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SectionEEEigenvalues and Eigenvectors

In this section, we will define the eigenvalues and eigenvectors of a matrix, and see how to compute them. More theoretical properties will be taken up in the next section.

SubsectionEEMEigenvalues and Eigenvectors of a Matrix

We start with the principal definition for this chapter.

DefinitionEEMEigenvalues and Eigenvectors of a Matrix

Suppose that \(A\) is a square matrix of size \(n\text{,}\) \(\vect{x}\neq\zerovector\) is a vector in \(\complex{n}\text{,}\) and \(\lambda\) is a scalar in \(\complexes\text{.}\) Then we say \(\vect{x}\) is an eigenvector of \(A\) with eigenvalue \(\lambda\) if \begin{equation*} A\vect{x}=\lambda\vect{x}\text{.} \end{equation*}

Before going any further, perhaps we should convince you that such things ever happen at all. Understand the next example, but do not concern yourself with where the pieces come from. We will have methods soon enough to be able to discover these eigenvectors ourselves.

Example SEE hints at a number of intriguing properties, and there are many more. We will explore the general properties of eigenvalues and eigenvectors in Section PEE, but in this section we will concern ourselves with the question of actually computing eigenvalues and eigenvectors. First we need a bit of background material on polynomials and matrices.

SubsectionPMPolynomials and Matrices

A polynomial is a combination of powers, multiplication by scalar coefficients, and addition (with subtraction just being the inverse of addition). We never have occasion to divide when computing the value of a polynomial. So it is with matrices. We can add and subtract matrices, we can multiply matrices by scalars, and we can form powers of square matrices by repeated applications of matrix multiplication. We do not normally divide matrices (though sometimes we can multiply by an inverse). If a matrix is square, all the operations constituting a polynomial will preserve the size of the matrix. So it is natural to consider evaluating a polynomial with a matrix, effectively replacing the variable of the polynomial by a matrix. We will demonstrate with an example.

SubsectionEEEExistence of Eigenvalues and Eigenvectors

Before we embark on computing eigenvalues and eigenvectors, we will prove that every matrix has at least one eigenvalue (and an eigenvector to go with it). Later, in Theorem MNEM, we will determine the maximum number of eigenvalues a matrix may have.

The determinant (Definition DM) will be a powerful tool in Subsection EE.CEE when it comes time to compute eigenvalues. However, it is possible, with some more advanced machinery, to compute eigenvalues without ever making use of the determinant. Sheldon Axler does just that in his book, Linear Algebra Done Right. Here and now, we give Axler's “determinant-free” proof that every matrix has an eigenvalue. The result is not too startling, but the proof is most enjoyable.

Proof

The proof of Theorem EMHE is constructive (it contains an unambiguous procedure that leads to an eigenvalue), but it is not meant to be practical. We will illustrate the theorem with an example, the purpose being to provide a companion for studying the proof and not to suggest this is the best procedure for computing an eigenvalue.

SubsectionCEEComputing Eigenvalues and Eigenvectors

Fortunately, we need not rely on the procedure of Theorem EMHE each time we need an eigenvalue. It is the determinant, and specifically Theorem SMZD, that provides the main tool for computing eigenvalues. Here is an informal sequence of equivalences that is the key to determining the eigenvalues and eigenvectors of a matrix, \begin{equation*} A\vect{x}=\lambda\vect{x}\iff A\vect{x}-\lambda I_n\vect{x}=\zerovector\iff \left(A-\lambda I_n\right)\vect{x}=\zerovector\text{.} \end{equation*}

So, for an eigenvalue \(\lambda\) and associated eigenvector \(\vect{x}\neq\zerovector\text{,}\) the vector \(\vect{x}\) will be a nonzero element of the null space of \(A-\lambda I_n\text{,}\) while the matrix \(A-\lambda I_n\) will be singular and therefore have zero determinant. These ideas are made precise in Theorem EMRCP and Theorem EMNS, but for now this brief discussion should suffice as motivation for the following definition and example.

DefinitionCPCharacteristic Polynomial

Suppose that \(A\) is a square matrix of size \(n\text{.}\) Then the characteristic polynomial of \(A\) is the polynomial \(\charpoly{A}{x}\) defined by \begin{equation*} \charpoly{A}{x}=\detname{A-xI_n}\text{.} \end{equation*}

The characteristic polynomial is our main computational tool for finding eigenvalues, and will sometimes be used to aid us in determining the properties of eigenvalues.

Proof

Let us now turn our attention to the computation of eigenvectors.

DefinitionEMEigenspace of a Matrix

Suppose that \(A\) is a square matrix and \(\lambda\) is an eigenvalue of \(A\text{.}\) Then the eigenspace of \(A\) for \(\lambda\text{,}\) \(\eigenspace{A}{\lambda}\text{,}\) is the set of all the eigenvectors of \(A\) for \(\lambda\text{,}\) together with the inclusion of the zero vector.

Example SEE hinted that the set of eigenvectors for a single eigenvalue might have some closure properties, and with the addition of the one eigenvector that is never an eigenvector, \(\zerovector\text{,}\) we indeed get a whole subspace.

Proof

Theorem EMS tells us that an eigenspace is a subspace (and hence a vector space in its own right). Our next theorem tells us how to quickly construct this subspace.

Proof

You might notice the close parallels (and differences) between the proofs of Theorem EMRCP and Theorem EMNS. Since Theorem EMNS describes the set of all the eigenvectors of \(A\) as a null space we can use techniques such as Theorem BNS to provide concise descriptions of eigenspaces. Theorem EMNS also provides a trivial proof for Theorem EMS.

SubsectionECEEExamples of Computing Eigenvalues and Eigenvectors

There are no theorems in this section, just a selection of examples meant to illustrate the range of possibilities for the eigenvalues and eigenvectors of a matrix. These examples can all be done by hand, though the computation of the characteristic polynomial would be very time-consuming and error-prone. It can also be difficult to factor an arbitrary polynomial, though if we were to suggest that most of our eigenvalues are going to be integers, then it can be easier to hunt for roots. These examples are meant to look similar to a concatenation of Example CPMS3, Example EMS3 and Example ESMS3. First, we will sneak in a pair of definitions so we can illustrate them throughout this sequence of examples.

DefinitionAMEAlgebraic Multiplicity of an Eigenvalue

Suppose that \(A\) is a square matrix and \(\lambda\) is an eigenvalue of \(A\text{.}\) Then the algebraic multiplicity of \(\lambda\text{,}\) \(\algmult{A}{\lambda}\text{,}\) is the highest power of \((x-\lambda)\) that divides the characteristic polynomial, \(\charpoly{A}{x}\text{.}\)

Since an eigenvalue \(\lambda\) is a root of the characteristic polynomial, there is always a factor of \((x-\lambda)\text{,}\) and the algebraic multiplicity is just the power of this factor in a factorization of \(\charpoly{A}{x}\text{.}\) So in particular, \(\algmult{A}{\lambda}\geq 1\text{.}\) Compare the definition of algebraic multiplicity with the next definition.

DefinitionGMEGeometric Multiplicity of an Eigenvalue

Suppose that \(A\) is a square matrix and \(\lambda\) is an eigenvalue of \(A\text{.}\) Then the geometric multiplicity of \(\lambda\text{,}\) \(\geomult{A}{\lambda}\text{,}\) is the dimension of the eigenspace \(\eigenspace{A}{\lambda}\text{.}\)

Every eigenvalue must have at least one eigenvector, so the associated eigenspace cannot be trivial, and so \(\geomult{A}{\lambda}\geq 1\text{.}\)

SubsectionReading Questions

1

Suppose \(A\) is the \(2\times 2\) matrix \begin{equation*} A=\begin{bmatrix} -5 & 8\\-4 & 7 \end{bmatrix}\text{.} \end{equation*} Find the eigenvalues of \(A\text{.}\)

2

For each eigenvalue of \(A\text{,}\) find the corresponding eigenspace.

3

For the polynomial \(p(x)=3x^2-x+2\) and \(A\) from above, compute \(p(A)\text{.}\)

SubsectionExercises

C10

Find the characteristic polynomial of the matrix \begin{equation*} A = \begin{bmatrix} 1 & 2\\ 3 & 4 \end{bmatrix}\text{.} \end{equation*}

Solution
C11

Find the characteristic polynomial of the matrix \begin{equation*} A = \begin{bmatrix} 3 & 2 & 1\\ 0 & 1 & 1\\ 1 & 2 & 0 \end{bmatrix}\text{.} \end{equation*}

Solution
C12

Find the characteristic polynomial of the matrix \begin{equation*} A = \begin{bmatrix} 1 & 2 & 1 & 0\\ 1 & 0 & 1 & 0\\ 2 & 1 & 1 & 0\\ 3 & 1 & 0 & 1 \end{bmatrix}\text{.} \end{equation*}

Solution
C19

Find the eigenvalues, eigenspaces, algebraic multiplicities and geometric multiplicities for the matrix below. It is possible to do all these computations by hand, and it would be instructive to do so. \begin{equation*} C= \begin{bmatrix} -1 & 2 \\ -6 & 6 \end{bmatrix}\text{.} \end{equation*}

Solution
C20

Find the eigenvalues, eigenspaces, algebraic multiplicities and geometric multiplicities for the matrix below. It is possible to do all these computations by hand, and it would be instructive to do so. \begin{equation*} B= \begin{bmatrix} -12&30\\-5&13 \end{bmatrix} \end{equation*}

Solution
C21

The matrix \(A\) below has \(\lambda=2\) as an eigenvalue. Find the geometric multiplicity of \(\lambda=2\) using your calculator only for row-reducing matrices. \begin{equation*} A= \begin{bmatrix} 18 & -15 & 33 & -15\\ -4 & 8 & -6 & 6\\ -9 & 9 & -16 & 9\\ 5 & -6 & 9 & -4 \end{bmatrix} \end{equation*}

Solution
C22

Without using a calculator, find the eigenvalues of the matrix \(B\text{.}\) \begin{equation*} B= \begin{bmatrix} 2 & -1 \\ 1 & 1 \end{bmatrix} \end{equation*}

Solution
C23

Find the eigenvalues, eigenspaces, algebraic and geometric multiplicities for \begin{equation*} A = \begin{bmatrix} 1 & 1\\ 1 & 1 \end{bmatrix}\text{.} \end{equation*}

Solution
C24

Find the eigenvalues, eigenspaces, algebraic and geometric multiplicities for \begin{equation*} A = \begin{bmatrix} 1 & -1 & 1\\ -1 & 1 & -1\\ 1 & -1 & 1 \end{bmatrix}\text{.} \end{equation*}

Solution
C25

Find the eigenvalues, eigenspaces, algebraic and geometric multiplicities for the \(3\times 3\) identity matrix \(I_3\text{.}\) Do your results make sense?

Solution
C26

For matrix \begin{equation*} A = \begin{bmatrix} 2 & 1 & 1\\ 1 & 2 & 1\\ 1 & 1 & 2 \end{bmatrix} \end{equation*} the characteristic polynomial of \(A\) is \(\charpoly{A}{x} = (4-x)(1-x)^2\text{.}\) Find the eigenvalues and corresponding eigenspaces of \(A\text{.}\)

Solution
C27

For the matrix \begin{align*} A = \begin{bmatrix} 0 & 4 & -1 & 1\\ -2 & 6 & -1 & 1\\ -2 & 8 & -1 & -1\\ -2 & 8 & -3 & 1 \end{bmatrix}, \end{align*} the characteristic polynomial of \(A\) is \(\charpoly{A}{x} = (x+2)(x-2)^2(x-4)\text{.}\) Find the eigenvalues and corresponding eigenspaces of \(A\text{.}\)

Solution
M60

Repeat Example CAEHW by choosing \(\vect{x}=\colvector{0\\8\\2\\1\\2}\) and then arrive at an eigenvalue and eigenvector of the matrix \(A\text{.}\) The hard way.

Solution
T10

A matrix \(A\) is idempotent if \(A^2=A\text{.}\) Show that the only possible eigenvalues of an idempotent matrix are \(\lambda=0\) and \(\lambda=1\text{.}\) Then give an example of a matrix that is idempotent and has both of these two values as eigenvalues.

Solution
T15

The characteristic polynomial of the square matrix \(A\) is usually defined as \(r_A(x)=\detname{xI_n-A}\text{.}\) Find a specific relationship between our characteristic polynomial, \(\charpoly{A}{x}\text{,}\) and \(r_A(x)\text{,}\) give a proof of your relationship, and use this to explain why Theorem EMRCP can remain essentially unchanged with either definition. Explain the advantages of each definition over the other. (Computing with both definitions, for a \(2\times 2\) and a \(3\times 3\) matrix, might be a good way to start.)

Solution
T20

Suppose that \(\lambda\) and \(\rho\) are two different eigenvalues of the square matrix \(A\text{.}\) Prove that the intersection of the eigenspaces for these two eigenvalues is trivial. That is, \(\eigenspace{A}{\lambda}\cap\eigenspace{A}{\rho}=\set{\zerovector}\text{.}\)

Solution