr}(A)\lambda + \det(A)$ , the discriminant is:

$\Delta = \operatorname{tr}(A)^2 - 4\det(A).$

$\Delta > 0$ : two distinct real eigenvalues.
$\Delta = 0$ : one repeated real eigenvalue.
$\Delta < 0$ : two complex conjugate eigenvalues (no real eigenvalues).

ExampleThree cases of the discriminant

$A = \begin{pmatrix} 3 & 1 \\ 0 & 1 \end{pmatrix}$ : $\Delta = 16 - 12 = 4 > 0$ . Eigenvalues: $3, 1$ .
$A = \begin{pmatrix} 2 & 1 \\ 0 & 2 \end{pmatrix}$ : $\Delta = 16 - 16 = 0$ . Eigenvalue: $2$ (repeated).
$A = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}$ : $\Delta = 0 - 4 = -4 < 0$ . Eigenvalues: $\pm i$ .

Summary

RemarkRole of the characteristic polynomial

The characteristic polynomial $p_A(\lambda) = \det(A - \lambda I)$ is the bridge between matrix algebra and polynomial algebra:

Its roots are the eigenvalues.
Its coefficients encode symmetric functions of eigenvalues (trace, determinant, etc.).
It is similarity-invariant, so it belongs to the linear map, not just the matrix.
It factors over $\mathbb{C}$ into linear factors (Fundamental Theorem of Algebra), guaranteeing that every complex matrix has at least one eigenvalue.
The Cayley--Hamilton theorem states that $p_A(A) = 0$ : every matrix satisfies its own characteristic polynomial.