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On this page:

• Introduction
• Eigenstuff
• Spectral Mapping Theorem
• Elementary Symmetric Polynomials and Eigenvalues
• Jordan Canonical Form
• Diagonalization
• Worked Exercises

Introduction

There are two ways I mainly think of (finite-dimensional) spectral theory at this point in time. The first is a naive geometric perspective, and the second is a completely algebraic perspective.

• From a geometric perspective, I think of eigenvectors as elements of a $1$-dimensional $T$-invariant subspace. In particular, lines that don’t “move around” with the transformation, in some sense. Accordingly, we can understand linear operators by studying the eigenspaces.
• From an algebraic perspective, I think of eigenvectors (and generalized eigenvectors) as elements of one of the direct summands in elementary divisor decomposition. In these particular cases, the elementary divisors are linear factors.

In any case, the point that I am trying to make here is that spectral theory tells us a lot about the operators we are working with, and often a lot about the underlying vector space as well.

Eigenstuff

Suppose that we have a $1$-dimensional subspace $S \subseteq V$ that is $T$-invariant. Then, for any nonzero $v \in S$, it follows that $Tv \in S$. Since $S$ is $1$-dimensional, it is spanned by $v$ and so we see that

for some $\lambda \in k$. Let us now define some terminology.

[definition] [deftitle]Definition %counter% (Eigenstuff)[/deftitle]

Let $T$ be a linear operator on $V$.

1. We say that $\lambda \in k$ is an eigenvalue of $T$ if there is some nonzero $v \in V$ for which $Tv = \lambda$. We call $v$ an eigenvector of $T$ associated with $\lambda$.
2. The set $E(\lambda, T)$ of all eigenvectors associated with the eigenvalue $\lambda$ of $T$ is called the eigenspace of $\lambda$.
3. The set $\Spec(T)$ of all eigenvalues of $T$ is called the spectrum of $T$.

All the above definitions apply to matrices in the obvious way. [/definition]

[example] [extitle]Remark %counter%[/extitle]

According to Wikipedia, the word “spectrum” comes from functional analysis in the study of operators. One can make the connection between the spectrum of a ring and the spectrum of an operator as well by appealing to commutative algebra/algebraic geometric language of varieties. [/example]

When it comes to actually computing the eigenvalues in elementary linear algebra, we are all taught to factor the characteristic polynomial and the roots are the eigenvalues. Let us prove this result now.

[theorem] [thmtitle]Lemma %counter% (Alternative characterization of eigenvalue)[/thmtitle]

Let $T$ be a linear operator on the finite-dimensional vector space $V$. Then, $\lambda \in k$ is an eigenvalue of $T$ if and only if $\lambda I - T$ is a singular operator. [/theorem]

[proof] ($\implies$) Say $v \in E(\lambda, T)$ is nonzero. Then

As $v \neq 0$, it follows that $\lambda I - T$ has nontrivial kernel and is therefore singular.

($\impliedby$) Same proof but run the steps in reverse. [/proof]

[example] [extitle]Remark %counter%[/extitle]

By the Rank-Nullity Theorem, since $\lambda I - T$ is an operator, the following are all equivalent:

• $\lambda I - T$ is not injective.
• $\lambda I - T$ is not surjective.
• $\lambda I - T$ is not invertible.

So it suffices to show any one of these conditions. [/example]

[theorem] [thmtitle]Theorem %counter% (Eigenvalues are the roots of $\chi_T(x)$)[/thmtitle]

Let $T$ be a linear operator on the finite-dimensional vector space $V$. Then $\lambda\in k$ is an eigenvalue of $T$ if $\lambda$ is a root of $\chi_T(x)$. [/theorem]

[proof] By the Lemma, $\lambda$ is an eigenvalue if and only if the operator $\lambda I - T$ is singular. This is the case if and only if $\det(\lambda I - T) = 0$. But since

it follows that $\lambda$ is a root of the characteristic polynomial of $T$ if and only if $\lambda$ is an eigenvalue of $T$ and we are done. [/proof]

[theorem] [thmtitle]Corollary %counter%[/thmtitle]

Let $T$ be a linear operator on the finite-dimensional vector space $V$. Then the number of eigenvalues of $T$, including multiplcity, is $\le \dim V$. [/theorem]

[proof] The eigenvalues of $T$ are the roots of the characteristic polynomial which has degree $\dim V$. So the claim follows imemdiately since we know a polynomial of degree $\dim V$ will have at most $\dim V$ roots. [/proof]

From geometric intuition of eigenspaces, it should be fairly obvious that eigenspaces have trivial intersection. Let us formalize this now.

[theorem] [thmtitle]Proposition %counter%[/thmtitle]

Let $\lambda_1, \ldots, \lambda_m$ be distinct eigenvalues of the linear operator $T$. Then the eigenvectors associated with distinct eigenvalues are linearly independent. [/theorem]

[proof] Call the aforementioned eigenvectors $v_1, \ldots, v_m$. Then we know that $Tv_i = \lambda v_i$ for each $i \in [m]$. Now say that (and reordering the $v_i$, if necessary)

For the sake of contradiction, assume that the expression above is the least number of eigenvectors needed to obtain a nontrivial linear combination of $0$ (that is, all $a_i$ are nonzero). Applying the operator $\lambda_j I - T$ to both sides gives

But this violates the minimality requiremenet that we made above. Thus, $v_1, \ldots, v_m$ must be linear independent and we are done. [/proof]

[theorem] [thmtitle]Corollary %counter%[/thmtitle]

Let $\lambda_1, \ldots, \lambda_m$ be distinct eigenvalues of the linear operator $T$. Then the sum of the associated eigenspace is a direct sum. [/theorem]

[proof] Immediate from the Proposition we just proved. [/proof]

Spectral Mapping Theorem

In functional analysis, there is a classical result in operator theory that goes something along the lines of this: Given a unital Banach algebra $A$ and $p(z) \in \CC[z]$, we have that

where $a\in A$. This is known as (version of) the spectral mapping theorem and it can be specialized to the case of finite-dimensional vector spaces.

[theorem] [thmtitle]Theorem %counter% (Spectral mapping theorem)[/thmtitle]

Let $V$ be a vector space over an algebraically closed field $k$, $T$ be a linear operator on $V$, and $p(x) \in k[x]$. Then

[/theorem]

[proof] ($\subseteq$) Let $\lambda \in \Spec(p(T))$. Then

where $v \in E(\lambda, p(T))$ is nonzero. Since $k$ is algebraically closed, we can factor

where $a \in k$. Thus,

This impleis that there is some $(T - r_m)$ for which

Thus, $T$ has eigenvector $\left(\prod_{i=m+1}^n(T - r_i)v\right)$ with eigenvalue $r_m$. Furthermore, we see that $p(r_m) = \lambda$ and so $\lambda\in p(\Spec(T))$.

($\supseteq$) This is the easy direction and actually does not require the algebraically closure assumption. Say that $\lambda \in \Spec(T)$ and write

Then

where $v \in E(\lambda, T)$. So it follows that $\lambda \in \Spec(p(T))$ as desired. [/proof]

[example] [extitle]Remark %counter%[/extitle]

As mentioned in the proof, the $\supseteq$ inclusion doesn’t actually require the algebraically closed field assumption and that direction (that $p(\lambda)$ is an eigenvalue of $p(T)$) is actually quite useful. When solving ordinary differential equations with constant coefficients, we can always characterize them via differential operators. For instance, if

where $D^i(f) \coloneqq \dfrac{d^i f}{dt^i}$ for some appropriately smooth function $f$, then

completely encodes any homogeneous ordinary differential equation with constant coefficients. Furthermore, we know that $Ce^{\lambda t}$ is an eigenfunction of this operator which tells us that

which is the characteristic equation. [/example]

Elementary Symmetric Polynomials and Eigenvalues

In symmetric function theory, one of the objects of interest are the symmetric polynomials: that is, polynomials that invariant under permutations of the variables. One of the bases (that is actually not too nice to work with in algebraic combinatorics), are the elementary symmetric polynomials. Here’s out setup: Let $x_1, \ldots, x_n$ be our variables. Then the elementary symmetric polynomials are

Our goal will be to connect these formulas to eigenvalues. Recall that the eigenvalues are the roots of the characteristic polynomial $\chi_T(x)$. Assume that $\chi_T(x)$ splits over $k$ so that

By Vieta’s formulas, we know that

This allows us to make connections between the eigenvalues and the trace of a matrix.

[theorem] [thmtitle]Theorem %counter% (Determinant is product of eigenvalues)[/thmtitle]

Let $V$ be a finite-dimensional $k$-vector space and let $T$ be a linear operator on $V$ whose characteristic polynomial $\chi_T(x)$ splits over $k$. Then, the determinant of $T$ is the product of the eigenvalues (including multiplicity). [/theorem]

[proof] Since $\chi_T(x)$ splits over $k$, we can write it as

Recall that

Letting $x = 0$, we obtain

Thus, $\det(T) = \prod_{i=1}^n \lambda_i$. [/proof]

[theorem] [thmtitle]Theorem %counter% (Trace is sum of eigenvalues)[/thmtitle]

Let $V$ be a finite-dimensional $k$-vector space and let $T$ be a linear operator on $V$ whose characteristic polynomial $\chi_T(x)$ splits over $k$. Then, the trace of $T$ is the sum of the eigenvalues (including multiplicity). [/theorem]

This latter theorem is not too difficult to prove from the Laplace expansion formula and bookkeeping which terms end up with an $x^{n-1}$. I do not think this is a particularly enlightening thing to see, so I will skip the proof.

Jordan Canonical Form

Now that we have some basic spectral theory in place, we can formulate another approach to matrix canonical forms — namely the Jordan canonical form. The Jordan canonical form arises naturally from the elementary divisor decomposition as we will see now.

Suppose the minimal polynomial $m_T(x)$ splits over $k$ as

and $V$ itself decomposes as

Then, we know that each $\generator{v_{i,j}}$ has basis

where $d_{i,j} = \deg(p_i^{e_{i,j}}(x))$. However, instead of choosing this basis, we will consider a different basis (the Jordan basis). Since $\mathcal{B}_{i,j}$ is a basis for $\generator{v_{i,j}}$, it follows that any

where $\deg p_\ell(x) = \ell$ will be a basis for $\generator{v_{i,j}}$. In our case, we may write each $p_{i}^{e_{i,j}} = (x - \lambda_i)^{e_{i,j}}$ and so

is a basis for $\generator{v_{i,j}}$. So, then, computing out the matrix for $T|_{\generator{v_{i,j}}}$ with respect to this basis:

for $0 \le m < e_{i,j} - 2$. In the case where $m = e_{i,j} - 1$:

where the last inequality follows from the fact that $(x - \lambda_i)^{e_{i,j}}$ is the minimal polynomial of $T|_{\generator{v_{i,j}}}$. Define

Then, we see that

So, then, we see that there is a canonical basis for $T$ given by

This basis is called a Jordan basis for $T$ and we see that

This matrix is called the Jordan canonical form of $T$.

[theorem] [thmtitle]Theorem %counter% (Jordan canonical form)[/thmtitle]

Let $T$ be a linear operator on the finite-dimensional $k$-vector space $V$. Assume that the minimal polynomial of $T$ splits over $k$. Then $T$ has a Jordan canonical form. [/theorem]

[example] [extitle]Remark %counter%[/extitle]

A remark that is fairly clear is that if $k$ is algebraically closed, then it follows that $A \sim B$ if and only if there is a Jordan basis for which $A$ and $B$ have the same Jordan canonical form. So the JCF forms a full set of representatives for the similarity classes of $M_n(k)$ when $k$ is algebraically closed. [/example]

Diagonalization

[definition] [deftitle]Definition %counter% (Diagonalizable operators)[/deftitle]

A linear operator $T$ is diagonalizable if there is a choice of basis $\mathcal{B}$ for which $[T]_B$ is diagonal. [/definition]

There is an equivalent formulation of the above definition which makes an immediate connection to spectral theory.

[theorem] [thmtitle]Theorem %counter% (Diagonalizable $\iff$ $V$ has eigenbasis)[/thmtitle]

A linear operator $T$ is diagonalizable if and only if it has a basis consisting of eigenvectors. [/theorem]

[proof] ($\implies$) Say $\mathcal{B} = \{v_i\}_{i=1}^n$ is a basis for which

is a diagonal matrix. Then

so $Tv_i = \lambda_i v_i$. Thus, the $v_i$ are eigenvectors and the claim follows.

($\impliedby$) Run the argument above in reverse. [/proof]

[theorem] [thmtitle]Corollary %counter%[/thmtitle]

A linear operator $T$ on $V$ is diagonalizable if and only if

where the $\lambda_i$’s are distinct eigenvalues of $T$. [/theorem]

[proof] ($\implies$) Say $\mathcal{B} = \{v_i\}_{i=1}^n$ is a basis for which

is a diagonal matrix. Then

so $Tv_i = \lambda_i v_i$. Thus, the $v_i$ are eigenvectors and the claim follows.

($\impliedby$) Run the argument above in reverse. [/proof]

[theorem] [thmtitle]Theorem %counter%[/thmtitle]

A linear operator $T$ on $V$ is diagonalizable if and only if the minimal polynomial splits into distinct factors over $k$. [/theorem]

[proof] Immediate from Jordan canonical form. [/proof]

Worked Exercises

[example] [extitle]Roman Advanced Linear Algebra Exercise 8.1[/extitle]

Let $J$ be the $n\times n$ matrix all of whose entries are equal to $1$. Find the minimal polynomial and characteristic polynomial of $J$ and the eigenvalues. [/example]

[solution] We claim that $J$ has eigenvalues $n$ (with multiplicity 1) and $0$ (with multiplicity $n-1$). Since $J$ having eigenvalue $0$ is equivalent to saying that $J$ has nontrivial null space, and said null space is dimension $n-1$ by the Rank-Nullity Theorem, $J$ has eigenvalue $0$ with multiplicity $n-1$. To see that $n$ is an eigenvalue, note

Thus, we have a collection of $n$-eigenvalues, so we have found all eigenvalues. It follows that the characteristic polynomial is

We claim that the minimal polynomial is $x(x - n)$. It is immediate to see that none of $x$, $x^2$, and $x - n$ are the minimal polynomial (just apply each operator to $e_1$). So minimality implies $x(x - n)$ is, in fact, the minimal polynomial. [/solution]

[example] [extitle]Roman Advanced Linear Algebra Exercise 8.6[/extitle]

Prove that the spectrum of a nilpotent operator is $\{0\}$ and find a nonnilpotent operator $T$ with spectrum $\{0\}$. [/example]

[solution] ($\Spec T = \{0\}$) Let $T$ be a nilpotent operator. Then we know that there is some positive integer $n$ for which $T^n = 0$. Thus, we can pick any nonzero vector $v$ and it follows that

So $T$ vanishes on some $T^mv \neq 0$. This tells us that $0 \in \Spec(T)$. To show that $T$ has no other eigenvalues, suppose that $T$ did have some other nonzero eigenvalue $\lambda$. Then

This implies that $\lambda^n = 0$. Since $\lambda$ is an element of a field, this constradicts the assumption that $\lambda$ is nonzero.

(Nonnilpotent operator $T$ with spectrum $\{0\}$) This depends on the choice of the base field (let us just stick with finite-dimensional vector spaces). If the minimal polynomial fully splits, then there are no such operators because the minimal polynomial guarantees that $T^n = 0$ where $n = \dim V$. On the other hand, if the minimal polynomial doesn’t split, such as if

is considered as a linear operator on $\RR^4$ as a real vector space, then $A$ fails to be nilpotent and its only eigenvalue is $0$. [/solution]

[example] [extitle]Roman Advanced Linear Algebra Exercise 8.7[/extitle]

Show that if $S$ and $T$ are linear operators on $V$ and one of $S$ and $T$ is invertible, $ST \sim TS$. Furthermore, $ST$ and $TS$ have the same eigenvalues, counting multiplicity. [/example]

[proof] To see that $ST$ and $TS$ are similar, assume WLOG that $S$ is invertible. Then we see that

which implies that $ST \sim TS$. Now, if $\lambda$ is an eigenvalue of $ST$ such that $STv = \lambda v$, then

Thus, $\lambda$ is an eigenvalue of $TS$. An identical argument shows that any eigenvalue of $TS$ is also an eigenvalue of $ST$. [/proof]

[example] [extitle]Roman Advanced Linear Algebra Exercise 8.8[/extitle]

Roman credits this problem to Halmos.
a) Find a linear operator $T$ that is not idempotent but for which $T^2(I - T) = 0$.
b) Find a linear operator $T$ that is not idempotent but for which $T(I - T)^2 = 0$.
c) Prove that if $T^2(I - T) = T(I - T)^2 = 0$, then $T$ is idempotent. [/example]

[solution] (a) Let $A$ be the matrix

By direct computation, we can show that the matrix $A$ satisfies the requested requirements of $A$.

(b) Let $B$ be the matrix

By direct computation, we can show that the matrix $B$ satisfies the requested requirements of $B$.

(c) Subtracting the equations given implies

Therefore, $T^2 = T$ as desired and we are done. [/solution]

[example] [extitle]Axler Linear Algebra Done Right Exercise 5B.1[/extitle]

Let $T$ be a linear operator on a vector space $V$ over a field of characteristic $0$. Prove that $9$ is eigenvalue of $T^2$ if and only if $3$ or $-3$ is an eigenvalue of $T$. [/example]

[proof] ($\implies$) Assume that $T^2v = 9v$ for some nonzero $v \in V$. Then it follows that

Since the operators $T - 3I$ and $T + 3I$ commute with each other, it follows that $3$ or $-3$ is an eigenvalue of $T$.

($\impliedby$) Obvious. [/proof]

[example] [extitle]Axler Linear Algebra Done Right Exercise 5B.6[/extitle]

Let $T\in\End_k(k^2)$ be defined by $T(w,z) = (-z, w)$. Find the minimal polynomial of $T$. [/example]

[solution] The matrix representation of $T$ with respect to the standard basis of $k^2$ is

The characteristic polynomial of this matrix is $x^2 + 1$. But since we know the characteristic and minimal polynomials have the same sets of roots, the characteristic polynomial must be the minimal polynomial. (Note that this also could have been seen by just realizing that $k^2$ is cyclic as a $k[x]$-module here.) [/solution]

[example] [extitle]Axler Linear Algebra Done Right Exercise 5B.12[/extitle]

Let $k$ be a field of characteristic 0 and $T\in\End_k(k^n)$ by defined by

Find the minimal polynomial of $T$. [/example]

[solution] The matrix representation of $T$ with respect to the standard basis of $k^n$ is

So the characteristic polynomial of $T$ is just

Since $k$ has characteristic $0$, each factor is distinct so the minimal and characteristic polynomials are actually the same. [/solution]

[example] [extitle]KU Algebra Qual August 2023 Problem 3[/extitle]

Suppose that $A$ is a square complex matrix such that $A$ is similar to $A^n$, for some $n > 1$. Prove that the eigenvalues of $A$ are either $0$ or roots of unity. [/example]

[proof] Let $\lambda$ be an eigenvalue of $A$. Then there is some nonzero $v$ for which

By induction, it follows that

So $\lambda^n$ is an eigenvalue of $A^n$. But since $A\sim A^n$, it follows that $A$ and $A^n$ have the same eigenvalues. For the sake of contradiction, suppose that $\lambda$ is a nonzero eigenvalue of $A$ that is not a root of unity. Then the list

consists of eigenvalues of $A$ that all nonzero and not roots of unity (if any of them were, then $\lambda$ itself would be a root of unity). Now, since $\lambda$ is not a root of unity, we have $|\lambda| \neq 1, 0$. If $0 < |\lambda| < 1$, then

is a strictly decreasing sequence. This implies that the $\lambda^{n^i}$’s are all distinct so $A$ has infinitely many eigenvalues. This would imply that the characteristic polynomial of $A$ has infinitely many roots, which is a contradiction. [/proof]

[example] [extitle]KU Algebra Qual January 2023 Problem 1[/extitle]

Suppose that $A$ and $B$ are $n\times n$ matrices over the complex numbers.

1. Prove that if $A$ and $B$ are similar, then $A$ and $B$ have the same minimal and characteristic polynomials.
2. Determine, with justification, whether the converse of part 1 holds when $n = 3$. What about when $n = 4$? [/example]

[proof] (Part 1) First we show that $A$ and $B$ have the same minimal polynomials. Let $m_A(x) = \sum_{i=0}^r a_i x^i$ and $m_B(x) = \sum_{i=0}^s b_i x^i$ be the minimal polynomials of $A$ and $B$, respectively. As $A$ and $B$ are similar, there is some invertible $n\times n$ matrix $P$ for which $A = PBP^{-1}.$ So it follows that

Thus, $m_A(B) = 0$. It follows that $m_B(x) \mid m_A(x)$. An identical argument shows that $m_A(x) \mid m_B(x)$. Since $m_A(x)$ and $m_B(x)$ are monic, it follows that $m_A(x) = m_B(x)$.

Now we show that $A$ and $B$ have the same characteristic polynomial. Recall that

Now, since

the claim follows.

(Part 2) Over the complex numbers, recall that $A$ and $B$ are similar if and only if they have the same Jordan canonical form. With some casework, we can show that the minimal polynomials of $A$ and $B$ agreeing and the characteristic polynomials of $A$ and $B$ agree implies $A \sim B$. On the other hand, it is easy to construct a counter example for $n = 4$. Consider

Both $A$ and $B$ have the same minimal/characteristic polynomials, but they have different Jordan canonical forms. Thus, $A$ and $B$ are not similar. [/proof]