---
title: The Characteristic Equation
module: Eigenvalues and Eigenvectors
moduleNumber: 5
lessonNumber: 2
order: 502
summary: >
  The eigenvalues of a matrix are the roots of its characteristic polynomial
  det(A minus lambda I). This degree-n polynomial carries an algebraic
  multiplicity at each repeated root, a nonzero determinant is equivalent to
  zero not being an eigenvalue, and similar matrices share a characteristic
  polynomial and hence the same eigenvalues.
topics: [Eigenvalues and Eigenvectors]
sources:
  - book: Lay
    ref: "§5.2 The Characteristic Equation"
---

For a matrix that is not triangular, the eigenvalues do not sit on the diagonal and
must be found some other way. A scalar $\lambda$ is an eigenvalue of $A$ exactly when
$A - \lambda I$ is singular, and singularity is detected by the determinant. This turns
the search for eigenvalues into a single polynomial equation in $\lambda$.

## The determinant condition

A matrix fails to be invertible exactly when its determinant is zero, so the eigenvalues
of $A$ are the scalars $\lambda$ satisfying $\det(A - \lambda I) = 0$.

> **Worked example.** Find the eigenvalues of
> $A = \begin{bmatrix} 2 & 3 \\ 3 & -6 \end{bmatrix}$.
>
> $$
> \det(A - \lambda I)
> = \det \begin{bmatrix} 2 - \lambda & 3 \\ 3 & -6 - \lambda \end{bmatrix}
> = (2 - \lambda)(-6 - \lambda) - (3)(3).
> $$
>
> Expanding and factoring,
> $$
> \det(A - \lambda I) = \lambda^2 + 4\lambda - 21 = (\lambda - 3)(\lambda + 7),
> $$
> which vanishes at $\lambda = 3$ and $\lambda = -7$. These are the eigenvalues.

The determinant has converted the matrix equation $(A - \lambda I)x = 0$, which carries
two unknowns $\lambda$ and $x$, into a scalar equation in the single unknown $\lambda$.

> **Definition (Characteristic equation and polynomial).** The **characteristic
> equation** of an $n \times n$ matrix $A$ is
> $$
> \det(A - \lambda I) = 0.
> $$
> The left side, $\det(A - \lambda I)$, is a polynomial of degree $n$ in $\lambda$,
> the **characteristic polynomial** of $A$. A scalar $\lambda$ is an eigenvalue of
> $A$ if and only if it satisfies the characteristic equation.

$$
% caption: The characteristic polynomial $\det(A - \lambda I)$, plotted here as
% $p(t)$, crosses zero at each eigenvalue; the roots sit at $-7$ and $3$.
\begin{tikzpicture}[>=stealth, scale=0.62, font=\footnotesize]
  \definecolor{acc}{HTML}{4A6FA5}
  \draw[->, black] (-8.8,0) -- (4.6,0) node[right, black!70] {$t$};
  \draw[->, black] (0,-3.4) -- (0,2.2) node[above, black!70] {$p(t)$};
  % parabola (t-3)(t+7)/9; roots at -7 and 3, vertex near (-2,-2.8)
  \draw[black, very thick] plot[domain=-8.4:4.4,samples=90] ({\x},{(\x-3)*(\x+7)/9});
  \fill[acc] (3,0) circle (3pt);
  \fill[acc] (-7,0) circle (3pt);
  \node[acc, anchor=north east] at (-6.9,-0.2) {left root};
  \node[acc, anchor=north west] at (3.2,-0.2) {right root};
\end{tikzpicture}
$$

## Determinant by row reduction

For matrices larger than $2 \times 2$, the determinant is computed by row reduction.
Reduce $A$ to an echelon form $U$ using only row replacements and row interchanges
(no scaling), and let $r$ be the number of interchanges. Then

$$
\det A =
\begin{cases}
(-1)^r \cdot (\text{product of the pivots in } U), & A \text{ invertible}, \\
0, & A \text{ not invertible}.
\end{cases}
$$

> **Worked example.** Compute $\det A$ for
> $A = \begin{bmatrix} 1 & 5 & 0 \\ 2 & 4 & -1 \\ 0 & -2 & 0 \end{bmatrix}$.
>
> Subtract $2$ times row $1$ from row $2$, then interchange rows $2$ and $3$:
> $$
> A \sim \begin{bmatrix} 1 & 5 & 0 \\ 0 & -6 & -1 \\ 0 & -2 & 0 \end{bmatrix}
> \sim \begin{bmatrix} 1 & 5 & 0 \\ 0 & -2 & 0 \\ 0 & -6 & -1 \end{bmatrix}
> \sim \begin{bmatrix} 1 & 5 & 0 \\ 0 & -2 & 0 \\ 0 & 0 & -1 \end{bmatrix} = U.
> $$
> One interchange was used ($r = 1$) and the pivots are $1, -2, -1$, so
> $\det A = (-1)^1 (1)(-2)(-1) = -2$.

> **Theorem (Properties of determinants).** For $n \times n$ matrices $A$ and $B$:
> - $A$ is invertible if and only if $\det A \neq 0$.
> - $\det(AB) = (\det A)(\det B)$.
> - $\det A^\top = \det A$.
> - If $A$ is triangular, $\det A$ is the product of its diagonal entries.
> - A row replacement leaves $\det A$ unchanged; a row interchange flips its sign;
>   a row scaling multiplies it by the same scalar.

The first property adds two more equivalent conditions to the Invertible Matrix
Theorem.

> **Invertible Matrix Theorem (continued).** Let $A$ be $n \times n$. Add to the
> [existing list](/linear-algebra/matrix-algebra/matrix-inverse-and-invertibility)
> the equivalent conditions: **the number $0$ is not an eigenvalue of $A$**, and
> **$\det A \neq 0$**.

## Multiplicity

The characteristic polynomial of an $n \times n$ matrix has degree $n$, so counting
multiplicities and allowing complex roots it has exactly $n$ roots. A root can repeat.

> **Definition (Algebraic multiplicity).** The **(algebraic) multiplicity** of an
> eigenvalue $\lambda$ is its multiplicity as a root of the characteristic equation,
> the number of times $(\lambda_0 - \lambda)$ divides the characteristic polynomial.

For the upper-triangular matrix

$$
A = \begin{bmatrix}
5 & -2 & 6 & -1 \\
0 & 3 & -8 & 0 \\
0 & 0 & 5 & 4 \\
0 & 0 & 0 & 1
\end{bmatrix},
$$

Part (d) of the determinant-properties theorem gives $\det(A - \lambda I) = (5 - \lambda)^2(3 - \lambda)(1 - \lambda)$.
The eigenvalue $5$ has multiplicity $2$; the eigenvalues $3$ and $1$ each have
multiplicity $1$. As another example, a $6 \times 6$ matrix with characteristic
polynomial

$$
\lambda^6 - 4\lambda^5 - 12\lambda^4 = \lambda^4(\lambda - 6)(\lambda + 2)
$$

has eigenvalues $0$ (multiplicity $4$), $6$, and $-2$, which can be listed with
repetition as $0, 0, 0, 0, 6, -2$.

$$
% caption: A simple root crosses the horizontal axis; a repeated root touches it
% without crossing, its multiplicity visible as tangency.
\begin{tikzpicture}[>=stealth, scale=0.85, font=\footnotesize]
  \definecolor{acc}{HTML}{4A6FA5}
  \definecolor{red}{HTML}{B0413E}
  \begin{scope}
    \draw[->, black] (-1.6,0) -- (1.9,0) node[right, black!70] {$t$};
    \draw[->, black] (0,-1.4) -- (0,1.7);
    \draw[black, very thick] plot[domain=-1.3:1.3,samples=40] ({\x},{1.1*\x});
    \fill[acc] (0,0) circle (2pt);
    \node[acc, anchor=north] at (0.95,-0.15) {simple root};
    \node[black, anchor=south] at (0,1.55) {crosses};
  \end{scope}
  \begin{scope}[xshift=5cm]
    \draw[->, black] (-1.6,0) -- (1.9,0) node[right, black!70] {$t$};
    \draw[->, black] (0,-1.4) -- (0,1.7);
    \draw[black, very thick] plot[domain=-1.25:1.25,samples=40] ({\x},{1.05*\x*\x});
    \fill[red] (0,0) circle (2pt);
    \node[red, anchor=north] at (0.95,-0.15) {double root};
    \node[black, anchor=south] at (0,1.55) {touches};
  \end{scope}
\end{tikzpicture}
$$

> **The characteristic equation is a theoretical tool, not a numerical one.** There is
> no finite formula solving a general polynomial of degree $5$ or more, and forming
> the characteristic polynomial then rooting it is numerically unstable. Production
> eigenvalue software avoids the characteristic polynomial entirely; see
> [iterative estimates](/linear-algebra/eigenvalues/power-method) and the
> [numerical eigenvalue problem](/linear-algebra/numerical-linear-algebra/numerical-eigenvalues-and-svd).

## Similarity

The characteristic polynomial is a property of the transformation, not of the
particular coordinates chosen to represent it. Similarity makes this precise.

> **Definition (Similarity).** Two $n \times n$ matrices $A$ and $B$ are **similar**
> if there is an invertible matrix $P$ with
> $$
> P^{-1} A P = B, \qquad \text{equivalently} \qquad A = P B P^{-1}.
> $$
> Passing from $A$ to $P^{-1} A P$ is a **similarity transformation**.

$$
% caption: A similarity transformation reads $A$ in the coordinate frame set by $P$;
% the change of frame does not disturb the eigenvalues.
\begin{tikzpicture}[>=stealth, font=\footnotesize,
  b/.style={draw, minimum width=17mm, minimum height=10mm, align=center}]
  \definecolor{acc}{HTML}{4A6FA5}
  \node[b] (A) at (0,0) {$A$};
  \node[b] (B) at (5,0) {$B = P^{-1} A P$};
  \draw[->, black, thick] (A) -- (B) node[midway, above] {change frame by $P$};
  \draw[->, black, thick] (B) to[bend left=38] node[midway, below] {same eigenvalues} (A);
\end{tikzpicture}
$$

> **Theorem (Similarity preserves the characteristic polynomial).** If $A$ and $B$ are similar, then they have the same characteristic
> polynomial, and therefore the same eigenvalues with the same multiplicities.

> **Proof.** If $B = P^{-1} A P$, then
>
> $$
> B - \lambda I = P^{-1} A P - \lambda P^{-1} P = P^{-1}(A - \lambda I) P.
> $$
>
> By the multiplicative property of determinants,
>
> $$
> \det(B - \lambda I)
> = \det(P^{-1}) \cdot \det(A - \lambda I) \cdot \det(P).
> $$
>
> Since $\det(P^{-1}) \det(P) = \det(P^{-1} P) = \det I = 1$, the two determinants are
> equal for every $\lambda$. $\blacksquare$

Two warnings:

- **Equal eigenvalues do not imply similarity.** The matrices
  $\begin{bmatrix} 2 & 1 \\ 0 & 2 \end{bmatrix}$ and
  $\begin{bmatrix} 2 & 0 \\ 0 & 2 \end{bmatrix}$ share the eigenvalue $2$ (twice) but
  are not similar; the second is $2I$, similar only to itself.
- **Similarity is not row equivalence.** Row operations correspond to
  left-multiplication $B = EA$, not to $P^{-1} A P$, and they usually change the
  eigenvalues.

The similarity-invariance theorem is the foundation of the practical eigenvalue algorithms: they generate a
sequence of matrices all similar to $A$ (hence sharing its eigenvalues) that drift
toward triangular form, whose diagonal then displays the eigenvalues.

## A dynamical system in closed form

Similarity and the characteristic equation combine to solve a difference equation in
closed form.

> **Worked example.** Solve $x_{k+1} = A x_k$ for
> $$
> A = \begin{bmatrix} .95 & .03 \\ .05 & .97 \end{bmatrix},
> \qquad
> x_0 = \begin{bmatrix} .60 \\ .40 \end{bmatrix}.
> $$
>
> The characteristic equation factors as
> $$
> 0 = \det(A - \lambda I) = \lambda^2 - 1.92\lambda + 0.92 = (\lambda - 1)(\lambda - 0.92),
> $$
> so the eigenvalues are $\lambda_1 = 1$ and $\lambda_2 = 0.92$, with eigenvectors
> $v_1 = \begin{bmatrix} 3 \\ 5 \end{bmatrix}$ and
> $v_2 = \begin{bmatrix} 1 \\ -1 \end{bmatrix}$. Writing $x_0 = c_1 v_1 + c_2 v_2$ and
> solving gives $c_1 = 0.125$, $c_2 = 0.225$. Since $v_1, v_2$ are eigenvectors,
> $$
> x_k = c_1 (\lambda_1)^k v_1 + c_2 (\lambda_2)^k v_2
>     = 0.125 \begin{bmatrix} 3 \\ 5 \end{bmatrix}
>     + 0.225 (0.92)^k \begin{bmatrix} 1 \\ -1 \end{bmatrix}.
> $$
> As $k \to \infty$, $(0.92)^k \to 0$, so $x_k \to 0.125\, v_1 = (0.375, 0.625)$. The
> eigenvalue $1$ fixes the steady state; the eigenvalue $0.92$ sets the rate of approach.

| Object | Role |
| --- | --- |
| $\det(A - \lambda I)$ | characteristic polynomial, degree $n$ |
| roots of $\det(A - \lambda I) = 0$ | the eigenvalues |
| multiplicity of a root | algebraic multiplicity of that eigenvalue |
| $\lambda = 1$ eigenvector | steady-state direction |
| $\vert\lambda\vert < 1$ eigenvalues | transient parts, decaying as $\lambda^k$ |

The decomposition $x_0 = c_1 v_1 + c_2 v_2$ works only when $A$ has a full set of
independent eigenvectors. That condition, restated as a matrix factorization, is
[diagonalization](/linear-algebra/eigenvalues/diagonalization).[^lay52]

[^lay52]: **Lay**, _Linear Algebra and Its Applications_, 5th ed., §5.2 — The Characteristic Equation: $\det(A - \lambda I) = 0$, the characteristic polynomial and algebraic multiplicity, the determinant/eigenvalue additions to the Invertible Matrix Theorem, and Theorem 4 on similar matrices sharing a characteristic polynomial.