Opening the book…
- Describe eigenvalues geometrically and algebraically.
- Find eigenvalues and eigenvectors for a square matrix.
Spectral Theory refers to the study of eigenvalues and eigenvectors of a matrix. It is of fundamental importance in many areas and is the subject of our study for this chapter.
Complex Numbers
In this section, we will work with the entire set of complex numbers, denoted by \(\mathbb{C}\).
The complex numbers, \(\mathbb{C}\), are numbers of the form \(a + bi\) where a and b are real numbers and i is a root of the equation \(x^2 = -1\) For the number \(z = a+bi\), \(a\) is called the real part of \(z\), and \(b\) is called the imaginary part of \(z\). \(\mathbb{R}\) is the subset of numbers in \(\mathbb{C}\) with imaginary part equal to \(0\).
The sum two complex numbers we get
\[(a+bi) + (c+di) = (a + c) + (b + d)i \nonumber\]
That is, the real part of the sum of two numbers is the sum of the real parts of the numbers, and the imaginary part of the sum of two numbers is the sum of the imaginary parts of the numbers. When computing the product of two complex numbers, we can use the fact that \(i^2 = -1\) and get
\[(a+bi)(c+di) = (ac -bd) + (ad +bc)i \nonumber\]
The conjugate of a complex number \(z=a+bi\), denoted by \(\bar{z}\), is the number \(\bar{z} = a-bi\). Notice that
\[z + \bar{z} = 2a \text{ and } z\bar{z} = a^2 – b^2 \nonumber\]
Hence the sum of a complex number and its conjugate is a real number and the product of a complex number with its conjugate is a real number.
An important property of the complex numbers is that every polynomial of degree \(n\) can be completely factored into n polynomials of degree \(1\). If \(p(x)\) is a polynomial of degree \(n\) and leading coefficient \(1\), \(p(x) = (x- a_1)(x- a_2) …(x-a_n)\). The roots of \(p(x) = 0\) are \(a_1, a_2… a_n\). Of course, the \(a_i\) may not all be distinct. If the value \(a_i\) is repeated m times, then \(a_i\) said to be a root of multiplicity \(m\). You can see from the factorization of \(p(x)\) into \(n\) factors that the total of the multiplicity of the distinct roots will be \(n\). If we consider only real numbers, we cannot always factor a polynomial into a product of first degree polynomials. For example \(p(x) = x^2 +1\) cannot be factored with real numbers and \(x^2 + 1 = 0\) has no real roots. However, if we allow complex numbers, \(x^2 + 1 = (x – i)(x + i)\) and \(x^2 +1 = 0\) has two solutions, \(i\), and \(-i\).
Definition of Eigenvectors and Eigenvalues
To illustrate the idea behind what will be discussed, consider the following example.
Let \[A = \left[ \begin{array}{rrr} 0 & 5 & -10 \\ 0 & 22 & 16 \\ 0 & -9 & -2 \end{array} \right]\nonumber\] Compute the product \(A\vec{x}\) for \[\vec{x} = \left[ \begin{array}{r} -5 \\ -4 \\ 3 \end{array} \right], \vec{x} = \left[ \begin{array}{r} 1 \\ 0 \\ 0 \end{array} \right]\nonumber\] What do you notice about \(A\vec{x}\) in each of these products?
Solution
First, compute \(A\vec{x}\) for \[\vec{x} =\left[ \begin{array}{r} -5 \\ -4 \\ 3 \end{array} \right]\nonumber\]
This product is given by \[A\vec{x} = \left[ \begin{array}{rrr} 0 & 5 & -10 \\ 0 & 22 & 16 \\ 0 & -9 & -2 \end{array} \right] \left[ \begin{array}{r} -5 \\ -4 \\ 3 \end{array} \right] = \left[ \begin{array}{r} -50 \\ -40 \\ 30 \end{array} \right] =10\left[ \begin{array}{r} -5 \\ -4 \\ 3 \end{array} \right]\nonumber\]
In this case, the product \(A\vec{x}\) resulted in a vector which is equal to \(10\) times the vector \(\vec{x}\). In other words, \(A\vec{x}=10\vec{x}\).
Let’s see what happens in the next product. Compute \(A\vec{x}\) for the vector \[\vec{x} = \left[ \begin{array}{r} 1 \\ 0 \\ 0 \end{array} \right]\nonumber\]
This product is given by \[A\vec{x} = \left[ \begin{array}{rrr} 0 & 5 & -10 \\ 0 & 22 & 16 \\ 0 & -9 & -2 \end{array} \right] \left[ \begin{array}{r} 1 \\ 0 \\ 0 \end{array} \right] = \left[ \begin{array}{r} 0 \\ 0 \\ 0 \end{array} \right] =0\left[ \begin{array}{r} 1 \\ 0 \\ 0 \end{array} \right]\nonumber\]
In this case, the product \(A\vec{x}\) resulted in a vector equal to \(0\) times the vector \(\vec{x}\)., \(A\vec{x}=0\vec{x}\).
Perhaps this matrix is such that \(A\vec{x}\) results in \(k\vec{x}\), for every vector \(\vec{x}\). However, consider \[\left[ \begin{array}{rrr} 0 & 5 & -10 \\ 0 & 22 & 16 \\ 0 & -9 & -2 \end{array} \right] \left[ \begin{array}{r} 1 \\ 1 \\ 1 \end{array} \right] = \left[ \begin{array}{r} -5 \\ 38 \\ -11 \end{array} \right]\nonumber \] In this case, \(A\vec{x}\) did not result in a vector of the form \(k\vec{x}\) for some scalar \(k\).
There is something special about the first two products calculated in Example \(\PageIndex{1}\). Notice that for each, \(A\vec{x}=k\vec{x}\) where \(k\) is some scalar. When this equation holds for some non-zero vector \(\vec{x}\) and scalar \(k\), we call the scalar \(k\) an eigenvalue of \(A\). We often use the special symbol \(\lambda\) instead of \(k\) when referring to eigenvalues. In Example \(\PageIndex{1}\), the values \(10\) and \(0\) are eigenvalues for the matrix \(A\) and we can label these as \(\lambda_1 = 10\) and \(\lambda_2 = 0\).
When \(A\vec{x} = \lambda \vec{x}\) for some \(\vec{x} \neq \vec{0}\), we call such an \(\vec{x}\) an eigenvector of the matrix \(A\). The eigenvectors of \(A\) are associated to an eigenvalue. Hence, if \(\lambda_1\) is an eigenvalue of \(A\) and \(A\vec{x} = \lambda_1 \vec{x}\), we can label this eigenvector as \(\vec{x}_1\). Note again that in order to be an eigenvector, \(\vec{x}\) must be nonzero.
There is also a geometric significance to eigenvectors. When you have a nonzero vector which, when multiplied by a matrix results in another vector which is parallel to the first or equal to \(\vec{0}\), this vector is called an eigenvector of the matrix. This is the meaning when the vectors are in \(\mathbb{R}^{n}.\)
The formal definition of eigenvalues and eigenvectors is as follows.
Let \(A\) be an \(n\times n\) matrix and let \(\vec{x} \in \mathbb{C}^{n}\) be a nonzero vector for which
\[A\vec{x}=\lambda \vec{x} \label{eigen1}\] for some scalar \(\lambda .\) Then \(\lambda\) is called an eigenvalue of the matrix \(A\) and \(\vec{x}\) is called an eigenvector of \(A\) associated with \(\lambda\), or a \(\lambda\)-eigenvector of \(A\).
The set of all eigenvalues of an \(n\times n\) matrix \(A\) is denoted by \(\sigma \left( A\right)\) and is referred to as the spectrum of \(A.\)
The eigenvectors of a matrix \(A\) are those vectors \(\vec{x}\) for which multiplication by \(A\) results in a vector in the same direction or opposite direction to \(\vec{x}\). Since the zero vector \(\vec{0}\) has no direction this would make no sense for the zero vector. As noted above, \(\vec{0}\) is never allowed to be an eigenvector.
Let’s look at eigenvectors in more detail. Suppose \(\vec{x}\) satisfies \(\eqref{eigen1}\). Then \[\begin{array}{c} A\vec{x} - \lambda \vec{x} = \vec{0} \\ \mbox{or} \\ \left( A-\lambda I\right) \vec{x} = \vec{0} \end{array}\nonumber\] for some \(\vec{x} \neq \vec{x}.\) Equivalently you could write \(\left( \lambda I-A\right)\vec{x} = \vec{0}\), which is more commonly used. Hence, when we are looking for eigenvectors, we are looking for nontrivial solutions to this homogeneous system of equations!
Recall that the solutions to a homogeneous system of equations consist of basic solutions, and the linear combinations of those basic solutions. In this context, we call the basic solutions of the equation \(\left( \lambda I - A\right) \vec{x} = \vec{x}\) basic eigenvectors. It follows that any (nonzero) linear combination of basic eigenvectors is again an eigenvector.
Suppose the matrix \(\left(\lambda I - A\right)\) is invertible, so that \(\left(\lambda I - A\right)^{-1}\) exists. Then the following equation would be true. \[\begin{aligned} \vec{x} &= I\vec{x} \\ &= \left( \left( \lambda I - A\right) ^{-1}\left(\lambda I - A \right) \right) \vec{x} \\ &=\left( \lambda I - A\right) ^{-1}\left( \left( \lambda I - A\right) \vec{x}\right) \\ &= \left( \lambda I - A\right) ^{-1}\vec{0}\\ &= \vec{0}\end{aligned}\] This claims that \(\vec{x}=\vec{0}\). However, we have required that \(\vec{x} \neq \vec{0}\). Therefore \(\left(\lambda I - A\right)\) cannot have an inverse!
Recall that if a matrix is not invertible, then its determinant is equal to \(0\). Therefore we can conclude that \[\det \left( \lambda I - A\right) =0 \label{eigen2}\] Note that this is equivalent to \(\det \left(A- \lambda I \right) =0\).
The expression \(\det \left( \lambda I-A\right)\) is a polynomial (in the variable \(\lambda\)) called the characteristic polynomial of \(A\), and \(\det \left( \lambda I-A\right) =0\) is called the characteristic equation. For this reason we may also refer to the eigenvalues of \(A\) as characteristic values, but the former is often used for historical reasons.
The following theorem claims that the roots of the characteristic polynomial are the eigenvalues of \(A\).
Let \(A\) be an \(n\times n\) matrix and suppose \(\det \left( \lambda I - A\right) =0\) for some \(\lambda \in \mathbb{C}\).
Then \(\lambda\) is an eigenvalue of \(A\) and thus there exists a nonzero vector \(\vec{x} \in \mathbb{C}^{n}\) such that \(A\vec{x}=\lambda \vec{x}\).
- Proof
-
For \(A\) an \(n\times n\) matrix, the method of Laplace Expansion demonstrates that \(\det \left( \lambda I - A \right)\) is a polynomial of degree \(n.\) As such, the equation \(\eqref{eigen2}\) has a solution \(\lambda \in \mathbb{C}\) by the Fundamental Theorem of Algebra. The fact that \(\lambda\) is an eigenvalue is left as an exercise.
While the characteristic polynomial is guaranteed to have roots in the complex numbers, it may not have real roots. Hence, \(A\) may not have eigenvectors if we restrict ourself to real numbers.
Finding Eigenvectors and Eigenvalues
Now that eigenvalues and eigenvectors have been defined, we will study how to find them for a matrix \(A\).
First, consider the following definition.
Let \(A\) be an \(n \times n\) matrix with characteristic polynomial given by \(\det \left( \lambda I - A\right)\). Then, the algebraic multiplicity of an eigenvalue \(\lambda\) of \(A\) is the number of times \(\lambda\) occurs as a root of that characteristic polynomial.
In section 6.3 we will explore another concept of multiplicity called geometric multiplicity. In general, if we just refer to the multiplicity of an eigenvalue, we mean the algebraic multiplicity defined in 6.1.2.
For example, suppose the characteristic polynomial of \(A\) is given by \(\left( \lambda - 2 \right)^2\). Solving for the roots of this polynomial, we set \(\left( \lambda - 2 \right)^2 = 0\) and solve for \(\lambda \). We find that \(\lambda = 2\) is a root that occurs twice. Hence, in this case, \(\lambda = 2\) is an eigenvalue of \(A\) of multiplicity equal to \(2\).
We will now look at how to find the eigenvalues and eigenvectors for a matrix \(A\) in detail. The steps used are summarized in the following procedure.
Let \(A\) be an \(n \times n\) matrix.
- First, find the eigenvalues \(\lambda\) of \(A\) by solving the equation \(\det \left( \lambda I -A \right) = 0\).
- For each \(\lambda\), find the basic eigenvectors \(\vec{x} \neq 0\) by finding the basic solutions to \(\left( \lambda I - A \right) \vec{x} = 0\).
To verify your work, make sure that \(A\vec{x}=\lambda \vec{x}\) for each \(\lambda\) and associated eigenvector \(\vec{x}\).
We will explore these steps further in the following example.
Let \(A = \left[ \begin{array}{rr} -5 & 2 \\ -7 & 4 \end{array} \right]\). Find its eigenvalues and eigenvectors.
Solution
We will use Procedure \(\PageIndex{1}\). First we find the eigenvalues of \(A\) by solving the equation \[\det \left( \lambda I - A \right) =0\nonumber \]
This gives \[\begin{aligned} \det \left( \lambda \left[ \begin{array}{rr} 1 & 0 \\ 0 & 1 \end{array} \right] - \left[ \begin{array}{rr} -5 & 2 \\ -7 & 4 \end{array} \right] \right) &= 0 \\ \\ \det \left[ \begin{array}{cc} \lambda +5 & -2 \\ 7 & \lambda -4 \end{array} \right] &= 0 \end{aligned}\]
Computing the determinant as usual, the result is \[\lambda ^2 + \lambda - 6 = 0\nonumber\]
Solving this equation, we find that \(\lambda_1 = 2\) and \(\lambda_2 = -3\).
Now we need to find the basic eigenvectors for each \(\lambda\). First we will find the eigenvectors for \(\lambda_1 = 2\). We wish to find all vectors \(\vec{x} \neq 0\) such that \(A\vec{x} = 2\vec{x}\). These are the solutions to \((2I - A)\vec{x} = 0\). \[\begin{aligned} \left( 2 \left[ \begin{array}{rr} 1 & 0 \\ 0 & 1 \end{array}\right] - \left[ \begin{array}{rr} -5 & 2 \\ -7 & 4 \end{array}\right] \right) \left[ \begin{array}{c} x \\ y \end{array}\right] &= \left[ \begin{array}{r} 0 \\ 0 \end{array} \right] \\ \\ \left[ \begin{array}{rr} 7 & -2 \\ 7 & -2 \end{array}\right] \left[ \begin{array}{c} x \\ y \end{array}\right] &= \left[ \begin{array}{r} 0 \\ 0 \end{array} \right] \end{aligned}\]
The augmented matrix for this system and corresponding reduced row-echelon form are given by \[\left[ \begin{array}{rr|r} 7 & -2 & 0 \\ 7 & -2 & 0 \end{array}\right] \rightarrow \cdots \rightarrow \left[ \begin{array}{rr|r} 1 & -\frac{2}{7} & 0 \\ 0 & 0 & 0 \end{array} \right]\nonumber\]
The solution is any vector of the form \[\left[ \begin{array}{c} \frac{2}{7}s \\ s \end{array} \right] = s \left[ \begin{array}{r} \frac{2}{7} \\ 1 \end{array} \right]\nonumber \]
Multiplying this vector by \(7\) we obtain a simpler description for the solution to this system, given by \[t \left[ \begin{array}{r} 2 \\ 7 \end{array} \right]\nonumber\]
This gives the basic eigenvector for \(\lambda_1 = 2\) as \[\left[ \begin{array}{r} 2\\ 7 \end{array} \right]\nonumber\]
To check, we verify that \(A\vec{x} = 2\vec{x}\) for this basic eigenvector.
\[\left[ \begin{array}{rr} -5 & 2 \\ -7 & 4 \end{array}\right] \left[ \begin{array}{r} 2 \\ 7 \end{array} \right] = \left[ \begin{array}{r} 4 \\ 14 \end{array}\right] = 2 \left[ \begin{array}{r} 2\\ 7 \end{array} \right]\nonumber\]
This is what we wanted, so we know this basic eigenvector is correct.
Next we will repeat this process to find the basic eigenvector for \(\lambda_2 = -3\). We wish to find all vectors \(\vec{x} \neq 0\) such that \(A\vec{x}= -3\vec{x}\). These are the solutions to \(((-3)I-A)\vec{x} = 0\). \[\begin{aligned} \left( (-3) \left[ \begin{array}{rr} 1 & 0 \\ 0 & 1 \end{array}\right] - \left[ \begin{array}{rr} -5 & 2 \\ -7 & 4 \end{array}\right] \right) \left[ \begin{array}{c} x \\ y \end{array}\right] &= \left[ \begin{array}{r} 0 \\ 0 \end{array} \right] \\ \left[ \begin{array}{rr} 2 & -2 \\ 7 & -7 \end{array}\right] \left[ \begin{array}{c} x \\ y \end{array}\right] &= \left[ \begin{array}{r} 0 \\ 0 \end{array} \right] \end{aligned}\]
The augmented matrix for this system and corresponding reduced row-echelon form are given by \[\left[ \begin{array}{rr|r} 2 & -2 & 0 \\ 7 & -7 & 0 \end{array}\right] \rightarrow \cdots \rightarrow \left[ \begin{array}{rr|r} 1 & -1 & 0 \\ 0 & 0 & 0 \end{array} \right]\nonumber \]
The solution is any vector of the form \[\left[ \begin{array}{c} s \\ s \end{array} \right] = s \left[ \begin{array}{r} 1 \\ 1 \end{array} \right]\nonumber\]
This gives the basic eigenvector for \(\lambda_2 = -3\) as \[\left[ \begin{array}{r} 1\\ 1 \end{array} \right]\nonumber\]
To check, we verify that \(A\vec{x} = -3\vec{x}\) for this basic eigenvector.
\[\left[ \begin{array}{rr} -5 & 2 \\ -7 & 4 \end{array}\right] \left[ \begin{array}{r} 1 \\ 1 \end{array} \right] = \left[ \begin{array}{r} -3 \\ -3 \end{array}\right] = -3 \left[ \begin{array}{r} 1\\ 1 \end{array} \right]\nonumber\]
This is what we wanted, so we know this basic eigenvector is correct.
The following is an example using Procedure \(\PageIndex{1}\) for a \(3 \times 3\) matrix.
Find the eigenvalues and eigenvectors for the matrix \[A=\left[ \begin{array}{rrr} 5 & -10 & -5 \\ 2 & 14 & 2 \\ -4 & -8 & 6 \end{array} \right]\nonumber \]
Solution
We will use Procedure \(\PageIndex{1}\). First we need to find the eigenvalues of \(A\). Recall that they are the solutions of the equation \[\det \left( \lambda I - A \right) =0\nonumber \]
In this case the equation is \[\det \left( \lambda \left[ \begin{array}{rrr} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array} \right] - \left[ \begin{array}{rrr} 5 & -10 & -5 \\ 2 & 14 & 2 \\ -4 & -8 & 6 \end{array} \right] \right) =0\nonumber \] which becomes \[\det \left[ \begin{array}{ccc} \lambda - 5 & 10 & 5 \\ -2 & \lambda - 14 & -2 \\ 4 & 8 & \lambda - 6 \end{array} \right] = 0\nonumber \]
Using Laplace Expansion, compute this determinant and simplify. The result is the following equation. \[\left( \lambda -5\right) \left( \lambda ^{2}-20\lambda +100\right) =0\nonumber \]
Solving this equation, we find that the eigenvalues are \(\lambda_1 = 5, \lambda_2=10\) and \(\lambda_3=10\). Notice that \(10\) is a root of multiplicity two due to \[\lambda ^{2}-20\lambda +100=\left( \lambda -10\right) ^{2}\nonumber \] Therefore, \(\lambda_2 = 10\) is an eigenvalue of multiplicity two.
Now that we have found the eigenvalues for \(A\), we can compute the eigenvectors.
First we will find the basic eigenvectors for \(\lambda_1 =5.\) In other words, we want to find all non-zero vectors \(\vec{x}\) so that \(A\vec{x} = 5\vec{x}\). This requires that we solve the equation \(\left( 5 I - A \right) \vec{x} = 0\) for \(\vec{x}\) as follows. \[\left( 5\left[ \begin{array}{rrr} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array} \right] - \left[ \begin{array}{rrr} 5 & -10 & -5 \\ 2 & 14 & 2 \\ -4 & -8 & 6 \end{array} \right] \right) \left[ \begin{array}{r} x \\ y \\ z \end{array} \right] =\left[ \begin{array}{r} 0 \\ 0 \\ 0 \end{array} \right]\nonumber \]
That is you need to find the solution to \[ \left[ \begin{array}{rrr} 0 & 10 & 5 \\ -2 & -9 & -2 \\ 4 & 8 & -1 \end{array} \right] \left[ \begin{array}{r} x \\ y \\ z \end{array} \right] =\left[ \begin{array}{r} 0 \\ 0 \\ 0 \end{array} \right]\nonumber \]
By now this is a familiar problem. You set up the augmented matrix and row reduce to get the solution. Thus the matrix you must row reduce is \[\left[ \begin{array}{rrr|r} 0 & 10 & 5 & 0 \\ -2 & -9 & -2 & 0 \\ 4 & 8 & -1 & 0 \end{array} \right]\nonumber \] The reduced row-echelon form is \[\left[ \begin{array}{rrr|r} 1 & 0 & - \frac{5}{4} & 0 \\ 0 & 1 & \frac{1}{2} & 0 \\ 0 & 0 & 0 & 0 \end{array} \right]\nonumber \] and so the solution is any vector of the form \[\left[ \begin{array}{c} \frac{5}{4}s \\ -\frac{1}{2}s \\ s \end{array} \right] =s\left[ \begin{array}{r} \frac{5}{4} \\ -\frac{1}{2} \\ 1 \end{array} \right]\nonumber \] where \(s\in \mathbb{R}\). If we multiply this vector by \(4\), we obtain a simpler description for the solution to this system, as given by \[t \left[ \begin{array}{r} 5 \\ -2 \\ 4 \end{array} \right] \label{basiceigenvect}\] where \(t\in \mathbb{R}\). Here, the basic eigenvector is given by \[\vec{x}_1 = \left[ \begin{array}{r} 5 \\ -2 \\ 4 \end{array} \right]\nonumber \]
Notice that we cannot let \(t=0\) here, because this would result in the zero vector and eigenvectors are never equal to \vec{0}! Other than this value, every other choice of \(t\) in \(\eqref{basiceigenvect}\) results in an eigenvector.
It is a good idea to check your work! To do so, we will take the original matrix and multiply by the basic eigenvector \(\vec{x}_1\). We check to see if we get \(5\vec{x}_1\). \[\left[ \begin{array}{rrr} 5 & -10 & -5 \\ 2 & 14 & 2 \\ -4 & -8 & 6 \end{array} \right] \left[ \begin{array}{r} 5 \\ -2 \\ 4 \end{array} \right] = \left[ \begin{array}{r} 25 \\ -10 \\ 20 \end{array} \right] =5\left[ \begin{array}{r} 5 \\ -2 \\ 4 \end{array} \right]\nonumber \] This is what we wanted, so we know that our calculations were correct.
Next we will find the basic eigenvectors for \(\lambda_2, \lambda_3=10.\) These vectors are the basic solutions to the equation, \[\left( 10\left[ \begin{array}{rrr} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array} \right] - \left[ \begin{array}{rrr} 5 & -10 & -5 \\ 2 & 14 & 2 \\ -4 & -8 & 6 \end{array} \right] \right) \left[ \begin{array}{r} x \\ y \\ z \end{array} \right] =\left[ \begin{array}{r} 0 \\ 0 \\ 0 \end{array} \right]\nonumber \] That is you must find the solutions to \[\left[ \begin{array}{rrr} 5 & 10 & 5 \\ -2 & -4 & -2 \\ 4 & 8 & 4 \end{array} \right] \left[ \begin{array}{c} x \\ y \\ z \end{array} \right] =\left[ \begin{array}{r} 0 \\ 0 \\ 0 \end{array} \right]\nonumber \]
Consider the augmented matrix \[\left[ \begin{array}{rrr|r} 5 & 10 & 5 & 0 \\ -2 & -4 & -2 & 0 \\ 4 & 8 & 4 & 0 \end{array} \right]\nonumber \] The reduced row-echelon form for this matrix is \[\left[ \begin{array}{rrr|r} 1 & 2 & 1 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{array} \right]\nonumber \] and so the eigenvectors are of the form \[\left[ \begin{array}{c} -2s-t \\ s \\ t \end{array} \right] =s\left[ \begin{array}{r} -2 \\ 1 \\ 0 \end{array} \right] +t\left[ \begin{array}{r} -1 \\ 0 \\ 1 \end{array} \right]\nonumber \] Note that you can’t pick \(t\) and \(s\) both equal to zero because this would result in the zero vector and eigenvectors are never equal to zero.
Here, there are two basic eigenvectors, given by \[\vec{x}_2 = \left[ \begin{array}{r} -2 \\ 1\\ 0 \end{array} \right] , \vec{x}_3 = \left[ \begin{array}{r} -1 \\ 0 \\ 1 \end{array} \right]\nonumber \]
Taking any (nonzero) linear combination of \(\vec{x}_2\) and \(\vec{x}_3\) will also result in an eigenvector for the eigenvalue \(\lambda =10.\) As in the case for \(\lambda =5\), always check your work! For the first basic eigenvector, we can check \(A\vec{x}_2 = 10 \vec{x}_2\) as follows. \[\left[ \begin{array}{rrr} 5 & -10 & -5 \\ 2 & 14 & 2 \\ -4 & -8 & 6 \end{array} \right] \left[ \begin{array}{r} -1 \\ 0 \\ 1 \end{array} \right] = \left[ \begin{array}{r} -10 \\ 0 \\ 10 \end{array} \right] =10\left[ \begin{array}{r} -1 \\ 0 \\ 1 \end{array} \right]\nonumber \] This is what we wanted. Checking the second basic eigenvector, \(\vec{x}_3\), is left as an exercise.
It is important to remember that for any eigenvector \(\vec{x}\), \(\vec{x} \neq 0\). However, it is possible to have eigenvalues equal to zero. This is illustrated in the following example.
Let \[A=\left[ \begin{array}{rrr} 2 & 2 & -2 \\ 1 & 3 & -1 \\ -1 & 1 & 1 \end{array} \right]\nonumber \] Find the eigenvalues and eigenvectors of \(A\).
Solution
First we find the eigenvalues of \(A\). We will do so using Definition \(\PageIndex{1}\).
In order to find the eigenvalues of \(A\), we solve the following equation. \[\det \left(\lambda I -A \right) = \det \left[ \begin{array}{ccc} \lambda -2 & -2 & 2 \\ -1 & \lambda - 3 & 1 \\ 1 & -1 & \lambda -1 \end{array} \right] =0\nonumber \]
This reduces to \(\lambda ^{3}-6 \lambda ^{2}+8\lambda =0\). You can verify that the solutions are \(\lambda_1 = 0, \lambda_2 = 2, \lambda_3 = 4\). Notice that while eigenvectors can never equal \(\vec{0}\), it is possible to have an eigenvalue equal to \(0\).
Now we will find the basic eigenvectors. For \(\lambda_1 =0\), we need to solve the equation \(\left( 0 I - A \right) \vec{x} = \vec{0}\). This equation becomes \(-A\vec{x}=\vec{0}\), and so the augmented matrix for finding the solutions is given by \[\left[ \begin{array}{rrr|r} -2 & -2 & 2 & 0 \\ -1 & -3 & 1 & 0 \\ 1 & -1 & -1 & 0 \end{array} \right]\nonumber \] The reduced row-echelon form is \[\left[ \begin{array}{rrr|r} 1 & 0 & -1 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{array} \right]\nonumber \] Therefore, the eigenvectors are of the form \(t\left[ \begin{array}{r} 1 \\ 0 \\ 1 \end{array} \right]\) where \(t\neq 0\) and the basic eigenvector is given by \[\vec{x}_1 = \left[ \begin{array}{r} 1 \\ 0 \\ 1 \end{array} \right]\nonumber \]
We can verify that this eigenvector is correct by checking that the equation \(A\vec{x}_1 = 0 \vec{x}_1\) holds. The product \(A\vec{x}_1\) is given by \[A\vec{x}_1=\left[ \begin{array}{rrr} 2 & 2 & -2 \\ 1 & 3 & -1 \\ -1 & 1 & 1 \end{array} \right] \left[ \begin{array}{r} 1 \\ 0 \\ 1 \end{array} \right] = \left[ \begin{array}{r} 0 \\ 0 \\ 0 \end{array} \right]\nonumber \]
This clearly equals \(0\vec{x}_1\), so the equation holds. Hence, \(A\vec{x}_1 = 0\vec{x}_1\) and so \(0\) is an eigenvalue of \(A\).
Computing the other basic eigenvectors is left as an exercise.
In the following section, we examine ways to simplify this process of finding eigenvalues and eigenvectors by using properties of special types of matrices.
This page titled 6.1: Eigenvalues and Eigenvectors of a Matrix was last modified on Thu, 10 Apr 2025 05:35:40 GMT and is shared under a CC BY 4.0 license and was authored, remixed, and/or curated by Doli Bambhania, Fatemeh Yarahmadi, and Bill Wilson.
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