The Characteristic Equation

Math 303: Section 17

Dr. Janssen

\[ \def\R{{\mathbb R}} \def\b{{\mathbf{b}}} \def\x{{\mathbf{x}}} \def\v{{\mathbf{v}}} \def\w{{\mathbf{w}}} \DeclareMathOperator{\null}{Nul} \]

Eigenvalues and Eigenvectors

Recall

Definition 1 (Eigenvalue and eigenvector) Let \(A\) be an \(n\times n\) matrix. A nonzero vector \(\v\) is an eigenvector of \(A\) if there is a scalar \(\lambda\) such that

\[ A \v = \lambda \v. \]

The scalar \(\lambda\) is called an eigenvalue for \(A\).

These are super useful. One of the main uses we’ll see is that if \(A\) is sufficiently nice, we can get a basis for \(\R^n\) out of eigenvectors of \(A\). Then when we multiply \(A\) by any vector in \(\R^n\) we can reduce to a bunch of scalar multiplications.

When we can use eigenvalues we also make it easier to do lots of multiplications by \(A\): \(A^k \x = \lambda^k \x\), so if \(A\) represents transitions between states of a dynamical system, we get easily get longterm behavior.

Typical process

• Given an \(n\times n\) matrix \(A\).
• Find an eigenvalue \(\lambda\)
• Solve the equation \((A- \lambda I_n) \x = \mathbf{0}\) to find eigenvector(s) corresponding to \(\lambda\)

How do we find \(\lambda\)?

Answer 1: The Characteristic Equation

Definition 2 Let \(A\) be \(n\times n\) matrix. The characteristic polynomial of \(A\) is the polynomial

\[ \det(A-\lambda I_n), \]

where \(I_n\) is the \(n\times n\) identity matrix. The characteristic equation of \(A\) is the equation

\[ \det(A-\lambda I_n) = 0. \]

Solving the characteristic equation gives the eigenvalues of \(A\).

Preview Activity 17.1

Activity 17.1

• Find the characteristic polynomial of the matrix

\[ A = \left[\begin{matrix} 3 & -2 & 5 \\ 1 & 0 & 7 \\ 0 & 0 & 1 \end{matrix}\right], \]

and use it to find the eigenvalues of \(A\).

• Verify that 1 and 2 are the only eigenvalues of the matrix

\[ \left[\begin{matrix} 1 & 0 & 0 & 1 \\ 1 & 2 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{matrix}\right]. \]

A = Matrix([[3,-2,5],[1,0,7],[0,0,1]])
B = Matrix([[1,0,0,1],[1,2,0,0],[0,0,1,0],[0,0,0,1]])

A.charpoly(), A.eigenvalues(), B.charpoly(), B.eigenvalues()

(x^3 - 4*x^2 + 5*x - 2, [2, 1, 1], x^4 - 5*x^3 + 9*x^2 - 7*x + 2, [2, 1, 1, 1])

Answer 2: Use sage

A = Matrix([[3,-2,5],[1,0,7],[0,0,1]])
B = Matrix([[1,0,0,1],[1,2,0,0],[0,0,1,0],[0,0,0,1]])

A.eigenvalues(), B.eigenvalues()

([2, 1, 1], [2, 1, 1, 1])

Algebraic Multiplicity

Definition 3 The (algebraic) multiplicity of an eigenvalue \(\lambda\) of a matrix \(A\) is the largest integer \(m\) so that \((x-\lambda)^m\) divides the characteristic polynomial of \(A\).

Summary

Theorem 1 (Facts about eigenvalues) Let \(A\) be \(n\times n\) with real entries. Then

• There are at most \(n\) eigenvalues of \(A\). If each eigenvalue (including complex eigenvalues) is counted with its multiplicity, there are exactly \(n\) eigenvalues.
• If \(A\) has a complex eigenvalue \(\lambda = a + bi\), the complex conjugate \(\overline{\lambda} = a-bi\) is also an eigenvalue of \(A\).
• If \(n\) is odd, \(A\) has at least one real eigenvalue.
• If \(A\) is upper/lower-triangular, the eigenvalues are the entries on the diagonal.

The Geometry of Eigenspaces

Recall

For each eigenvalue \(\lambda\) of \(A\), the eigenspace of \(A\) corresponding to the eigenvalue \(\lambda\) is \(\null(A-\lambda I_n)\).

That is, given a matrix \(A\) and eigenvalue \(\lambda\) of \(A\):

• Solve \((A-\lambda I_n)\x = \mathbf{0}\)
• The resulting vectors are linearly independent, and form a basis for the subspace \(E_\lambda\) of \(\R^n\) that they span.
• The space \(E_\lambda\) is called the eigenspace corresponding to \(\lambda\).

Dimensions of Eigenspaces: Activity 17.2

• Find the dimension of the eigenspace for each eigenvalue of \(A = \left[\begin{matrix} 3 & -2 & 5 \\ 1 & 0 & 7 \\ 0 & 0 & 1 \end{matrix}\right]\).

• Find the dimension of the eigenspace for each eigenvalue of \(A = \left[\begin{matrix} 1 & 0 & 0 & 1 \\ 1 & 2 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{matrix}\right]\).

• Consider a \(3\times 3\) matrix with three distinct eigenvalues \(\lambda_1, \lambda_2, \lambda_3\).

  • Recall that a polynomial of degree 3 can have at most three distinct roots. What does that say about the multiplicities of \(\lambda_1, \lambda_2, \lambda_3\)?
  • Use the fact that eigenvectors corresponding to distinct eigenvalues are linearly independent to find the dimensions of the eigenspaces for \(\lambda_1,\lambda_2,\lambda_3\).

A = Matrix([[3,-2,5],[1,0,7],[0,0,1]])
B = Matrix([[1,0,0,1],[1,2,0,0],[0,0,1,0],[0,0,0,1]])

A.eigenspaces_right(), B. eigenspaces_right()

([
 (2, Vector space of degree 3 and dimension 1 over Rational Field
 User basis matrix:
 [  1 1/2   0]),
 (1, Vector space of degree 3 and dimension 1 over Rational Field
 User basis matrix:
 [1 1 0])
 ],
 [
 (2, Vector space of degree 4 and dimension 1 over Rational Field
 User basis matrix:
 [0 1 0 0]),
 (1, Vector space of degree 4 and dimension 2 over Rational Field
 User basis matrix:
 [ 1 -1  0  0]
 [ 0  0  1  0])
 ])

Geometric Multiplicity

Definition 4 (Geometric Multiplicity) The geometric multiplicity of an eigenvalue of an \(n\times n\) matrix \(A\) is the dimension of the corresponding eigenspace \(\null(A-\lambda I_n)\).

Exploration

Exercise 1 Find the eigenvalues of the following matrices. For each eigenvalue, determine its algebraic and geometric multiplicity.

• \(A = \left[\begin{matrix} 5 & -8 & 1 \\ 0 & 0 & 7 \\ 0 & 0 & -2 \end{matrix}\right]\)

• \(B = \left[\begin{matrix} 2 & 4 & 3 \\ -4 & -6 & -3 \\ 3 & 3 & 1 \end{matrix}\right]\)

A = Matrix([[5,-8,1],[0,0,7],[0,0,-2]])
B = Matrix([[2,4,3],[-4,-6,-3],[3,3,1]])

A.eigenvalues(), A.eigenspaces_right(format='all'), B.eigenvalues(), B.eigenspaces_right(format='all')

([5, 0, -2],
 [
 (5, Vector space of degree 3 and dimension 1 over Rational Field
 User basis matrix:
 [1 0 0]),
 (0, Vector space of degree 3 and dimension 1 over Rational Field
 User basis matrix:
 [  1 5/8   0]),
 (-2, Vector space of degree 3 and dimension 1 over Rational Field
 User basis matrix:
 [    1 49/58 -7/29])
 ],
 [1, -2, -2],
 [
 (1, Vector space of degree 3 and dimension 1 over Rational Field
 User basis matrix:
 [ 1 -1  1]),
 (-2, Vector space of degree 3 and dimension 1 over Rational Field
 User basis matrix:
 [ 1 -1  0])
 ])