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7.2: Eigenvalues

( \newcommand{\kernel}{\mathrm{null}\,}\)

Definition 7.2.1. Let T in L(V,V). Then λ in F is an eigenvalue of T if there exists a nonzero vector uV such that

Tu=λu.

The vector u is called an eigenvector of T corresponding to the eigenvalue λ.

Finding the eigenvalues and eigenvectors of a linear operator is one of the most important problems in Linear Algebra. We will see later that this so-called ``eigen-information'' has many uses and applications. (As an example, quantum mechanics is based upon understanding the eigenvalues and eigenvectors of operators on specifically defined vector spaces. These vector spaces are often infinite-dimensional, though, and so we do not consider them further in these notes.)

Example 7.2.2.

  • Let T be the zero map defined by T(v)=0 for all vV. Then every vector u0 is an eigenvector of T with eigenvalue 0.
  • Let I be the identity map defined by I(v)=v for all vV. Then every vector u0 is an eigenvector of T with eigenvalue 1.
  • The projection map P:R3R3 defined by P(x,y,z)=(x,y,0) has eigenvalues 0 and 1. The vector (0,0,1) is an eigenvector with eigenvalue 0, and both (1,0,0) and (0,1,0) are eigenvectors with eigenvalue 1.
  • Take the operator R:F2 to F2 defined by R(x,y)=(y,x). When F=R,

R can be interpreted as counterclockwise rotation by 900. From this interpretation, it is clear that no non-zero vector in R2 is mapped to a scalar multiple of itself. Hence, for F=R, the operator R has no eigenvalues.

For F=C, though, the situation is significantly different! In this case, λC is an eigenvalue of R if

R(x,y)=(y,x)=λ(x,y)

so that y=λx and x=λy. This implies that y=λ2y, i.e., that λ2=1. The solutions are hence λ=±i. One can check that (1,i) is an eigenvector with eigenvalue i and that (1,i) is an eigenvector with eigenvalue i.

Eigenspaces are important examples of invariant subspaces. Let TL(V,V), and let λF be an eigenvalue of T. Then

Vλ={vVTv=λv}

is called an eigenspace of T. Equivalently,

Vλ=null(TλI).

Note that Vλ{0} since λ is an eigenvalue if and only if there exists a nonzero vector u in V such that Tu=λu. We can reformulate this as follows:

λF is an eigenvalue of T if and only if the operator TλI is not injective.

Since the notion of injectivity, surjectivity, and invertibility are equivalent for operators on a finite-dimensional vector space, we can equivalently say either of the following:

  • λF is an eigenvalue of T if and only if the operator TλI is not surjective.
  • λF is an eigenvalue of T if and only if the operator TλI is not invertible.

We close this section with two fundamental facts about eigenvalues and eigenvectors.

Theorem 7.2.3. Let TL(V,V), and let λ1,,λmF be m distinct eigenvalues of T with corresponding nonzero eigenvectors v1,,vm. Then (v1,,vm) is linearly independent.

Proof. Suppose that (v1,,vm) is linearly dependent. Then, by the Linear Dependence Lemma, there exists an index k{2,,m} such that

vkspan(v1,,vk1)

and such that (v1,,vk1) is linearly independent. This means that there exist scalars a1,,ak1 in F such that

vk=a1v1++ak1vk1.

Applying T to both sides yields, using the fact that vj is an eigenvector with eigenvalue λj,

λkvk=a1λ1v1++ak1λk1vk1.

Subtracting λk times Equation (7.2.1) from this, we obtain

0=(λkλ1)a1v1++(λkλk1)ak1vk1.

Since (v1,,vk1) is linearly independent, we must have (λkλj)aj=0 for all j=1,2,,k1. By assumption, all eigenvalues are distinct, so λkλj0, which implies that aj=0 for all j=1,2,,k1. But then, by Equation (7.2.1), vk=0, which contradicts the assumption that all eigenvectors are nonzero. Hence (v1,,vm) is linearly independent.

Corollary 7.2.4. Any operator TL(V,V) has at most dim(V) distinct eigenvalues.

Proof. Let λ1,,λm be distinct eigenvalues of T, and let v1,,vm be corresponding nonzero eigenvectors. By Theorem 7.2.3, the list (v1,,vm) is linearly independent. Hence mdim(V).


This page titled 7.2: Eigenvalues is shared under a not declared license and was authored, remixed, and/or curated by Isaiah Lankham, Bruno Nachtergaele, & Anne Schilling.

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