What is the characteristic polynomial?

For a square matrix $A$, the standard characteristic polynomial is $p_A(t)=\det(tI-A)$. It is a polynomial whose roots are the eigenvalues of $A$.

How does the eigenvector equation produce the characteristic equation?

From $A\mathbf{v}=\lambda\mathbf{v}$ with $\mathbf{v}\ne\mathbf{0}$, we get $(\lambda I-A)\mathbf{v}=\mathbf{0}$. This requires $\lambda I-A$ to be singular, so $\det(\lambda I-A)=0$.

What does it mean for a polynomial to be monic?

A polynomial is monic when its leading coefficient is $1$. The standard characteristic polynomial of an $n\times n$ matrix is monic and has degree $n$.

How do the two common characteristic-polynomial conventions differ?

The standard convention is $\det(tI-A)$; another is $\det(A-tI)$. The latter equals $(-1)^n$ times the former, so both have the same roots.

What is the characteristic equation?

It is the equation formed by setting the characteristic polynomial equal to zero, such as $\det(tI-A)=0$. Its solutions are the eigenvalues.

Linear algebraLesson 3 of 5 in the course Eigenvalues and eigenvectors

Characteristic Polynomials: Coefficients, Invariance, and Matrix Functions

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Summary

For a square matrix $A$ , an eigenvector $\mathbf{v}\ne\mathbf{0}$ with eigenvalue $\lambda$ satisfies $A\mathbf{v}=\lambda\mathbf{v}$ . Rearranging gives $(\lambda I-A)\mathbf{v}=\mathbf{0}$ . A square matrix has a nonzero vector in its null space precisely when it is singular, so an eigenvalue must satisfy $\det(\lambda I-A)=0$ .

Replacing $\lambda$ with a variable $t$ gives a polynomial whose roots are the eigenvalues of $A$ . The equation obtained by setting this polynomial equal to zero is called the characteristic equation, or sometimes the determinantal equation.

All flashcards

What is the characteristic polynomial?: For a square matrix $A$ , the standard characteristic polynomial is $p_A(t)=\det(tI-A)$ . It is a polynomial whose roots are the eigenvalues of $A$ .
How does the eigenvector equation produce the characteristic equation?: From $A\mathbf{v}=\lambda\mathbf{v}$ with $\mathbf{v}\ne\mathbf{0}$ , we get $(\lambda I-A)\mathbf{v}=\mathbf{0}$ . This requires $\lambda I-A$ to be singular, so $\det(\lambda I-A)=0$ .
What does it mean for a polynomial to be monic?: A polynomial is monic when its leading coefficient is $1$ . The standard characteristic polynomial of an $n\times n$ matrix is monic and has degree $n$ .
How do the two common characteristic-polynomial conventions differ?: The standard convention is $\det(tI-A)$ ; another is $\det(A-tI)$ . The latter equals $(-1)^n$ times the former, so both have the same roots.
What is the characteristic equation?: It is the equation formed by setting the characteristic polynomial equal to zero, such as $\det(tI-A)=0$ . Its solutions are the eigenvalues.

Quiz questions

What is the standard definition of the characteristic polynomial of an $n\times n$ matrix $A$ ?Answer: $p_A(t)=\det(tI-A)$
Why must an eigenvalue $\lambda$ make $\det(\lambda I-A)$ equal to zero?Answer: It has a nonzero null vector, so it is singular.
What is the degree of the characteristic polynomial of a $4\times4$ matrix?Answer: Four
For $A=\begin{pmatrix}2&1\\-1&0\end{pmatrix}$ , what is $\det(tI-A)$ ?Answer: $t^2-2t+1$
If a $3\times3$ matrix uses the convention $\det(A-tI)$ , how does its polynomial compare with $\det(tI-A)$ ?Answer: It is the negative, with the same roots.

Based on the Wikipedia article Characteristic polynomial, licensed CC BY-SA 4.0. Adapted into a summary, flashcards and quizzes.

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