In the previous lecture, we learned how to determine whether a given vector is an eigenvector for a matrix, and whether a given scalar is an eigenvalue for that matrix. We also learned how to find eigenvectors for a given eigenvalue. However, we still don't know how to find the eigenvalues for a matrix in the first place.
Given a square matrix \( A \), we know that \( \lambda \) is an eigenvalue for \( A \) if and only if the equation \( (A-\lambda I)\bbm x = \bbm 0 \) has nontrivial solutions. By the Invertible Matrix Theorem, this means that \( \lambda \) is an eigenvalue for \( A \) whenever the matrix \( A - \lambda I \) is singular. So, \( \lambda \) is an eigenvalue for \( A \) if and only if \( \det (A - \lambda I) = 0 \).
Example 1. Find the eigenvalues of \( A = \begin{bmatrix} 2 & 3 \\ 3 & -6 \end{bmatrix} \).
We first find \( \det (A-\lambda I) \): \[ \det (A-\lambda I) = \det \left( \begin{bmatrix} 2 & 3 \\ 3 & -6 \end{bmatrix} - \begin{bmatrix} \lambda & 0 \\ 0 & \lambda \end{bmatrix} \right) = \det \begin{bmatrix} 2-\lambda & 3 \\ 3 & -6-\lambda \end{bmatrix} = (2-\lambda)(-6-\lambda) - (3)(3) = \lambda^2 +4\lambda - 21. \]
Setting this equal to zero and solving gives two solutions: \( \lambda = 7 \) and \( \lambda = -3 \), which are the eigenvalues of \( A \). \( \Box \)
If \( A \) is an \( n\times n\) matrix, then \( \det (A-\lambda I) \) is a polynomial with degree \( n \). The roots of this polynomial, which are the solutions of the equation \( \det (A-\lambda I) = 0\), are the eigenvalues of \( A \).
Definition. Given an \( n \times n\) matrix \( A \), the characteristic polynomial of \( A \) is \( \det (A-\lambda I) \). The characteristic equation of \( A \) is \( \det (A-\lambda I) = 0 \).
Example 2. Let \( B = \begin{bmatrix} 1 & 2 & -1 \\ 0 & 0 & 2 \\ 0 & 1 & 1 \end{bmatrix} \). Find the characteristic polynomial of \( B \). Solve the characteristic equation of \( B \) to find its eigenvalues.
First, the characteristic polynomial of \( B \) is \[ \det (B - \lambda I) = \det \begin{bmatrix} 1-\lambda & 2 & -1 \\ 0 & -\lambda & 2 \\ 0 & 1 & 1-\lambda \end{bmatrix} = -\lambda^3 +2\lambda^2 +\lambda -2. \]
Setting this equal to zero and solving gives \( \lambda=-1\), \(\lambda=1\), and \( \lambda = 2\). \( \Box \)
Example 3. Let \( C = \begin{bmatrix} 1 & 3 & -6 & 1 \\ 0 & 4 & 10 & 0 \\ 0 & 0 & 4 & 5 \\ 0 & 0 & 0 & 0 \end{bmatrix} \). Find the eigenvalues of \( C \).
Again, we need to find the characteristic polynomial: \[ \det (C - \lambda I) = \det \begin{bmatrix} 1-\lambda & 3 & -6 & 1 \\ 0 & 4-\lambda & 10 & 0 \\ 0 & 0 & 4-\lambda & 5 \\ 0 & 0 & 0 & -\lambda \end{bmatrix} \]
While this \( 4\times 4\) determinant may seem daunting at first, we recognize that \( C - \lambda I \) is a triangular matrix, which means that its determinant is the product of its diagonal entries: \( \det (C-\lambda I) = (1-\lambda)(4-\lambda)(4-\lambda)(-\lambda) \). Setting this equal to zero gives three solutions: \( \lambda = 1\), \(\lambda =4\), and \(\lambda =0\). \( \Box \)
We saw in Example 3 that the eigenvalue \( \lambda = 4 \) appeared twice as a root of the characteristic polynomial. We say that this eigenvalue has "multiplicity 2."
Definition. The multiplicity of an eigenvalue is the number of times that eigenvalue appears as a root of the characteristic polynomial.
For an \( n\times n\) matrix \( A \), the characteristic polynomial of \( A \) has degree \( n\). Since eigenvalues can have multiplicities greater than 1, this doesn't guarantee that we always have \( n \) eigenvalues. Are we guaranteed to have \( n \) eigenvalues if we account for multiplicities? The following example illustrates that the answer is still no.
Example 4. Let \( M = \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix} \). Find the eigenvalues of \( M \).
We find the characteristic polynomial: \[ \det (M-\lambda I) = \det \begin{bmatrix} -\lambda & -1 \\ 1 & -\lambda \end{bmatrix} = \lambda^2 + 1. \]
The characteristic equation is \( \lambda^2 + 1 = 0 \), which has no real solutions. So, \( M \) has no real eigenvalues. \( \Box \)
If we allow ourselves to consider complex eigenvalues, then it is true that an \( n\times n\) matrix has \( n \) complex eigenvalues, counting multiplicities. However, for the purposes of this course, we will stick to only considering real eigenvalues.
Definition. Two \( n\times n\) matrices \( A \) and \( B \) are similar, written \( A \sim B \), if there exists an invertible matrix \( P \) for which \( B = PAP^{-1} \).
Theorem (Eigenvalues of Similar Matrices). If \( A \) and \( B \) are similar matrices, then \( A \) and \( B \) have the same eigenvalues with the same multiplicities.
We will prove this theorem by showing that \( A \) and \( B \) have the same characteristic polynomial. First, we will need to recall some facts about determinants from Lecture 26.
If \( X \) and \( Y \) are square matrices, then \( \det (XY) = \det X \cdot \det Y \). It follows that, if \( P \) is invertible, then since \( \det(I) = 1 \) and \( \det (PP^{-1}) = \det P \cdot \det P^{-1} \), we have \( \det P^{-1} = \frac 1 {\det P} \).
Proof of the Eigenvalues of Similar Matrices Theorem. Let \( A \) and \( B \) be similar, with \( B = PAP^{-1} \) for some invertible matrix \( P \). Then, \[ \begin{eqnarray*} \det (B-\lambda I) & = & \det(PAP^{-1} - \lambda I) \\ & = & \det(PAP^{-1} - P(\lambda I)P^{-1}) \\ & = & \det(P (A-\lambda I) P^{-1}) \\ & = & \det P \cdot \det (A-\lambda I) \cdot \det P^{-1} \\ & = & \det(A-\lambda I). \end{eqnarray*} \]
Since \( A \) and \( B \) have the same characteristic polynomial, they must have the same eigenvalues with the same multiplicities. \( \Box \)
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