Characteristic Polynomial

Definition and Basic Concepts

To find the eigenvalues of a matrix, we need a very important mathematical tool in linear algebra. Imagine we want to find all values $\lambda$ that make the matrix $A - \lambda I$ become singular (not invertible).

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

This function is a polynomial of degree $n$ in $t \in \mathbb{K}$, which we call the characteristic polynomial of $A$.

with coefficients $a_0, \ldots, a_{n-1}, a_n \in \mathbb{K}$.

In fact, $\chi_A(t)$ is indeed a polynomial of degree $n$ for every matrix $A \in \mathbb{K}^{n \times n}$.

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:

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:

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 most important concept of the characteristic polynomial is its relationship with eigenvalues.

Now, let's look at a very important relationship: $\lambda \in \mathbb{K}$ is an eigenvalue of $A$ if and only if $\det(A - \lambda \cdot I) = 0$.

In other words, eigenvalues are the roots of the characteristic polynomial.

The equation for $t \in \mathbb{K}$:

is what we call 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 $\lambda$, it holds:

Relationship between Geometric and Algebraic Multiplicity

One important result in eigenvalue theory is the relationship between geometric and algebraic multiplicity.

For any matrix $A \in \mathbb{K}^{n \times n}$ and $\lambda \in \mathbb{K}$, we have an interesting relationship:

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

Why does this happen? This can be explained using basis transformation and Jordan block form.

Examples of Characteristic Polynomial Calculation

After understanding the basic concepts, let's see how to apply them in concrete examples of characteristic polynomial calculation:

3x3 Matrix Example

Let $A = \begin{pmatrix} 3 & 2 & -1 \\ 1 & 0 & -4 \\ 3 & 0 & 1 \end{pmatrix} \in \mathbb{K}^{3 \times 3}$. The characteristic polynomial of $A$ is:

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.

Simple Example

The characteristic polynomial of matrix $A = \begin{pmatrix} 1 & 2 \\ 0 & 1 \end{pmatrix}$ is:

This matrix has the root $\lambda = 1$ with algebraic multiplicity $\mu_A(1) = 2$. $\lambda = 1$ is the only eigenvalue of $A$. We have calculated that $\dim \text{Eig}_A(1) = 1$.

Geometric Transformation Examples

Now, let's explore something fascinating: how the characteristic polynomial works on common geometric transformations we often encounter in $\mathbb{R}^2 \to \mathbb{R}^2$:

Rotation

Rotation with $A = \begin{pmatrix} \cos(\alpha) & -\sin(\alpha) \\ \sin(\alpha) & \cos(\alpha) \end{pmatrix}$ has the characteristic polynomial:

This has real roots if and only if $\cos(\alpha)^2 - 1 \geq 0$, that is $\cos(\alpha)^2 = 1$, so $\alpha = 0$ or $\alpha = \pi$.

Reflection

Reflection along axes with $A = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}$ has the characteristic polynomial:

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

Scaling

Scaling with $A = \begin{pmatrix} s & 0 \\ 0 & s \end{pmatrix}$ has the characteristic polynomial:

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

Shearing

Shearing with $A = \begin{pmatrix} 1 & s \\ 0 & 1 \end{pmatrix}$ with $s \neq 0$ has the characteristic polynomial:

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

Properties of Similar Matrices

Similar matrices have a fascinating property: they have the same characteristic polynomial, and therefore have the same eigenvalues, the same trace, and the same determinant.

Let's see why this is true. Let $S \in \mathbb{K}^{n \times n}$ be invertible and $B = S^{-1} \cdot A \cdot S$. Then:

Eigenvector Properties of Similar Matrices

Now, what about the eigenvectors of similar matrices? Let $A, B \in \mathbb{K}^{n \times n}$ be similar matrices with $B = S^{-1} \cdot A \cdot S$ and invertible matrix $S \in \mathbb{K}^{n \times n}$. If $\lambda \in \mathbb{K}$ is an eigenvalue of both $A$ and $B$, and $v \in \mathbb{K}^n$ is an eigenvector of $A$ for eigenvalue $\lambda$, then $w = S^{-1} \cdot v$ is an eigenvector of $B$ for eigenvalue $\lambda$.

Let's see why this is true. Let $A \cdot v = \lambda \cdot v$ and $w = S^{-1} \cdot v$. Then:

This shows that similarity transformation not only preserves eigenvalues, but also provides a systematic way to transform eigenvectors.