Definition. Given a square matrix \( A \) and a nonzero vector \( \bbm v \in \mathbb R^n \), we say that \( \bbm v \) is an eigenvector for \( A \) if \( A \bbm v = \lambda \bbm v \) for some scalar \( \lambda \). In this situation, we say that \( \lambda \) is the eigenvalue corresponding to \( \bbm v\).
Example 1. Let \( A = \begin{bmatrix} 1 & 6 \\ 5 & 2 \end{bmatrix} \) and \( \bbm v = \vectwo 6 {-5} \). Show that \( \bbm v\) is an eigenvector for \( A \) and find the corresponding eigenvalue.
We multiply \( A \bbm v = \begin{bmatrix} 1 & 6 \\ 5 & 2 \end{bmatrix} \vectwo 6 {-5} = \vectwo {-24} {20} \), which is equal to \( -4\bbm v \). So, \( \bbm v\) is an eigenvector for \( A \) corresponding to the eigenvalue \( -4 \). \( \Box \)
Example 2. Let \( B = \begin{bmatrix} -7 & 0 & -9 \\ 0 & -1 & 0 \\ 6 & 0 & 8 \end{bmatrix} \), \( \bbm p = \vecthree {-5} 0 5 \), and \( \bbm q = \vecthree 123 \). Determine whether \( \bbm p \) or \( \bbm q \) are eigenvectors for \( B \) and, if so, find the corresponding eigenvalue.
For \(\bbm p\), we multiply \(B\bbm p = \vecthree {-10} 0 {10} \), which equals \(2\bbm p\). So, \(\bbm p\) is an eigenvector for \(B\) corresponding to the eigenvalue 2.
For \(\bbm q\), we multiply \(B\bbm q = \vecthree {-34} {-2} {30}\), which is not a scalar multiple of \(\bbm q\). So, \(\bbm q\) is not an eigenvector for \( B \). \( \Box \)
Eigenvectors are important for when we want to repeatedly multiply a matrix by a vector as we did in Lecture 12 when we studied input-output models. If \( \bbm v \) is an eigenvector for a matrix \( A \) with \( A \bbm v = \lambda \bbm v\), then \[ A^k \bbm v = \underbrace{AA\cdots A}_{k\ {\rm times}}\bbm v = \lambda^k \bbm v. \] It takes much less computational power to raise a scalar to the \( k^{\rm th} \) power than to repeatedly multiply a matrix by a vector \( k \) times.
For a matrix \( A \) to have any chance of having eigenvectors, it must be square. If \( A \) is not square, then \( A \bbm v \) and \( \bbm v\) have different numbers of entries, and so there would be no chance for \( A \bbm v \) to ever equal \( \lambda \bbm v\).
Note that, in the definition of eigenvector, the vector \( \bbm v\) must be nonzero. If \( \bbm v = \bbm 0 \), then \( A\bbm v = \bbm 0 \) and \(\lambda \bbm v = \bbm 0 \) so matter what \( \lambda \) is. The situation is only interesting or notable when \( \bbm v \) is nonzero. Also notice that the same vector \( \bbm v \) cannot be an eigenvector for two different eigenvalues. If we had \( A \bbm v = \lambda_1 \bbm v\) and \( A \bbm v = \lambda_2 \bbm v \), then subtracting gives \( (\lambda_1 - \lambda_2) \bbm v = \bbm 0 \). Since \( \bbm v \neq \bbm 0\), this implies \( \lambda_1 = \lambda_2 \).
Given a square matrix \( A \), how can we tell whether a given value \( \lambda \) is an eigenvalue for \( A \)? We would need to consider whether the equation \( A \bbm x = \lambda \bbm x\) has any nonzero solutions. We can rewrite this equation by expressing \( \lambda \bbm x \) as \( (\lambda I)\bbm x\) and subtracting this from both sides to get the homogeneous equation \( (A-\lambda I)\bbm x = \bbm 0 \). If this equation has nonzero solutions, then those solutions are eigenvectors for \( A \).
Example 3. Show that \( \lambda = 7\) is an eigenvalue for \( A = \begin{bmatrix} 1 & 6 \\ 5 & 2 \end{bmatrix} \) and find the corresponding eigenvectors.
First, we construct \( A - 7I \) and row-reduce it to find the solutions of \( (A-7I)\bbm x = \bbm 0 \): \[ A-7I = \begin{bmatrix} 1 & 6 \\ 5 & 2 \end{bmatrix} - \begin{bmatrix} 7 & 0 \\ 0 & 7 \end{bmatrix} = \begin{bmatrix} -6 & 6 \\ 5 & -5 \end{bmatrix} \longrightarrow \begin{bmatrix} 1 & -1 \\ 0 & 0 \end{bmatrix} \]
This gives the solution \( x_1 = x_2 \) and \( x_2 \) is free. In parametric form, the solution is \( \bbm x= x_2 \vectwo 11 \). Any vector of this form is an eigenvector for \( A \) corresponding to the eigenvalue \( \lambda = 7 \). \( \Box \)
For a given value of \(\lambda\), the set of all solutions of \( (A-\lambda I)\bbm x = \bbm 0 \) is just the null space of the matrix \( A - \lambda I \). This means that the collection of eigenvectors corresponding to a given eigenvalue, plus the zero vector, forms a subspace of \( \mathbb R^n \).
Definition. Given an \(n\times n\) matrix \( A \) with eigenvalue \( \lambda \), the eigenspace of \( A \) corresponding to \(\lambda \) is the set of all eigenvectors of \( A \) corresponding to \( \lambda \), together with the zero vector.
Example 4. Let \( C = \begin{bmatrix} -3 & 2 & 3 \\ 2 & 0 & 6 \\ 1 & 2 & -1 \end{bmatrix} \). Show that \( \lambda = -4 \) is an eigenvalue for \( C \) and find a basis for the corresponding eigenspace.
Remember that the eigenspace is just the null space of \( C - (-4)I \). As we learned in Lecture 30, to find a basis for this null space we find the parametric solution of \( (C-(-4)I)\bbm x = \bbm 0 \): \[ C-(-4)I = C+4I = \begin{bmatrix} -3 & 2 & 3 \\ 2 & 0 & 6 \\ 1 & 2 & -1 \end{bmatrix} + \begin{bmatrix} 4 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & 4 \end{bmatrix} = \begin{bmatrix} 1 & 2 & 3 \\ 2 & 4 & 6 \\ 1 & 2 & 3 \end{bmatrix} \longrightarrow \begin{bmatrix} 1 & 2 & 3 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}. \]
The parametric solution is \( \bbm x = x_2 \vecthree {-2} 1 0 + x_3 \vecthree {-3} 0 1 \), so a basis for the \( \lambda = -4 \) eigenspace is \( \left\{ \vecthree {-2} 1 0, \vecthree {-3} 0 1 \right\}\). \( \Box \)
Given a square matrix \( A \), how are eigenvectors for different eigenvalues of \( A \) related? The next theorem describes an important relationship.
Theorem (Eigenvectors for Distinct Eigenvalues). Let \( A \) be an \( n \times n\) matrix and let \( \bbm v_1, \bbm v_2, \ldots, \bbm v_r \) be eigenvectors corresponding to distinct eigenvalues \( \lambda_1, \lambda_2, \ldots, \lambda_r \). The set \( \{ \bbm v_1, \bbm v_2, \ldots, \bbm v_r \} \) is linearly independent.
Proof. Suppose that the set \( \{ \bbm v_1, \bbm v_2, \ldots, \bbm v_r \} \) is linearly dependent. By the Characterization of Linearly Dependent Sets Theorem, one of the vectors in this set is a linear combination of the preceding vectors. Let \( p \) be the first vector in the set that is a linear combination of the preceding vectors. This means that the set \( \{ \bbm v_1, \ldots, \bbm v_{p-1} \} \) is linearly independent.
Now we have \( \bbm v_p = c_1 \bbm v_1 + c_2 \bbm v_2 + \cdots + c_{p-1} \bbm v_{p-1} \) for some scalars \( c_1, c_2, \ldots, c_{p-1} \). Call this equation (*).
Multiplying both sides of (*) by \( A \) and remembering that each \( \bbm v_i \) is an eigenvector corresponding to \( \lambda_i \) gives \[ \begin{eqnarray*} A \bbm v_p & = & Ac_1 \bbm v_1 + Ac_2 \bbm v_2 + \cdots + Ac_{p-1} \bbm v_{p-1} \\ \lambda_p \bbm v_p & = & \lambda_1 c_1 \bbm v_1 + \lambda_2 c_2 \bbm v_2 + \cdots + \lambda_{p-1} c_{p-1} \bbm v_{p-1}. \end{eqnarray*} \]
Multiplying both sides of (*) by \( \lambda_p \) gives \[ \lambda_p \bbm v_p = \lambda_p c_1 \bbm v_1 + \lambda_p c_2 \bbm v_2 + \cdots + \lambda_p c_{p-1} \bbm v_{p-1}. \]
Now we subtract these two equations to get \[ \bbm 0 = c_1 (\lambda_1 - \lambda_p) \bbm v_1 + c_2 (\lambda_2 - \lambda_p) \bbm v_2 + \cdots + c_{p-1} (\lambda_{p-1} - \lambda_p) \bbm v_{p-1}. \]
Since the set \( \{ \bbm v_1, \ldots, \bbm v_{p-1} \} \) is linearly independent, it must be that all the scalars \( c_i (\lambda_i - \lambda_p)\) equal zero. This means that either \( c_i = 0 \) or \( \lambda_i - \lambda_p = 0\) for each \( i \). However, since all the \( \lambda \)'s are distinct, we cannot have \( \lambda_i - \lambda_p = 0 \). Thus, it must be that \( c_i = 0 \) for all \( 1 \le i \le p-1 \).
Going back to (*), we now have \[ \begin{eqnarray*} \bbm v_p & = & c_1 \bbm v_1 + c_2 \bbm v_2 + \cdots + c_{p-1} \bbm v_{p-1} \\ \bbm v_p & = & 0 \bbm v_1 + 0 \bbm v_2 + \cdots + 0 \bbm v_{p-1} \\ \bbm v_p & = & \bbm 0. \end{eqnarray*} \]
However, \( \bbm v_p \) is an eigenvector for \( A \), which means that \( \bbm v_p \) is nonzero. This contradiction shows that our assumption that \( \{ \bbm v_1, \bbm v_2, \ldots, \bbm v_r \} \) is a linearly dependent set was wrong, which completes the proof. \( \Box \)
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