Diagonalization

Math 303: Section 18

Dr. Janssen

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

Intro to Diagonalization

Preview Activity

Computing powers of a matrix is often desirable.

x0 = Matrix([[1],[0]])
A = Matrix([[0.7,0.4],[0.3,0.6]])
A^10*x0, A^20*x0, A^30*x0

(
[0.571431102100000]  [0.571428571443514]  [0.571428571428571]
[0.428568897900000], [0.428571428556485], [0.428571428571428]
)

Activity 18.1

Let \(D = \left[\begin{matrix} 2 & 0 \\ 0 & 3 \end{matrix}\right]\).

Explain in general why

\[ D^n = \left[\begin{matrix} 2^n & 0 \\ 0 & 3^n \end{matrix}\right]. \]

This will be true in general!

Activity 18.2

Let \(D\) be any matrix, \(P\) an invertible matrix, and \(A = P^{-1} D P\).

• Show that \(A^2 = P^{-1} D^2 P\).
• Show that \(A^3 = P^{-1} D^3 P\).
• Explain in general why \(A^n = P^{-1} D^n P\).

thus, we only need to determine powers of \(D\). If \(D\) is diagonal, this is straightforward!

Similarity

Definition

Definition 1 The \(n\times n\) matrix \(A\) is similar to the \(n\times n\) matrix \(B\) if there is an invertible matrix \(P\) such that \(A = P^{-1} B P\).

Activity 18.3

Let \(A= \left[\begin{matrix} 1 & 1 \\ 2 & 0 \end{matrix}\right]\) and \(B = \left[\begin{matrix} 2 & 2 \\ 0 & -1 \end{matrix}\right]\).

• Let \(P = \left[\begin{matrix} 1 & -1/2 \\ -1 & 1 \end{matrix}\right]\); then \(P^{-1} = \left[\begin{matrix} 2 & 1 \\ 2 & 2 \end{matrix}\right]\). Verify that \(A = P^{-1} B P\) (perhaps with sage).
• Calculate \(\det(A)\) and \(\det(B)\), and record your observations.
• Find the characteristic polynomials of \(A\) and \(B\). What do you notice?
• What can you say about the eigenvalues of \(A\) and \(B\)? Why?
• Explain why \(\x = \left[\begin{matrix} 1 \\ 1 \end{matrix}\right]\) is an eigenvector for \(A\) with eigenvalue 2. Is \(\x\) an eigenvector for \(B\) with eigenvalue 2? Why or why not?

Code

A = Matrix([[1,1],[2,0]])
B = Matrix([[2,2],[0,-1]])
P = Matrix([[2,1],[2,2]])
Pinv = P.inverse()

print("A and B are similar:")
A == P*B*Pinv

A and B are similar:

True

print("Determinants:")
det(A), det(B)

Determinants:

(-2, -2)

print("Characteristic polynomials:")
A.charpoly(), B.charpoly()

Characteristic polynomials:

(x^2 - x - 2, x^2 - x - 2)

print("eigenvalues:")

A.eigenvalues(), B.eigenvalues()

eigenvalues:

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

x = Matrix([[1],[1]])

print("x is an eigenvector of A with eigenvalue 2, but not B:")
A*x == 2*x, B*x == 2*x

x is an eigenvector of A with eigenvalue 2, but not B:

(True, False)

Takeaways

• Similar matrices share some, but not all, properties
• Why don’t similar matrices have to have the same eigenvectors?

Activity 18.4

Let \(A\) and \(B\) be similar matrices with \(A = P^{-1} B P\).

• Use the multiplicative property of the determinant to explain why \(\det(A) = \det(B)\).
• Use the fact that \(P^{-1} I P = I\) to show that \(A - \lambda I\) is similar to \(B - \lambda I\).
• Explain why it follows from the previous two parts that

\[ \det (A - \lambda I) = \det(B -\lambda I). \]

Summary

Theorem 1 (Properties of Similar Matrices) Let \(A\) and \(B\) be similar \(n\times n\) matrices and \(I\) the \(n\times n\) identity matrix. Then:

• \(\det(A) = \det(B)\)
• \(A-\lambda I\) is similar to \(B - \lambda I\)
• \(A\) and \(B\) have the same characteristic polynomial
• \(A\) and \(B\) have the same eigenvalues.

Similarity and Matrix Transformations

Recall

A matrix transformation via an \(m\times n\) matrix \(A\) takes \(\R^n\) to \(\R^m\) via left multiplication.

Another use of similar matrices is to rewrite a given matrix transformation (with respect to the standard basis) using a “nicer” matrix.

Example

Let \(A = \left[\begin{matrix} 7 & 2 \\ -4 & 1 \end{matrix}\right]\) define a transformation \(T: \R^2 \to \R^2\). We can verify that the eigenvalues of \(A\) are \(5\) and \(3\), corresponding to eigenvectors \(\v_1 = \left[\begin{matrix} 1 \\ -1 \end{matrix}\right]\) and \(\v_2 = \left[\begin{matrix} 1 \\ -2 \end{matrix}\right]\). If we let \(P = [ \v_1 \ \v_2]\) and \(D = \left[\begin{matrix} 5 & 0 \\ 0 & 3 \end{matrix}\right]\), then

\[ A = P D P^{-1}. \]

It is a fact that \(P^{-1} \mathbf{e}_j = \v_j\). That is, \(P^{-1}\) “rewrites” the standard basis vectors in an eigenbasis. Then we apply the transformation via multiplication by \(D\), and then transform back via multiplication by \(P\).

A = Matrix([[7,2],[-4,1]])
D = Matrix([[5,0],[0,3]])
P = Matrix([[1, 1],[-1, -2]])
Pinv = P.inverse()

A == P*D*Pinv

True

Diagonalization in General

The definition

Definition 2 Let \(A\) be \(n\times n\). We say \(A\) is diagonalizable if there is an invertible \(n\times n\) matrix \(P\) such that \(P^{-1} A P\) is a diagonal matrix.

The process

• Let \(A\) be \(n\times n\).
• Find the eigenvalues of \(A\), and the corresponding eigenvectors.
• If there are \(n\) linearly independent eigenvectors \(\v_1, \ldots, \v_n\), then \(P = [ \v_1 \cdots \v_n]\), and \(D\) is formed by putting the eigenvalues on the main diagonal in order, repeating as necessary. Congratulations! \(A\) is diagonalizable.
• If there are not \(n\) linearly independent eigenvectors, then \(A\) is not diagonalizable.

Exploration

Find, if possible, an invertible matrix \(P\) that diagonalizes \(A\), and write \(A = P D P^{-1}\).

• \(A = \left[\begin{matrix} 1 & 1 \\ 0 & 2 \end{matrix}\right]\)
• \(A = \left[\begin{matrix} 3 & 2 & 4 \\ 2 & 0 & 2 \\ 4 & 2 & 3 \end{matrix}\right]\)
• \(A = \left[\begin{matrix} 2 & 4 & 3 \\ -4 & -6 & -3 \\ 3 & 3 & 1 \end{matrix}\right]\)

A1 = Matrix([[1,1],[0,2]])
A2 = Matrix([[3,2,4],[2,0,2],[4,2,3]])
A3 = Matrix([[2,4,3],[-4,-6,-3],[3,3,1]])

A1.eigenspaces_left(), A2.eigenspaces_left(), A3.eigenspaces_left()

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