Characteristic Polynomial

Deriving the Polynomial from the Matrix Equation

An eigenvalue $\lambda$ of a square matrix $A$ has a nonzero eigenvector $v$ satisfying

Av=\lambda v

Rearranging gives $(A-\lambda I)v=0$ . A nonzero solution exists exactly when $A-\lambda I$ is singular, so $its determinant must vanish$ . The characteristic polynomial records this condition for every possible scalar at once.

Let $A \in \mathbb{K}^{n \times n}$ . We can form a special function:

\chi_A(t) = \det(A - t \cdot I)

This degree- $n$ polynomial in $t$ , with coefficients $a_0,\ldots,a_n\in\mathbb{K}$ , is the $characteristic polynomial$ of $A$ . Here, $\chi_A(t)$ uses the convention $\det(A-tI)$ . The equally common convention $\det(tI-A)$ differs by the factor $(-1)^n$ but has the same roots.

Matrix Trace and Polynomial Coefficients

Let $A \in \mathbb{K}^{n \times n}$ be a square matrix. The $trace$ of $A$ is the sum of the diagonal elements:

\operatorname{tr}(A) = \sum_{i=1}^n a_{ii}

The matrix trace has a close relationship with the coefficients of the characteristic polynomial.

Relationship of Coefficients with Trace and Determinant

In the characteristic polynomial $\chi_A(t)$ of $A$ , its coefficients have special meaning:

\begin{aligned} &a_n = (-1)^n, \\ &a_{n-1} = (-1)^{n-1}\operatorname{tr}(A), \\ &a_0 = \det A \end{aligned}

This means:

The highest coefficient is always $(-1)^n$
The second highest coefficient is related to the matrix trace
The $constant term$ is the determinant of the matrix

Eigenvalues as Polynomial Roots

The value $\lambda \in \mathbb{K}$ is an eigenvalue of $A$ if and only if $\det(A - \lambda \cdot I) = 0$ .

The eigenvalues are the $roots of the characteristic polynomial$ .

For $t \in \mathbb{K}$ , consider the following equation.

\det(A - t \cdot I) = 0

This equation is the $characteristic equation$ of $A$ .

Algebraic Multiplicity

Now, what happens if an eigenvalue appears multiple times as a root of the characteristic polynomial? Let $A \in \mathbb{K}^{n \times n}$ and $\lambda \in \mathbb{K}$ . The multiplicity of the root $t = \lambda$ of the characteristic polynomial $\chi_A(t)$ is called the $algebraic multiplicity$ $\mu_A(\lambda)$ of the eigenvalue $\lambda$ of $A$ . We say $\lambda$ is an eigenvalue with multiplicity $\mu_A(\lambda)$ of $A$ .

Multiplicity Bounds

For every eigenvalue $\lambda$ , its $algebraic multiplicity satisfies$

1 \leq \mu_A(\lambda) \leq n

Relationship between Geometric and Algebraic Multiplicity

Geometric and algebraic multiplicity satisfy the following bound.

For an eigenvalue $\lambda$ of $A \in \mathbb{K}^{n \times n}$ , its geometric and algebraic multiplicities satisfy

1 \leq \dim \operatorname{Eig}_A(\lambda) \leq \mu_A(\lambda) \leq n

The geometric multiplicity of every eigenvalue is always less than or equal to its algebraic multiplicity.

The geometric multiplicity $counts independent eigenvectors$ in the eigenspace, whereas the algebraic multiplicity counts how many times the root occurs in the characteristic polynomial. Jordan form shows that every independent eigenvector requires at least one Jordan block, so the geometric count cannot exceed the algebraic count.

Computing Characteristic Polynomials

The definition is applied to concrete matrices now, and the $coefficient identities$ serve as checks on the expansion. Follow each determinant step by step, then compare the roots with the trace and the determinant to confirm that the polynomial is correct.

Third Order Matrix

This third order matrix produces a characteristic polynomial of $degree three$ :

A = \begin{pmatrix} 3 & 2 & -1 \\ 1 & 0 & -4 \\ 3 & 0 & 1 \end{pmatrix} \in \mathbb{K}^{3 \times 3}

Its characteristic polynomial is:

\chi_A(t) = \det \begin{pmatrix} 3-t & 2 & -1 \\ 1 & -t & -4 \\ 3 & 0 & 1-t \end{pmatrix}

For $\mathbb{K} = \mathbb{R}$ , $\chi_A(t)$ only has the root $\lambda_1 \approx -1.8003$ with algebraic multiplicity $\mu_A(\lambda_1) = 1$ .

For $\mathbb{K} = \mathbb{C}$ , $\chi_A(t)$ has roots $\lambda_1 \approx -1.8003$ , $\lambda_2 \approx 2.9001 + 2.4559i$ , and $\lambda_3 \approx 2.9001 - 2.4559i$ with algebraic multiplicities $\mu_A(\lambda_1) = \mu_A(\lambda_2) = \mu_A(\lambda_3) = 1$ respectively.

The roots pass two useful checks: their sum is the trace, and their product is the determinant.

\lambda_1+\lambda_2+\lambda_3=4=\operatorname{tr}(A)

Repeated Eigenvalue in an Upper Triangular Matrix

This matrix has the $same diagonal entry twice$ , namely $1$ :

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

Its characteristic polynomial is:

\begin{aligned} \chi_A(t) &= \det(A - t \cdot I) \\ &= \det \begin{pmatrix} 1-t & 2 \\ 0 & 1-t \end{pmatrix} \end{aligned}

This matrix has the root $\lambda = 1$ with algebraic multiplicity $\mu_A(1) = 2$ . $\lambda = 1$ is the only eigenvalue of $A$ .

(A-I)\begin{pmatrix}x\\y\end{pmatrix}=\begin{pmatrix}2y\\0\end{pmatrix}=0\iff y=0,\qquad E_A(1)=\operatorname{span}\left\{\begin{pmatrix}1\\0\end{pmatrix}\right\}

The eigenspace has $dimension 1$ , even though the polynomial root occurs twice.

Characteristic Polynomials of Geometric Transformations

The calculation now runs on familiar maps of $\mathbb{R}^2$ , so the algebraic result and its $geometric meaning$ appear together. Rotation, reflection, scaling, and shearing each produce a polynomial whose roots describe what that map does to directions.

Rotation

The rotation is represented by the matrix:

A = \begin{pmatrix} \cos(\alpha) & -\sin(\alpha) \\ \sin(\alpha) & \cos(\alpha) \end{pmatrix}

Its characteristic polynomial is:

\begin{aligned} \chi_A(t) &= (\cos(\alpha) - t)^2 + \sin(\alpha)^2 \\ &= t^2 - 2 \cdot \cos(\alpha) \cdot t + 1 \end{aligned}

$Its discriminant$ is $-4\sin^2(\alpha)$ . The roots are real exactly when $\sin(\alpha)=0$ , so, modulo a full turn, $\alpha\equiv0$ or $\alpha\equiv\pi\pmod{2\pi}$ .

Reflection

$Reflection across$ the $x$ -axis has the matrix:

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

Its characteristic polynomial is:

\chi_A(t) = (1-t) \cdot (-1-t) = t^2 - 1

The eigenvalues are $\lambda_1 = 1$ and $\lambda_2 = -1$ with $\mu_A(1) = \mu_A(-1) = 1$ .

Scaling

$Scaling has$ the matrix:

A = \begin{pmatrix} s & 0 \\ 0 & s \end{pmatrix}

Its characteristic polynomial is:

\chi_A(t) = (s-t)^2 = t^2 - 2 \cdot s \cdot t + s^2

The eigenvalue is $\lambda = s$ with $\mu_A(s) = 2$ .

Shearing

$For a shear$ with $s \neq 0$ , the matrix is:

A = \begin{pmatrix} 1 & s \\ 0 & 1 \end{pmatrix}

Its characteristic polynomial is:

\chi_A(t) = (1-t)^2 = t^2 - 2 \cdot t + 1

The eigenvalue is $\lambda = 1$ with $\mu_A(1) = 2$ .

Properties of Similar Matrices

Similar matrices describe the $same linear transformation$ in different bases. They therefore have the same characteristic polynomial, eigenvalues, trace, and determinant.

Let $S \in \mathbb{K}^{n \times n}$ be invertible and $B = S^{-1} \cdot A \cdot S$ . Then:

\begin{aligned} &\chi_B(t) \\ &= \det(B - t \cdot I) \\ &= \det(S^{-1} \cdot A \cdot S - t \cdot S^{-1} \cdot I \cdot S) \end{aligned}

Eigenvector Properties of Similar Matrices

The eigenvectors change with the basis. If $v \in \mathbb{K}^n$ is an eigenvector of $A$ for $\lambda$ , then $w = S^{-1}v$ is an eigenvector of $B$ for the same eigenvalue.

Indeed, using $Av=\lambda v$ and $w=S^{-1}v$ gives

\begin{aligned} B \cdot w &= S^{-1} \cdot A \cdot S \cdot S^{-1} \cdot v \\ &= S^{-1} \cdot A \cdot v \end{aligned}

Similarity preserves eigenvalues while the $inverse basis change$ converts the corresponding eigenvectors.

Deriving the Polynomial from the Matrix Equation

An eigenvalue $\lambda$ of a square matrix $A$ has a nonzero eigenvector $v$ satisfying

Av=\lambda v

Let $A \in \mathbb{K}^{n \times n}$ . We can form a special function:

\chi_A(t) = \det(A - t \cdot I)

Matrix Trace and Polynomial Coefficients

Let $A \in \mathbb{K}^{n \times n}$ be a square matrix. The $trace$ of $A$ is the sum of the diagonal elements:

\operatorname{tr}(A) = \sum_{i=1}^n a_{ii}

The matrix trace has a close relationship with the coefficients of the characteristic polynomial.

Relationship of Coefficients with Trace and Determinant

In the characteristic polynomial $\chi_A(t)$ of $A$ , its coefficients have special meaning:

\begin{aligned} &a_n = (-1)^n, \\ &a_{n-1} = (-1)^{n-1}\operatorname{tr}(A), \\ &a_0 = \det A \end{aligned}

This means:

The highest coefficient is always $(-1)^n$
The second highest coefficient is related to the matrix trace
The $constant term$ is the determinant of the matrix

Eigenvalues as Polynomial Roots

The value $\lambda \in \mathbb{K}$ is an eigenvalue of $A$ if and only if $\det(A - \lambda \cdot I) = 0$ .

The eigenvalues are the $roots of the characteristic polynomial$ .

For $t \in \mathbb{K}$ , consider the following equation.

\det(A - t \cdot I) = 0

This equation is the $characteristic equation$ of $A$ .

Algebraic Multiplicity

Multiplicity Bounds

For every eigenvalue $\lambda$ , its $algebraic multiplicity satisfies$

1 \leq \mu_A(\lambda) \leq n

Relationship between Geometric and Algebraic Multiplicity

Geometric and algebraic multiplicity satisfy the following bound.

For an eigenvalue $\lambda$ of $A \in \mathbb{K}^{n \times n}$ , its geometric and algebraic multiplicities satisfy

1 \leq \dim \operatorname{Eig}_A(\lambda) \leq \mu_A(\lambda) \leq n

The geometric multiplicity of every eigenvalue is always less than or equal to its algebraic multiplicity.

Computing Characteristic Polynomials

Third Order Matrix

This third order matrix produces a characteristic polynomial of $degree three$ :

A = \begin{pmatrix} 3 & 2 & -1 \\ 1 & 0 & -4 \\ 3 & 0 & 1 \end{pmatrix} \in \mathbb{K}^{3 \times 3}

Its characteristic polynomial is:

\chi_A(t) = \det \begin{pmatrix} 3-t & 2 & -1 \\ 1 & -t & -4 \\ 3 & 0 & 1-t \end{pmatrix}

For $\mathbb{K} = \mathbb{R}$ , $\chi_A(t)$ only has the root $\lambda_1 \approx -1.8003$ with algebraic multiplicity $\mu_A(\lambda_1) = 1$ .

The roots pass two useful checks: their sum is the trace, and their product is the determinant.

\lambda_1+\lambda_2+\lambda_3=4=\operatorname{tr}(A)

Repeated Eigenvalue in an Upper Triangular Matrix

This matrix has the $same diagonal entry twice$ , namely $1$ :

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

Its characteristic polynomial is:

\begin{aligned} \chi_A(t) &= \det(A - t \cdot I) \\ &= \det \begin{pmatrix} 1-t & 2 \\ 0 & 1-t \end{pmatrix} \end{aligned}

This matrix has the root $\lambda = 1$ with algebraic multiplicity $\mu_A(1) = 2$ . $\lambda = 1$ is the only eigenvalue of $A$ .

(A-I)\begin{pmatrix}x\\y\end{pmatrix}=\begin{pmatrix}2y\\0\end{pmatrix}=0\iff y=0,\qquad E_A(1)=\operatorname{span}\left\{\begin{pmatrix}1\\0\end{pmatrix}\right\}

The eigenspace has $dimension 1$ , even though the polynomial root occurs twice.

Characteristic Polynomials of Geometric Transformations

Rotation

The rotation is represented by the matrix:

A = \begin{pmatrix} \cos(\alpha) & -\sin(\alpha) \\ \sin(\alpha) & \cos(\alpha) \end{pmatrix}

Its characteristic polynomial is:

\begin{aligned} \chi_A(t) &= (\cos(\alpha) - t)^2 + \sin(\alpha)^2 \\ &= t^2 - 2 \cdot \cos(\alpha) \cdot t + 1 \end{aligned}

$Its discriminant$ is $-4\sin^2(\alpha)$ . The roots are real exactly when $\sin(\alpha)=0$ , so, modulo a full turn, $\alpha\equiv0$ or $\alpha\equiv\pi\pmod{2\pi}$ .

Reflection

$Reflection across$ the $x$ -axis has the matrix:

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

Its characteristic polynomial is:

\chi_A(t) = (1-t) \cdot (-1-t) = t^2 - 1

The eigenvalues are $\lambda_1 = 1$ and $\lambda_2 = -1$ with $\mu_A(1) = \mu_A(-1) = 1$ .

Scaling

$Scaling has$ the matrix:

A = \begin{pmatrix} s & 0 \\ 0 & s \end{pmatrix}

Its characteristic polynomial is:

\chi_A(t) = (s-t)^2 = t^2 - 2 \cdot s \cdot t + s^2

The eigenvalue is $\lambda = s$ with $\mu_A(s) = 2$ .

Shearing

$For a shear$ with $s \neq 0$ , the matrix is:

A = \begin{pmatrix} 1 & s \\ 0 & 1 \end{pmatrix}

Its characteristic polynomial is:

\chi_A(t) = (1-t)^2 = t^2 - 2 \cdot t + 1

The eigenvalue is $\lambda = 1$ with $\mu_A(1) = 2$ .

Properties of Similar Matrices

Similar matrices describe the $same linear transformation$ in different bases. They therefore have the same characteristic polynomial, eigenvalues, trace, and determinant.

Let $S \in \mathbb{K}^{n \times n}$ be invertible and $B = S^{-1} \cdot A \cdot S$ . Then:

\begin{aligned} &\chi_B(t) \\ &= \det(B - t \cdot I) \\ &= \det(S^{-1} \cdot A \cdot S - t \cdot S^{-1} \cdot I \cdot S) \end{aligned}

Eigenvector Properties of Similar Matrices

The eigenvectors change with the basis. If $v \in \mathbb{K}^n$ is an eigenvector of $A$ for $\lambda$ , then $w = S^{-1}v$ is an eigenvector of $B$ for the same eigenvalue.

Indeed, using $Av=\lambda v$ and $w=S^{-1}v$ gives

\begin{aligned} B \cdot w &= S^{-1} \cdot A \cdot S \cdot S^{-1} \cdot v \\ &= S^{-1} \cdot A \cdot v \end{aligned}

Similarity preserves eigenvalues while the $inverse basis change$ converts the corresponding eigenvectors.