(I) Determinants (Review)

Theorem (The Invertible Matrix Theorem #19)

Let \(A\) be an \(n \times n\) matrix. Then \(A\) is invertible if and only if \(0\) is not an eigenvalue of \(A\).

Theorem 3 (Properties of Determinants)

Let \(A\) and \(B\) be \(n \times n\) matrices.

  1. \(A\) is invertible if and only if \(\det A \neq 0\).
  2. \(\det(AB) = (\det A)(\det B)\).
  3. \(\det A^T = \det A\).
  4. If \(A\) is triangular, then \(\det A\) is the product of the entries on the main diagonal of \(A\).
  5. A row replacement operation on \(A\) doesn't change the determinant. A row interchange changes the sign of the determinant. A row scaling also scales the determinant by the same scalar.

(II) The Characteristic Equation

We want to find all scalars \(\lambda\) such that \((A - \lambda I)\vec{x} = \vec{0}\) has a nontrivial solution.

Example 1

Find the eigenvalues of \( A = \begin{bmatrix} 1 & 3 & 3 \\ -3 & -5 & -3 \\ 3 & 3 & 1 \end{bmatrix} \).

Solution: Form the matrix \(A - \lambda I\) and compute its determinant:

\( \det(A - \lambda I) = \begin{vmatrix} 1-\lambda & 3 & 3 \\ -3 & -5-\lambda & -3 \\ 3 & 3 & 1-\lambda \end{vmatrix} \)

Perform row operations to simplify:

\( \xrightarrow{R_2' = R_1 + R_2, R_3' = -R_1 + R_3} \begin{vmatrix} 1-\lambda & 3 & 3 \\ -2-\lambda & -2-\lambda & 0 \\ 2+\lambda & 0 & -2-\lambda \end{vmatrix} \)

Factor out \(-(2+\lambda)\) from the second and third rows:

\( = (-2-\lambda)(-2-\lambda) \begin{vmatrix} 1-\lambda & 3 & 3 \\ 1 & 1 & 0 \\ -1 & 0 & 1 \end{vmatrix} = (-2-\lambda)^2 \begin{vmatrix} 1-\lambda & 3 & 3 \\ 1 & 1 & 0 \\ -1 & 0 & 1 \end{vmatrix} \)

Expanding gives:

\( = (-2-\lambda)^2 (-3+2+\lambda) = -(\lambda+2)^2(\lambda-1) = 0 \)

The eigenvalues are \(\lambda = 1, -2\) (with multiplicity 2).

Definitions

The Characteristic Equation of \(A\) is: \(\det(A - \lambda I) = 0\).

The Characteristic Polynomial: If \(A\) is an \(n \times n\) matrix, \(\det(A - \lambda I)\) is a polynomial of degree \(n\) called the characteristic polynomial of \(A\).

The characteristic equation has exactly \(n\) roots, counting multiplicities. (Complex roots will be discussed in Section 5.5).

The algebraic multiplicity of an eigenvalue is its multiplicity as a root of the characteristic equation.

Example 2

The characteristic polynomial of a \(6 \times 6\) matrix is \(\lambda^6 - 4\lambda^5 - 12\lambda^4\). Find the eigenvalues and their multiplicities.

Solution: Factor the polynomial:

\[ \lambda^6 - 4\lambda^5 - 12\lambda^4 = \lambda^4(\lambda^2 - 4\lambda - 12) = \lambda^4(\lambda + 2)(\lambda - 6) \]

The eigenvalues are:

  • \(\lambda = 0\) (multiplicity 4)
  • \(\lambda = -2\) (multiplicity 1)
  • \(\lambda = 6\) (multiplicity 1)
Example 3

Find \(h\) in matrix \(A\) below such that the eigenspace for \(\lambda = 6\) is 2-dimensional.

\[ A = \begin{bmatrix} 6 & -2 & 6 & -2 \\ 0 & 4 & h & 0 \\ 0 & 0 & 6 & 5 \\ 0 & 0 & 0 & 1 \end{bmatrix} \]

Solution: We know \(\lambda = 6\) is an eigenvalue. Compute \(A - 6I\):

\( A - 6I = \begin{bmatrix} 0 & -2 & 6 & -2 \\ 0 & -2 & h & 0 \\ 0 & 0 & 0 & 5 \\ 0 & 0 & 0 & -5 \end{bmatrix} \sim \begin{bmatrix} 0 & -2 & 6 & -2 \\ 0 & 0 & h-6 & 2 \\ 0 & 0 & 0 & 5 \\ 0 & 0 & 0 & 0 \end{bmatrix} \sim \begin{bmatrix} 0 & -2 & 6 & -2 \\ 0 & 0 & h-6 & 2 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 \end{bmatrix} \)

For the eigenspace to be 2-dimensional, we need exactly 2 free variables. The first column corresponds to a free variable. To get a second one, we need the third column to be free, which requires \(h - 6 = 0\).

Therefore, \(h = 6\).

(III) Similarity

Definition

If \(A\) and \(B\) are \(n \times n\) matrices, then \(A\) is similar to \(B\) if there is an invertible matrix \(P\) such that:

\[ A = PBP^{-1} \quad \text{or} \quad B = P^{-1}AP \]

The transformation \(A \mapsto P^{-1}AP\) is called a similarity transformation.

Theorem 4

If \(n \times n\) matrices \(A\) and \(B\) are similar, then they have the same characteristic polynomial and hence the same eigenvalues (with the same multiplicities).

Proof:

If \(B = P^{-1}AP\), then:

\[ B - \lambda I = P^{-1}AP - \lambda P^{-1}P = P^{-1}(A - \lambda I)P \]

Taking the determinant of both sides:

\[ \det(B - \lambda I) = \det(P^{-1}(A - \lambda I)P) = \det(P^{-1}) \cdot \det(A - \lambda I) \cdot \det(P) \]

Since \(\det(P^{-1}) = \frac{1}{\det P}\), we have:

\[ \det(B - \lambda I) = \left(\frac{1}{\det P}\right) \cdot \det(P) \cdot \det(A - \lambda I) = \det(A - \lambda I) \]

Warnings

  1. The opposite statement is false. Matrices with the same eigenvalues are not necessarily similar.
  2. Similarity is different from row equivalence. Row operations on a matrix usually change its eigenvalues.
Example 4

Show that \(A\) and \(A^T\) have the same characteristic polynomial.

Solution:

\[ \det(A - \lambda I) = \det((A - \lambda I)^T) = \det(A^T - (\lambda I)^T) = \det(A^T - \lambda I) \]

Thus, they share the same characteristic polynomial.

Application

Several common algorithms (e.g., QR algorithm / Jacobi method) compute a sequence of matrices \(A_1 = A\) and \(A_{k+1} = P_k^{-1}A_kP_k\). As \(k\) increases, the nondiagonal entries of \(A_{k+1}\) approach \(0\), and the diagonal entries approach the eigenvalues of \(A\).