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Section 5.1 Eigenvectors, Eigenvalues, and Eigenspaces

Handout 5.1 Eigenvectors, Eigenvalues, and Eigenspaces

 

Definition 5.1 Eigenvector and Eigenvalue.

Let \(A\in\mathbb{R}^{n\times n}\text{.}\) If there exists a nonzero vector \(\vec v\in\mathbb{R}^n\) and a scalar \(\lambda\in\mathbb{C}\) such that
\begin{equation*} A\vec v = \lambda \vec v, \end{equation*}
then \(\vec v\) is an eigenvector of \(A\) and \(\lambda\) is the corresponding eigenvalue.
Notes:
  • We consider only square matrices.
  • If \(\lambda\in\mathbb{R}\text{,}\) then:
    • for \(\lambda>0\text{,}\) vectors \(A\vec v\) and \(\vec v\) point in the same direction;
    • for \(\lambda<0\text{,}\) they point in opposite directions.
  • Even when entries of \(A\) and \(\vec v\) are real, \(\lambda\) can be complex (e.g., a rotation by \(90^\circ\) counterclockwise in \(\mathbb{R}^2\) has no real eigenvalues).
  • Complex eigenvalues will be explored in Section 5.5.
 

Example 5.2.

Which of the following are eigenvectors of
\begin{equation*} A=\begin{bmatrix} 1 \amp 1\\ 1 \amp 1 \end{bmatrix}? \end{equation*}
What are the corresponding eigenvalues?

(a)

\(\vec v_1 = \begin{bmatrix} 1 \\ 1 \end{bmatrix}\)
Solution.
Note that
\begin{equation*} A\vec v_1 = \begin{bmatrix} 1 \amp 1 \\ 1 \amp 1 \end{bmatrix}\begin{bmatrix} 1 \\ 1 \end{bmatrix} = \begin{bmatrix} 2 \\ 2 \end{bmatrix} = 2 \begin{bmatrix} 1 \\ 1 \end{bmatrix} = 2\vec v_1\text{.} \end{equation*}
Since \(A \vec v_1 = 2 \vec v_1\text{,}\) we have that \(\vec v_1\) is an eigenvector of \(A\) corresponding to an eigenvalue of \(\lambda_1 = 2\text{.}\)

(b)

\(\vec v_2 = \begin{bmatrix} 1 \\ -1 \end{bmatrix}\)
Solution.
Note that
\begin{equation*} A\vec v_2 = \begin{bmatrix} 1 \amp 1 \\ 1 \amp 1 \end{bmatrix}\begin{bmatrix} 1 \\ -1 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix} = 0 \begin{bmatrix} 1 \\ -1 \end{bmatrix} = 0\vec v_2\text{.} \end{equation*}
Since \(A \vec v_2 = 0 \vec v_2\text{,}\) we have that \(\vec v_2\) is an eigenvector of \(A\) corresponding to an eigenvalue of \(\lambda_2 = 0\text{.}\) Note that it is okay if the result of \(A\vec v\) is \(\vec 0\text{.}\) That does not break the definition of eigenvector. The problems is when \(\vec v = \vec 0\text{.}\)

(c)

\(\vec v_3 = \begin{bmatrix} 0 \\ 0 \end{bmatrix}\)
Solution.
Since \(\vec v_3 = \vec 0\text{,}\) Definition 5.1 forbids \(\vec v_3\) from being an eigenvector. The reason why is that \(A\vec 0 = \lambda \vec 0\) for every single \(\lambda \in \mathbb{R}\text{,}\) thus making every single real number an eigenvalue of \(A\) no matter the square matrix \(A\text{.}\) We don’t want this to happen since we want eigenvalues to be rare. That is why we forbid \(\vec 0\) from being considered an eigenvector.
 

Example 5.3.

Confirm that \(\lambda=3\) is an eigenvalue of
\begin{equation*} A=\begin{bmatrix} 2 \amp -4\\ -1 \amp -1 \end{bmatrix}. \end{equation*}
Solution.
To show that \(\lambda = 3\) is an eigenvalue of \(A\text{,}\) we need to find a nonzero vector \(\vec v\) so that \(A\vec v = 3\vec v\text{.}\) Subtracting \(3\vec v \) from both sides and factoring yields
\begin{equation*} A\vec v = 3\vec v \quad \implies \quad A\vec v - 3I\vec v = \vec 0 \quad \implies \quad (A - 3I)\vec v = \vec 0\text{.} \end{equation*}
Therefore, every \(\vec v\) which satisfies \(A\vec v = 3\vec v\) must live in \(\Nul(A - 3I)\text{.}\)
Note: This should look very similar to the process for finding steady-state vectors for stochastic matrices. See Example 4.4. This is because steady-state vectors of a stochastic matrix \(P\) are actually just probability eigenvectors of \(P\) which correspond to an eigenvalue of \(\lambda = 1\text{.}\)
Note that
\begin{equation*} A - 3I = \begin{bmatrix} 2 \amp -4 \\ -1 \amp -1 \end{bmatrix} - \begin{bmatrix} 3 \amp 0 \\ 0 \amp 3 \end{bmatrix} = \begin{bmatrix} -1 \amp -4 \\ -1 \amp -4 \end{bmatrix} \sim \begin{bmatrix} 1 \amp 4 \\ 0 \amp 0 \end{bmatrix}\text{.} \end{equation*}
By observation, we see that \(\vec v = \begin{bmatrix} -4 \\ 1 \end{bmatrix}\) is a nonzero vector in \(\Nul(A - 3I).\) Since \(A - 3I\) only had a single free varible, we know that \(\Nul(A - 3I)\) is 1-dimensional. Thus, we have that the eigenvectors of \(A\) which correspond to an eigenvalue of \(\lambda = 3\) are the nonzero vectors in
\begin{equation*} \Nul(A - 3I) = \Span\left\{\begin{bmatrix} -4 \\ 1 \end{bmatrix}\right\}\text{.} \end{equation*}
Let’s check this for a particular nonzero vector in
\begin{equation*} \Nul(A - 3I) = \Span\left\{\begin{bmatrix} -4 \\ 1 \end{bmatrix}\right\}\text{.} \end{equation*}
Let \(\vec w = \begin{bmatrix} -8 \\ 2 \end{bmatrix} \in \Nul(A - 3I).\) Note that
\begin{equation*} A\vec w = \begin{bmatrix}2 \amp -4 \\ -1 \amp -1 \end{bmatrix}\begin{bmatrix} -8 \\ 2 \end{bmatrix} = \begin{bmatrix} -24 \\ 6 \end{bmatrix} = 3\begin{bmatrix} -8 \\ 2 \end{bmatrix} = 3 \vec w\text{.} \end{equation*}
Thus, \(\vec w\) is an eigenvector of \(A\) corresponding to an eigenvalue of \(\lambda = 3\text{.}\)
 

Definition 5.4 Eigenspace.

Suppose \(A\in\mathbb{R}^{n\times n}\text{.}\) For an eigenvalue \(\lambda\) of \(A\text{,}\) the set of all eigenvectors associated with \(\lambda\) together with the zero vector forms a subspace of \(\mathbb{R}^n\) called the \(\lambda\)-eigenspace of \(A\text{.}\)
Note: The \(\lambda\)-eigenspace is precisely \(\operatorname{Null}(A-\lambda I)\text{.}\)
 

Example 5.5.

Construct a basis for the eigenspaces of
\begin{equation*} A=\begin{bmatrix} 5 \amp -6\\ 3 \amp -4 \end{bmatrix}, \qquad \lambda=-1,\;2. \end{equation*}
Solution.
Similar to the solution of Example 5.3, for each suspected eigenvalue \(\lambda\text{,}\) we need to find a nonzero vector in \(\Nul(A - \lambda I)\text{.}\)
\(\lambda_1 = -1:\) Note that
\begin{equation*} A - \lambda I = A + I = \begin{bmatrix} 6 \amp -6 \\ 3 \amp -3 \end{bmatrix} \sim \begin{bmatrix} 1 \amp -1 \\ 0 \amp 0 \end{bmatrix}\text{.} \end{equation*}
By observation, we see that \(\vec v_1 = \begin{bmatrix} 1 \\ 1 \end{bmatrix} \in \Nul(A + I)\text{.}\) Since \(A + I\) had a single free variable, we have that \(\dim(\Nul(A+I)) = 1\text{.}\) Therefore, the \(-1\text{-eigenspace}(A)\) is 1-dimensional, so \(\mathcal{B}_1 = \left\{\vec v_1 \right\}\) is a basis for \(-1\text{-eigenspace}(A)\text{.}\) Furthermore,
\begin{equation*} -1\text{-eigenspace}(A) = \Nul(A + I) = \Span\left\{\begin{bmatrix} 1 \\ 1 \end{bmatrix}\right\}\text{.} \end{equation*}
Before moving on to the next eigenvalue, it is always a good idea to double-check that \(\vec v_1\) really is an eigenvector of \(A\) corresponding to \(\lambda_1 = -1\text{.}\) Note that
\begin{equation*} A\vec v_1 = \begin{bmatrix} 5 \amp -6 \\ 3 \amp -4 \end{bmatrix}\begin{bmatrix} 1 \\ 1 \end{bmatrix} = \begin{bmatrix} -1 \\ -1 \end{bmatrix} = -1\begin{bmatrix} 1 \\ 1 \end{bmatrix} = -1\vec v_1\text{,} \end{equation*}
which confirms that \(\vec v_1\) is an eigenvector of \(A\) corresponding to an eigenvalue of \(\lambda_1 = -1\text{.}\)
\(\lambda_2 = 2:\) Note that
\begin{equation*} A - \lambda I = A - 2I = \begin{bmatrix} 3 \amp -6 \\ 3 \amp -6 \end{bmatrix} \sim \begin{bmatrix} 1 \amp -2 \\ 0 \amp 0 \end{bmatrix}\text{.} \end{equation*}
By observation, we see that \(\vec v_2 = \begin{bmatrix} 2 \\ 1 \end{bmatrix} \in \Nul(A - 2I)\text{.}\) Since \(A - 2I\) had a single free variable, we have that \(\dim(\Nul(A- 2I)) = 1\text{.}\) Therefore, the \(2\text{-eigenspace}(A)\) is 1-dimensional, so \(\mathcal{B}_2 = \left\{\vec v_2 \right\}\) is a basis for \(2\text{-eigenspace}(A)\text{.}\) Furthermore,
\begin{equation*} 2\text{-eigenspace}(A) = \Nul(A - 2I) = \Span\left\{\begin{bmatrix} 2 \\ 1 \end{bmatrix}\right\}\text{.} \end{equation*}
Again, it is wise to double check that \(\vec v_2\) is an eigenvector for \(A\) corresponding to \(\lambda_2 = 2\text{.}\) Note that
\begin{equation*} A \vec v_2 = \begin{bmatrix} 5 \amp -6 \\ 3 \amp -4 \end{bmatrix}\begin{bmatrix} 2 \\ 1 \end{bmatrix} = \begin{bmatrix} 4 \\ 2 \end{bmatrix} = 2 \begin{bmatrix} 2 \\ 1 \end{bmatrix} = 2 \vec v_2\text{,} \end{equation*}
which confirms that \(\vec v_2\) is an eigenvector for \(A\) corresponding to an eigenvalue of \(\lambda_2 = 2\text{.}\)
 
 
 
 
 

Note 5.10 Warning.

You cannot determine the eigenvalues of a matrix from its reduced row-echelon form. Row operations generally change eigenvalues.

Example 5.11.

Let
\begin{equation*} A=\begin{bmatrix} 1 \amp 1\\ 1 \amp 1 \end{bmatrix}, \end{equation*}
which has eigenvalues \(\lambda_1=2\) and \(\lambda_2=0\text{.}\)

(a)

Find eigenvectors for each eigenvalue of \(A\text{.}\)
Solution.
\(\lambda_1 = 2:\) Note that
\begin{equation*} A - \lambda_1 I = A - 2I = \begin{bmatrix} -1 \amp 1 \\ 1 \amp -1 \end{bmatrix} \sim \begin{bmatrix} 1 \amp -1 \\ 0 \amp 0 \end{bmatrix}\text{.} \end{equation*}
By inspection, we see that \(\vec v_1 = \begin{bmatrix} 1 \\ 1 \end{bmatrix}\) is an eigenvector of \(A\) corresponding to an eigenvalue of \(\lambda_1 = 2\text{.}\) Furthermore, we have that
\begin{equation*} 2\text{-eigenspace}(A) = \Nul(A-2I) = \Span\left\{\begin{bmatrix} 1 \\ 1 \end{bmatrix}\right\}. \end{equation*}
\(\lambda_2 = 0:\) Note that
\begin{equation*} A - \lambda_2 I = A - 0I = A = \begin{bmatrix} 1 \amp 1 \\ 1 \amp 1 \end{bmatrix} \sim \begin{bmatrix} 1 \amp 1 \\ 0 \amp 0 \end{bmatrix}\text{.} \end{equation*}
By inspection, we see that \(\vec v_2 = \begin{bmatrix} 1 \\ -1 \end{bmatrix}\) is an eigenvector of \(A\) corresponding to an eigenvalue of \(\lambda_2 = 0\text{.}\) Furthermore, we have that
\begin{equation*} 0\text{-eigenspace}(A) = \Nul(A) = \Span\left\{\begin{bmatrix} 1 \\ -1 \end{bmatrix}\right\}. \end{equation*}
These calculations confirm what we saw in Example 5.2.

(b)

Find the RREF of \(A\text{.}\) What are it’s eigenvalues?
Solution.
Note that
\begin{equation*} A = \begin{bmatrix} 1 \amp 1 \\ 1 \amp 1 \end{bmatrix} \sim \begin{bmatrix} 1 \amp 1 \\ 0 \amp 0 \end{bmatrix} = \text{RREF}(A)\text{.} \end{equation*}
From Theorem 5.6, we see that the eigenalues of \(\text{RREF}(A)\) are \(\lambda = 1,0\text{.}\) Note that \(\lambda_1 = 2\) is an eigenvalue of \(A\text{,}\) but is not an eigenvalue of \(\text{RREF}(A)\text{.}\)
 

Note 5.12 Converse of Theorem 5.9.

The converse of Theorem 5.9 is the following:
If \(\{\vec v_1, \vec v_2, \dots, \vec v_k\}\) are linearly independent eigenvectors of \(A\text{,}\) then \(\vec v_1, \vec v_2, \dots, \vec v_k\) must correspond to different eigenvalues.
This statement is false. See Example 5.13 below.

Example 5.13.

Let
\begin{equation*} A = \begin{bmatrix} 4 \amp -1 \amp 6 \\ 2 \amp 1 \amp 6 \\ 2 \amp -1 \amp 8 \end{bmatrix}. \end{equation*}
One of the eigenvalues of \(A\) is \(\lambda = 2\text{.}\) Find a basis for the \(2\)-eigenspace of \(A\text{.}\)
Solution.
Note that
\begin{equation*} A - 2I = \begin{bmatrix} 2 \amp -1 \amp 6 \\ 2 \amp -1 \amp 6 \\ 2 \amp -1 \amp 6 \end{bmatrix} \sim \begin{bmatrix} 2 \amp -1 \amp 6 \\ 0 \amp 0 \amp 0 \\ 0 \amp 0 \amp 0 \end{bmatrix}\text{.} \end{equation*}
Since \(A - 2I\) has two free variables, we have that
\begin{equation*} \dim(\Nul(A - 2I)) = 2\text{.} \end{equation*}
Therefore, \(2\text{-eigenspace}(A)\) is two-dimensional, so we need to find two linearly independent vectors in \(\Nul(A - 2I)\) in order to find a basis for the eigenspace. You could find two such vectors by guessing, which isn’t necessarily easy but does become easier with time. Or, you could find two vectors by using our typical process of setting \(x_3 = t \in \mathbb{R}\) and \(x_2 = r \in \mathbb{R}\) which yield \(x_1 = \frac{1}{2}r - 3t\) and
\begin{equation*} \vec x = \begin{bmatrix} \frac{1}{2}r - 3t \\ r \\ t \end{bmatrix} = r\begin{bmatrix} \frac{1}{2} \\ 1 \\ 0 \end{bmatrix} + t\begin{bmatrix} -3 \\ 0 \\ 1 \end{bmatrix} = \frac{r}{2} \begin{bmatrix} 1 \\ 2 \\ 0 \end{bmatrix} + t\begin{bmatrix} -3 \\ 0 \\ 1 \end{bmatrix}\text{.} \end{equation*}
Either way, we find that a basis for \(2\text{-eigenspace}(A)\) is
\begin{equation*} \mathcal{B} = \left\{\begin{bmatrix} 1 \\ 2 \\ 0 \end{bmatrix}, \begin{bmatrix} -3 \\ 0 \\ 1\end{bmatrix}\right\}\text{.} \end{equation*}
You should double-check that these two vectors are linearly independent and belong to \(2\text{-eigenspace}(A) = \Nul(A - 2I)\text{.}\) It follows that
\begin{equation*} 2\text{-eigenspace}(A) = \Nul(A - 2I) = \Span\left\{\begin{bmatrix} 1 \\ 2 \\ 0 \end{bmatrix}, \begin{bmatrix} -3 \\ 0 \\ 1\end{bmatrix}\right\}. \end{equation*}
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