📂Linear Algebra

Characteristics Polynomial of Linear Transformation

Overview

The characteristic polynomial of linear transformation is defined. From the theorem below, it can be seen that solving equation $\det(A - \lambda I) = 0$ is equivalent to finding the eigenvalues. Therefore, it is quite natural to name $\det(A - \lambda I)$ the characteristic polynomial.

Theorem¹

Let’s say $F$ is any field, and $A \in M_{n\times n}(F)$. That $\lambda \in F$ is an eigenvalue of $A$ is equivalent to $\det (A-\lambda I) = 0$.

Proof

Assume $\lambda$ is an eigenvalue of $A$. Then,

$$ \begin{align*} \lambda \text{ is eigenvalue of } A &\iff \exist \text{non-zero } v \text{ such that } Av = \lambda v \\ &\iff \exist \text{non-zero } v \text{ such that } (A - \lambda I)v = 0 \end{align*} $$

Conditions Equivalent to Invertibility

Let’s say $A$ is a square matrix of size $n\times n$. Then, the following propositions are all equivalent:

• $A$ is an invertible matrix.
• Homogeneous linear system $A\mathbf{x}=\mathbf{0}$ has only the trivial solution.
• $\det{A} \ne 0$

By the conditions equivalent to invertibility, $A - \lambda I$ is not invertible, and $\det (A - \lambda I) = 0$.

Definition

Let’s say $A \in M_{n \times n}(F)$. The polynomial $f(t) = \det(A - tI)$ is called the characteristic polynomial of $A$. $f(t) = 0$ is called the characteristic equation.

Let $V$ be a vector space of dimension $n$. Let $T : V \to V$ be a linear transformation. Let $\beta$ be an ordered basis of $V$. The characteristic polynomial $f(t)$ of $T$ is defined as the characteristic polynomial of the matrix representation of $T$. In other words, $f(t)$ is as follows.

$$ f(t) = \det\left( \begin{bmatrix} T \end{bmatrix}_{\beta} - t I \right) $$

Explanation

According to the definition, the roots of the characteristic polynomial of $T : V \to V$ are precisely the eigenvalues, and if the characteristic polynomial is factorable, $T$ has $n = \dim(V)$ eigenvalues (not said to be distinct).

By definition, it might seem that the characteristic polynomial of $T$ depends on how the ordered basis $\beta$ is chosen, but in reality, it does not. For this reason, the characteristic polynomial of the linear transformation $T$ is sometimes denoted as follows.

$$ \det (T - \lambda I) $$

Let’s check. If $\beta$, $\beta^{\prime}$ are the ordered bases of $V$, and $Q$ is the change of basis matrix that converts $\beta$ coordinates into $\beta^{\prime}$ coordinates, then,

$$ \begin{align*} \det( \begin{bmatrix} T \end{bmatrix}_{\beta} - tI) &= \det( \begin{bmatrix} T \end{bmatrix}_{\beta} - tI ) \det Q^{-1} \det Q \\ &= \det Q^{-1} \det( \begin{bmatrix} T \end{bmatrix}_{\beta} - tI ) \det Q \\ &= \det \left( Q^{-1} (\begin{bmatrix} T \end{bmatrix}_{\beta} - tI) Q \right) \\ &= \det \left( Q^{-1}\begin{bmatrix} T \end{bmatrix}_{\beta}Q - tI \right) \\ &= \det \left( \begin{bmatrix} T \end{bmatrix}_{\beta^{\prime}} - tI \right) \end{align*} $$