Section 6.1 Eigenvalues and eigenvectors: preliminaries
Eigenvalues and eigenvectors are defined for a square matrix \(A.\)
Definition 6.1.1. The eigenvalue of a matrix.
A number \(\lambda\) is an eigenvalue of a square matrix \(A\) if
for some \(\vec x\not=\vec0\text{.}\)
Definition 6.1.2. The eigenvector of a matrix.
If
for some \(\vec x\not=\vec0\text{,}\) then \(\vec x\) is called an eigenvector of \(A\) corresponding to the eigenvalue \(\lambda\text{.}\)
Notice that if \(\vec x=\vec 0\text{,}\) then \(A\vec x=\lambda\vec x\) is simply the equation \(\vec0=\vec0\) for any value of \(\lambda\text{.}\) This is not too interesting, and so we always have the restriction \(\vec x\not=\vec0\text{.}\)
Example 6.1.3. Eigenvalues of \(A= \begin{bmatrix} 5\amp -1\amp -2\\ 1\amp 3\amp -2\\ -1\amp -1\amp 4 \end{bmatrix} \).
which makes \(\vec x=\begin{bmatrix}1\\1\\1\end{bmatrix}\) an eigenvector with eigenvalue \(\lambda=2\text{.}\)
which makes makes \(\vec x=\begin{bmatrix} 1\\-1\\1 \end{bmatrix}\) an eigenvector with eigenvalue \(\lambda=4\text{.}\)
which makes makes \(\vec x=\begin{bmatrix}-1\\-1\\1\end{bmatrix}\) an eigenvector with eigenvalue \(\lambda=6\text{.}\)
We have now computed eigenvalues of \(A\text{:}\) \(\lambda=2\text{,}\) \(\lambda=4\) and \(\lambda=6\text{.}\) With a little more theory, we will see that there are no others.
Example 6.1.4. Eigenvalues of \(\begin{bmatrix}2\amp 1\amp 4\\ 0\amp 3\amp 0\\ 2\amp -2\amp -5 \end{bmatrix}\).
Let \(A=\begin{bmatrix}2\amp 1\amp 4\\ 0\amp 3\amp 0\\ 2\amp -2\amp -5 \end{bmatrix}\) Then we have
which makes \(\vec x=\begin{bmatrix}1\\1\\0\end{bmatrix}\) an eigenvector with eigenvalue \(\lambda=3\text{.}\) Also
which makes \(\vec x=\begin{bmatrix}4\\0\\1\end{bmatrix}\) an eigenvector with eigenvalue \(\lambda=3\) Finally
which makes \(\vec x=\begin{bmatrix}-1\\0\\2\end{bmatrix}\) an eigenvector with eigenvalue \(\lambda=-6\) Hence the demonstrated eigenvalues are \(\lambda=3\) and \(\lambda=-6\text{.}\) We will soon see that there are no more.
Example 6.1.5. Eigenvalues of \(\begin{bmatrix} 0\amp -1\\ 1\amp 0\end{bmatrix}\).
Let \(A=\begin{bmatrix} 0\amp -1\\ 1\amp 0\end{bmatrix}\) and consider the equation \(A\vec x=\lambda\vec x\text{.}\) Setting \(\vec x=\begin{bmatrix} x_1\\x_2\end{bmatrix}\text{,}\)
This is equivalent to the system of equations
This implies that \((\lambda^2+1)x_1 = \lambda^2 x_1 + x_1 = -\lambda x_2 +\lambda x_2=0 \) and \((\lambda^2+1)x_2 = \lambda^2 x_2 + x_2 = \lambda x_1 -\lambda x_21=0.\) Since \(\vec x \not=\vec 0\text{,}\) either \(x_1\not=0\) or \(x_2\not=0\text{,}\) and consequently \(\lambda^2=-1\text{.}\) Since no real number satisfies this equation, we conclude that there are no eigenvalues or eigenvectors for \(A\text{.}\) An aside: if we consider complex numbers, then \(i\) and \(-i\) are both eigenvalues of \(A\) with \(\begin{bmatrix} 1\\-i\end{bmatrix}\) and \(\begin{bmatrix} 1\\i\end{bmatrix}\) as corresponding eigenvectors.
Example 6.1.6. Eigenvalues of \(\begin{bmatrix} 1 \amp 1 \amp 1 \amp 1\\ 0 \amp 2 \amp 2 \amp 2\\ 0 \amp 0 \amp 3 \amp 3\\ 0 \amp 0 \amp 0 \amp 4 \end{bmatrix}\).
Consider the matrix
Then it's easy to verify that
and so we have \(1\text{,}\) \(2\text{,}\) \(3\) and \(4\) as eigenvalues. Notice that in this case the eigenvalues are just the diagonal elements and that the matrix is upper triangular. We shall see in Theorem 6.5.1 that for every upper (or lower) triangular matrix, the eigenvalues are the diagonal entries.
Definition 6.1.7. Eigenspaces.
Suppose that \(A\) is a square matrix of order \(n\text{.}\) Then for any real number \(\lambda\text{,}\) we define the eigenspace \(E_\lambda\) by
Clearly \(\vec0\) is in \(E_\lambda\) for any value of \(\lambda\text{,}\) and \(\lambda\) is an eigenvalue if and only if there is some \(\vec x\not=\vec0\) in \(E_\lambda\text{.}\)
Proposition 6.1.8. An eigenspace is a subspace.
Let \(A\) be an \(n\times n\) matrix and let \(\lambda\) be a real number. Then the set of vectors \(E_\lambda=\{\vec x \mid A\vec x=\lambda \vec x\}\) is a subspace of \(\R^n\text{.}\)
Proof.
From Definition 5.9.17 it is sufficient to show that two properties of closure under addition and of closure under scalar multiplication are satisfied. Suppose \(\vec x\) and \(\vec y\) are in \(E_\lambda\text{.}\)
-
Closure under addition:
\begin{align*} A(\vec x+\vec y)\amp=A(\vec x)+A(\vec y)\\ \amp=\lambda \vec x + \lambda \vec y \\ \amp= \lambda (\vec x + \vec y) \end{align*} -
Closure under scalar multiplication:
\begin{align*} A(r\vec x)\amp=rA(\vec x)\\ \amp=r(\lambda \vec x)\\ \amp= \lambda (r\vec x) \end{align*}