Functional Calculus in Linear Algebra, the Jordan Decomposition Reloaded and Cayley-Hamilton’s Theorem.

Abstract.
We explain the aims of functional calculus and specialize to polynomials evaluated at endomorphisms. We reconsider the Jordan decomposition and prove it with more generality. Then, we discuss Taylor expansion in the nilpotent part for endomorphisms with separable minimal polynomials, and prove Cayley-Hamilton again for arbitrary fields.

Let $V$ be a $K$ -vector space, $\varphi : V \to V$ an $K$ -endomorphism of $V$ and $f : K \to K$ a function. Here, we want to make sense of $f(\varphi)$ ; this should be another endomorphism of $V$ which is somehow related to both $\varphi$ and $f$ .

Let us make this more precise. For that, let $A$ be a subalgebra of the $K$ -algebra $\End_K(V)$ of endomorphisms of $V$ , containing the identity $\id_V$ , and let $F$ be a $K$ -subalgebra of the $K$ -algebra $Fun(K)$ of functions $K \to K$ , containing the identity $\id_K$ and the constant functions. We say that $\Psi : F \times A \to A$ is a functional calculus if $\Psi$ satisfies the following conditions:

for a fixed $\varphi \in A$ , the map $\Psi(\bullet, \varphi) : F \to A$ , $f \mapsto \Psi(f, \varphi)$ is a $K$ -algebra homomorphism with $\Psi(1, \varphi) = \id_V$ and $\Psi(\id_K, \varphi) = \varphi$ ;
for a fixed $f \in F$ , the map $\Psi(f, \bullet) : A \to A$ , $\varphi \mapsto \Psi(f, \varphi)$ is $K$ -algebra with $\Psi(f, \id_V) = f(1) \id_V$ .

We usually write $f(\varphi)$ for $\Psi(f, \varphi)$ if it is clear which $\Psi$ is meant.

Note that $F$ contains all polynomial functions $K \to K$ , i.e. the functions of the type $\lambda \mapsto f(\lambda)$ , where $f \in K[x]$ is a polynomial. Note that for polynomial functions $f$ , the value of $\Psi(f(\id_K), \varphi)$ for $f \in K[x]$ is completely determined by the fact that $\Psi(\bullet, \varphi)$ is an $K$ -algebra homomorphism with $\Psi(1, \varphi) = \id_V$ and $\Psi(\id_K, \varphi) = \varphi$ , as $\Psi(\lambda, \varphi) = \lambda \Psi(1, \varphi) = \lambda \id_V$ : if $f = \sum_{i=0}^n a_i x^i$ , then $\Psi(f(\id_K), \varphi) ={} & \Psi\biggl(\sum_{i=0}^n a_i (\id_K)^n, \varphi\biggr) \\ {}={} & \sum_{i=0}^n a_i \Psi(\id_K, \varphi)^n = \sum_{i=0}^n a_i \varphi^i = f(\varphi).$

In particular, this gives a canonical functional calculus $K[\id_K] \times \End_K(V) \to \End_K(V)$ , where $K[\id_K]$ is the image of the canonical map $K[x] \to Fun(K)$ . (In case you are curious, $K[x] \cong K[\id_K] \subsetneqq Fun(K)$ if, and only if, $K$ is infinite; in the other case, $Fun(K) = K[\id_K] \cong K[x] / (x^q - x)$ , where $q = \abs{K} < \infty$ .)

What about functions which are not polynomial? In case $\varphi$ is diagonalizable, i.e. $V$ has a basis consisting of eigenvectors of $\varphi$ , one can define $f(\varphi)$ for an arbitrary function $f : K \to K$ by defining $f(\varphi)$ as the linear map which maps an eigenvector $v$ of $\varphi$ with eigenvalue $\lambda$ to $f(\lambda) v$ . If one sets $A_\varphi := \{ f(\varphi) \mid f \in Fun(K) \}$ , one obtains a functional calculus $Fun(K) \times A_\varphi \to A_\varphi$ .

In Functional Analysis, one is interested in such functional calculi with $K = \R$ or $\C$ , and one obtains ones for holomorphic functions $f$ , for continuous functions $f$ and even for certain Borel-measureable functions $f$ . But for today, we want to stick to the situation of an arbitrary $K$ . We will use $A = \End_K(V)$ and $F = K[\id_K]$ , i.e. the canonical functional calculus.

Definition.

Let $\varphi \in \End_K(V)$ . In case the canonical map $K[x] \to \End_K(V)$ , $f \mapsto f(\varphi)$ is not injective, the unique normed generator of $K[x] \to \End_K(V)$ is called the minimal polynomial of $\varphi$ and denoted by $\mu_f$ .

In case $\dim_K V < \infty$ , every $\varphi \in \End_K(V)$ has a minimal polynomial, as $\dim_K \End_K(V) = (\dim_K V)^2 < \infty$ and $\dim_K K[x] = \infty$ . In case $\dim_K V = \infty$ , certain elements of $\End_K(V)$ do have a minimal polynomial; for example, $\varphi = \id_V$ has the minimal polynomal $x - 1$ ; other elements of $\End_K(V)$ do not possess a minimal polynomial, for example any endomorphism with infinitely many different eigenvalues.

Lemma.

Assume that $\varphi$ possesses a minimal polynomial. Then $\lambda \in K$ is an eigenvalue of $\varphi$ if, and only if, $\mu_\varphi(\lambda) = 0$ .

Proof.

In case $\lambda$ is an eigenvalue, let $v$ be an corresponding eigenvector and let $W := \gen{v}$ . Then $\End_K(W) \cong K$ , and $\varphi|_W$ corresponds to $\lambda$ . Clearly, $0 = \mu_\varphi(\varphi)|_W = \mu_\varphi(\varphi|_W) = \mu_\varphi(\lambda \id_W) = \mu_\varphi(\lambda) \id_W$ , whence $\mu_\varphi(\lambda) = 0$ .
Conversely, assume that $\mu_\varphi(\lambda) = 0$ . Write $\mu_\varphi = (x - \lambda)^n f$ , where $n \in \N$ and $f \in K[x]$ satisfies $f(\lambda) \neq 0$ . As $\mu_\varphi(\lambda) = 0$ , $n > 0$ . Now $0 = \mu_\varphi(\varphi) = (\varphi - \lambda \id_V)^n \circ f(\varphi)$ . In case $\ker (\varphi - \lambda \id_V)^n \neq 0$ , $\lambda$ is an eigenvalue of $\varphi$ (let $v \in \ker (\varphi - \lambda \id_V)^n \setminus 0$ and choose $i \in \N$ maximal with $w := (\varphi - \lambda \id_V)^i v \neq 0$ ; then $\varphi(w) = \lambda w$ ); hence, assume $\ker (\varphi - \lambda \id_V)^n = 0$ . In that case, we must have $f(\varphi) = 0$ . But as $f$ is a proper divisor of $\mu_\varphi$ , this cannot be.
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The minimal polynomial is a rather powerful tool. In case it exists, one gets an $\varphi$ -invariant decomposition of $V$ as follows:

Lemma.

Let $f$ be a prime divisor of $\mu_\varphi$ . Then $\displaystyle \GEig(\varphi, f) := \{ v \in V \mid \exists n : f(\varphi)^n(v) = 0 \}$ is an $\varphi$ -invariant subspace of $V$ . If $g$ is another prime divisor of $\mu_\varphi$ coprime to $f$ , then $g(\varphi)$ is $\GEig(\varphi, f)$ -invariant and $g(\varphi)|_{\GEig(\varphi, f)}$ is an monomorphism.
If $f$ is an arbitrary prime polynomial, $\GEig(\varphi, f) \neq 0$ if, and only if, $f \mid \mu_\varphi$ .

Proof.

Clearly, $\GEig(\varphi, f) = \bigcup_{n=0}^\infty \ker f(\varphi)^n$ . As $\ker f(\varphi)^n \subseteq \ker f(\varphi)^{n+1}$ , this is a subspace of $V$ . As $\ker f(\varphi)^n$ is $\varphi$ -invariant (as $f(\varphi) \circ \varphi = (f x)(\varphi) = (x f)(\varphi) = \varphi \circ f(\varphi)$ ), it follows that $\GEig(\varphi, f)$ is $\varphi$ -invariant as well.
As $g(\varphi) f(\varphi) = (f g)(\varphi) = f(\varphi) g(\varphi)$ , $\GEig(\varphi, f)$ is $g(\varphi)$ -invariant as well. Let $v \in \GEig(\varphi, f) \cap \ker g(\varphi)$ and let $n \in \N$ be minimal with $f(\varphi)^n(v) = 0$ . As $f^n, g$ are coprime, there exist $h, h' \in K[x]$ with $1 = h f^n + h' g$ . Then $0 = h(\varphi) f(\varphi)^n(v) + h'(\varphi) g(\varphi)(v) = (h f^n + h' g)(v) = v$ . Therefore, $g(\varphi)|_{\GEig(\varphi, f)}$ is injective.
Finally, the last statement can be proven in exactly the same way as the previous lemma.
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Lemma.

Let $\mu_\varphi = \prod_{i=1}^n f_i^{e_i}$ , where $f_1, \dots, f_n$ is a set of pairwise distinct monic prime polynomials and $e_i \in \N_{\ge 1}$ . Then $\bigoplus_{i=1}^n \GEig(\varphi, f_i)$ is a direct sum.

Proof.

Assume that this is not a direct sum. Then there exists $v_i \in \GEig(\varphi, f_i)$ , not all zero, such that $0 = \sum_{i=1}^n v_i$ . Assume that the number of non-zero $v_i$ is minimal under this condition. Let $i$ be with $v_i \neq 0$ , and let $n \in \N$ satisfy $f_i(\varphi)^n(v_i) = 0$ . Then $0 = \sum_{j=1}^n f_i(\varphi)^n(v_j)$ , and $f_i(\varphi)^n(v_i) = 0$ . If $j \neq i$ , then $f_i(\varphi)^n(v_j) \neq 0$ as $f_i(\varphi)|_{\GEig(\varphi, f_j)}$ is injective and so is its $n$ -th power. But this is only possible by the minimality assumption of $v_j = 0$ for all $j \neq i$ , i.e. $0 = v_i$ , contradicting the choice of $i$ . Therefore, the sum is a direct sum.
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Lemma.

Let $f$ be a prime factor of $\mu_\varphi$ and let $e \in \N$ be the maximal exponent of $f$ appearing in $\mu_\varphi$ , i.e. $f^e \mid \mu_\varphi$ and $f^{e+1} \nmid \mu_\varphi$ . Then $\ker f(\varphi)^e = \GEig(\varphi, f)$ .

Proof.

Let $v \in \GEig(\varphi, f)$ , and write $\mu_\varphi = f^e \cdot g$ with $g \in K[x]$ ; then $g$ and $f$ are coprime. We have to show $f(\varphi)^e(v) = 0$ . Clearly $w := f(\varphi)^e(v)$ lies in the kernel of $g(\varphi)$ , as $g f^e = \mu_\varphi$ . Let $n \in \N$ be such that $w \in \ker f(\varphi)^n$ ; as $f^n$ and $g$ are coprime, there exist $h, h' \in K[x]$ with $f^n h + g h' = 1$ . Therefore, $0 ={} & h(\varphi) f(\varphi)^n(w) + h'(\varphi) g(\varphi)(w) \\ {}={} & (h f^n + h' g)(\varphi)(w) = w = f(\varphi)^e(v).$
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Note that one can in fact show that $\image f(\varphi)^{e+1} = \image f(\varphi)^e$ .

Lemma.

Let $f$ be a prime factor of $\mu_\varphi$ . Then there exists a $\varphi$ -invariant subspace $W \subseteq V$ with $V = W \oplus \GEig(\varphi, f)$ .

Proof.

Write $\mu_\varphi = f^e \cdot g$ with $e \in \N$ , $g \in K[x]$ and $f \nmid g$ . Then $\GEig(\varphi, f) = \ker f(\varphi)^e$ and, in particular, $\ker f(\varphi)^e = \ker f(\varphi)^{e+i}$ for every $i \in \N$ . Let $v \in \image f(\varphi)^e \cap \ker f(\varphi)^e = 0$ ; we can write $v = f(\varphi)^e(w)$ for some $w \in V$ . Now $f(\varphi)^{2 e}(w) = 0$ , whence $w \in \ker f(\varphi)^{2 e} = \ker f(\varphi)^e$ , i.e. $v = f(\varphi)^e(w) = 0$ . Therefore, $\image f(\varphi)^e \cap \GEig(\varphi, f) = 0$ .
Let $h, h' \in \N$ such that $1 = h f^e + h' g$ . Let $v \in V$ ; then $v = f^e(\varphi) (h(\varphi)(v)) + g(\varphi) (h'(\varphi)(v)) = f(\varphi)^e(w_1) + g(\varphi)(w_2)$ with $w_1, w_2 \in V$ . Now $f^e g = \mu_\varphi$ , whence $0 = \mu_\varphi(w_2) = f(\varphi)^e g(\varphi)(w_2)$ , i.e. $g(\varphi)(w_2) \in \ker f(\varphi)^e$ . As $f(\varphi)^e(w_1) \in \image f(\varphi)^e$ , we see $v \in \image f(\varphi)^e + \ker f(\varphi)^e = \image f(\varphi)^e + \GEig(\varphi, f)$ .
Hence, we get $V = \image f(\varphi)^e \oplus \GEig(\varphi, f)$ . Finally, note that $W := \image f(\varphi)^e$ is clearly $\varphi$ -invariant.
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Theorem (Generalized Jordan Decomposition).

Let $\mu_\varphi = \prod_{i=1}^n f_i^{e_i}$ , where $f_1, \dots, f_n$ is a set of pairwise distinct monic prime polynomials and $e_i \in \N_{\ge 1}$ . Then $V = \bigoplus_{i=1}^n \GEig(\varphi, f_i)$ .

Proof.

We show this by induction on $n$ . In case $n = 0$ , we have $\mu_\varphi = 1$ , which is only possible if $V = 0$ . In that case, the statement is obvious. Hence, assume $n > 0$ .
Let $W_1 := \GEig(\varphi, f_n)$ and $W_2$ be an $\varphi$ -invariant direct complement of $W_1$ . Clearly $f_n(\varphi)^{e_n}$ is injective on $W_2$ , whence $f_n(\varphi)^{e_n} \circ \prod_{i=1}^{n-1} f_i(\varphi)^{e_i} = \mu_\varphi(\varphi|_{W_2}) = 0$ implies $\prod_{i=1}^{n-1} f_i(\varphi)^{e_i} = 0$ . Therefore, $\mu_{f|_{W_2}}$ has strictly less than $n$ distinct prime factors, whence $W_2 = \bigoplus_{i=1}^{n-1} \GEig(\varphi|_{W_2}, f_i)$ . In particular, $V = W_1 \oplus W_2 ={} & \GEig(\varphi, f_n) \oplus \bigoplus_{i=1}^{n-1} \GEig(\varphi|_{W_2}, f_i) \\ {}\subseteq{} & \GEig(\varphi, f_n) + \sum_{i=1}^{n-1} \GEig(\varphi, f_i),$ whence the claim follows.
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Note that this is a generalization of the Jordan decomposition. Note that in fact, $\bigoplus_{i=1}^n \GEig(\varphi, f_i)$ is the minimal $K[\varphi]$ -decomposition of $V$ in case $\mu_\varphi = \prod_{i=1}^n f_i^{e_i}$ . This completes the task started in my post on such decompositions, namely finding minimal $K[\varphi]$ -decompositions in case the characteristic polynomial of $\varphi$ (assuming $\dim_K V < \infty$ ) does not splits into linear factors.

Lemma.

Assume that $\varphi$ has a minimal polynomial $\mu_\varphi$ of the form $f^e$ , where $f$ is prime and $e \in \N$ . Let $\varphi_n := f(\varphi)$ and $\varphi_d := \varphi - \varphi_n$ . Then $\varphi_d \varphi_n = \varphi_n \varphi_d$ and $\varphi_n$ is nilpotent of index $e$ . Moreover, $\varphi_d$ is diagonalizable if, and only if $\deg f = 1$ . Finally, $\varphi$ is diagonalizable if, and only if, $\deg f = 1$ and $e = 1$ .

Proof.

Clearly, $\varphi_d \varphi_n = \varphi_n \varphi_d$ as both are elements of $K[\varphi] \cong K[x]/(\mu_\varphi)$ . Moreover, $\varphi_n = f(\varphi)$ , whence $\varphi_n$ is nilpotent of index $e$ .
If $\deg f = 1$ , $f = x - \lambda$ for some $\lambda \in K$ . In that case, $\varphi_d = \varphi - \varphi_n = \lambda \id_V$ . Conversely, if $\varphi_d$ is diagonalizable, any eigenvalue of $\varphi_d$ must be a zero of $f^e$ . This is only possible if $\deg f = 1$ .
Finally, assume that $\varphi$ is diagonalizable. Hence, $\varphi_n$ is diagonalizable as well; but the only diagonalizable and nilpotent endomorphism is 0, whence $e = 1$ and $\varphi_d = \varphi$ is diagonalizable, i.e. $\deg f = 1$ . Conversely, assume $\deg f = 1$ and $e = 1$ ; then $\varphi_n = 0$ and $\varphi = \varphi_d$ is diagonalizable.
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Corollary.

Assume that $\varphi$ has a minimal polynomial. Then $\varphi$ is diagonalizable if, and only if, $\mu_\varphi$ is squarefree and splits over $K$ .

Proof.

Write $\mu_\varphi = \prod_{i=1}^n f_i^{e_i}$ with $e_i \in \N$ and pairwise distinct, monic prime polynomials $f_i$ . Then $V = \bigoplus_{i=1}^n \GEig(\varphi, f_i)$ by the generalized Jordan decomposition. Hence $\varphi$ is diagonalizable if, and only if, $\varphi|_{\GEig(\varphi, f_i)}$ is diagonalizable for every $i$ . For a fixed $i$ , we have that $\mu_{\varphi|_{\GEig(\varphi, f_i)}} = f_i^{e_i}$ , whence by the previous lemma, $\varphi|_{\GEig(\varphi, f_i)}$ is diagonalizable if, and only if, $\deg f_i = 1$ and $e_i = 1$ .
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Lemma.

Assume that $\varphi$ has a minimal polynomial. Then there exist polynomials $f_d, f_n \in K[x]$ such that $\varphi_n = f_n(\varphi)$ and $\varphi_d = f_d(\varphi)$ , where $\varphi_n, \varphi_d$ are the endomorphisms from the previous corollary.

Proof.

As $\varphi_n + \varphi_d = \varphi$ , it suffices to show the existence of $f_n$ . Write $\mu_\varphi = \prod_{i=1}^n f_i^{e_i}$ with $e_i \in \N$ and pairwise distinct monic primes $f_i$ , and set $V_i := \GEig(\varphi, f_i)$ . We want a polynomial $f \in K[x]$ such that $f(\varphi)|_{V_i} = f_i(\varphi)|_{V_i}$ . Now the minimal polynomial of $\varphi|_{V_i}$ is $f_i^{e_i}$ , whence $Formula does not parse: f(\varphi)|_{V_i} = (f \mymod f_i^{e_i})(\varphi)|_{V_i}$ , i.e. it suffices to solve the congruences $f \equiv f_i \pmod{f_i^{e_i}}$ , $i = 1, \dots, n$ . But since $f_i^{e_i}$ , $1 \le i \le n$ , are pairwise coprime, such an $f$ exists by the Chinese Remainder Theorem.
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Corollary (Generalized Jordan Decomposition).

Assume that $\varphi$ has a minimal polynomial which is separable (i.e. its prime factors do not have multiple roots in their splitting field). Then there exist unique endomorphisms $\varphi_d, \varphi_n \in \End_K(V)$ such that

$\varphi = \varphi_n + \varphi_d$ and $\varphi_d \varphi_n = \varphi_n \varphi_d$ ;

$\varphi_n$ is nilpotent;

if $L$ is a splitting field of $\mu_\varphi$ over $L$ , $\varphi_n \otimes_K L \in \End_L(V \otimes_K L)$ is diagonalizable.

Proof.

By the previous lemma and corollary, there exist polynomials $f_n, f_d \in K[x]$ with $\varphi_n = f_n(\varphi)$ , $\varphi_d = f_d(\varphi)$ such that $\varphi_n, \varphi_d$ satisfy the conditions. (Note that $\mu_{\varphi_d \otimes_K L} = \mu_{\varphi_d} = \prod_{i=1}^n f_i$ , and since $L/K$ is separable, $\prod_{i=1}^n f_i$ is squarefree and splits into linear factors over $L$ . Hence, by the second-previous lemma, $\varphi_d \otimes_K L$ is diagonalizable.) Now let $\varphi'_n, \varphi'_d$ be any two endomorphisms which satisfy the conditions above. As $\varphi'_n + \varphi'_d = \varphi$ and $\varphi'_n \varphi'_d = \varphi'_d \varphi'_n$ , all of $\varphi'_n$ , $\varphi'_d$ , $\varphi_n$ and $\varphi_d$ commute with each other. Hence, we have $\varphi'_n - \varphi_n = \varphi_d - \varphi'_d$ , and $\varphi'_n - \varphi_n$ is nilpotent and $(\varphi_d - \varphi'_d) \otimes_K L$ is diagonalizable. But this is possible if, and only if, $\varphi'_n - \varphi_n = \varphi_d - \varphi'_d = 0$ , i.e. if $Formula does not parse: \arphi_n = \varphi'_n$ and $\varphi_d = \varphi'_d$ .
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Let us now return to the original idea of functional calculus. The generalized Jordan decomposition allows us to do a Taylor expansion in the nilpotent part:

Theorem (Taylor expansion in the nilpotent part).

Assume that $\varphi$ has a minimal polynomial which is separable. Let $\varphi = \varphi_d + \varphi_n$ be the generalized Jordan decomposition of $\varphi$ , and let $f \in K[x]$ . Finally, let $e$ be the nilpotence index of $\varphi_n$ , i.e. let $e$ satisfy $\varphi_n^e = 0$ . Then $\displaystyle f(\varphi) = \sum_{i=0}^{e-1} \frac{f^{(i)}}{i!}(\varphi_d) \varphi_n^i.$

Proof.

Consider $L := K(x)$ , the rational function field. The Taylor expansion of $f(t) \in L[t]$ around $\lambda = x \in K(x)$ is given by $f(t) = \sum_{i=0}^{\deg f} \frac{f^{(i)}}{i!}(x) (t - x)^i$ . Here, we have in fact $\frac{f^{(i)}}{i!}(x) \in K[x]$ . As $\varphi_n$ , $\varphi$ commute, we can plug in $x = \varphi_d$ and $t = \varphi$ and obtain $\displaystyle f(\varphi) = \sum_{i=0}^{\deg f} \frac{f^{(i)}}{i!}(\varphi_d) (\varphi - \varphi_d)^i = \sum_{i=0}^{\deg f} \frac{f^{(i)}}{i!}(\varphi_d) \varphi_n^i.$ Now $\varphi_n^i = 0$ for $i \ge e$ gives the formula.
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Note that in case $K = \R$ or $\C$ , this formula holds also for arbitrary analytic functions $f : K \to K$ . In fact, the function only needs to be analytic on an open set which contains the complex eigenvalues of $\varphi$ . The most important example is the exponential function $\exp : \C \to \C$ , $z \mapsto \sum_{i=0}^\infty \frac{z^i}{i!}$ . The above shows that every $\varphi \in \End_\C(V)$ possessing a minimal polynomial can be decomposed into a diagonalizable part $\varphi_d$ and a nilpotent part $\varphi_n$ of finite index $e$ , and in that case, $\displaystyle \exp(\varphi) = \sum_{i=0}^{e-1} \frac{\exp(\varphi_d)}{i!} \varphi_n^i = \exp(\varphi_d) \sum_{i=0}^{e-1} \frac{\varphi_n^i}{i!}.$

Now let $\dim_K V < \infty$ . Recall the the characteristic polynomial of $\varphi \in \End_K(V)$ is defined as $c_\varphi := \det(\varphi - t \id_V) \in K[t]$ . So far, we have not used Cayley-Hamilton’s Theorem. In fact, we can use the above stuff to prove the theorem. For that, we first relate the minimal polynomial to the characteristic polynomial.

Lemma.

If $f$ is an irreducible prime, then $f$ divides $\mu_\varphi$ if, and only if, $f$ divides $c_\varphi$ .

Proof.

Let $L$ be a splitting field of $f$ over $K$ , and consider $\varphi_L := \varphi \otimes_K L \in \End_L(V \otimes_K L)$ . We have $\mu_{\varphi_L} = \mu_\varphi$ and $c_{\varphi_L} = c_\varphi$ , whence it suffices to show that $c_{\varphi_L}(\lambda) = 0$ if, and only if, $\mu_{\varphi_L}(\lambda) = 0$ for every $\lambda \in L$ .
For that, note that $\mu_{\varphi_L}(\lambda) = 0$ if, and only if, $\lambda$ is an eigenvalue of $\varphi_L$ . But this is equivalent to $\varphi_L - \lambda \id_{V \otimes_K L}$ not being injective, which in turn is equivalent (as $\dim_L (V \otimes_K L) = \dim_K V < \infty$ ) to that $\varphi_L - \lambda \id_{V \otimes_K L}$ is not invertible, which is the case if, and only if, $\det(\varphi_L - \lambda \id_{V \otimes_K L}) = 0$ , i.e. $c_{\varphi_L}(\lambda) = 0$ .
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In fact, we can show that $\mu_\varphi$ divides $c_\varphi$ , which implies the Cayley-Hamilton theorem as $\mu_\varphi(\varphi) = 0$ . For that, we show that $\dim \GEig(\varphi, f) = \nu_f(c_\varphi) \deg f$ for every prime polynomial $f$ , where $\nu_f : K[x] \setminus \{ 0 \} \to \N$ gives the exponent of $f$ in the prime factor decomposition of a non-zero element of $K[x]$ .

Lemma.

Assume $\dim_K V < \infty$ . If $\varphi = \varphi_d + \varphi_n$ with $\varphi_n = f(\varphi)$ being nilpotent, where $f \in K[x]$ , then $c_\varphi = c_{\varphi_d}$ .

Proof.

Let $L$ be a splitting field of $c_\varphi$ over $K$ , and let $\varphi_L := \varphi \otimes_K L$ , $\varphi_{d,L} := \varphi_d \otimes_K L$ and $\varphi_{n,L} := \varphi_n \otimes L$ . It then suffices to show that the statement holds for these $L$ -endomorphisms of $V \otimes_K L$ . Hence, we can assume that $c_\varphi$ splits over $K$ . In that case, there exists a basis $B$ of $V$ such that the representation matrix $M_B(\varphi)$ of $\varphi$ with respect to $B$ is in upper triangular form. Then $c_\varphi = \prod (x - \lambda)$ , where $\lambda$ ranges over the diagonal elements of $M_B(\varphi)$ . Now $\varphi_n = f(\varphi)$ , whence $M_B(\varphi_n)$ is in upper triangular form as well. As $\varphi_n$ is nilpotent, the diagonal elements of $M_B(\varphi_n)$ must all be zero. As $M_B(\varphi) = M_B(\varphi_d) + M_B(\varphi_n)$ , we see that $c_{\varphi_d} = c_\varphi$ .
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Lemma.

Assume that $\varphi$ has a minimal polynomial. Let $f := \prod_{i=1}^n f_i^{e_i}$ , where $f_1, \dots, f_n$ are pairwise distinct irreducible polynomials. Set $\displaystyle W := \{ v \in V \mid \exists n \in \N : f(\varphi)^n(v) = 0 \}.$ Then $\displaystyle W = \bigoplus_{i=1}^n \GEig(\varphi, f_i).$

Proof.

Let $v_i \in \GEig(\varphi, f_i)$ and let $t_i \in \N$ with $f_i(\varphi)^{t_i}(v_i) = 0$ ; then, if $t := \max\{ t_1, \dots, t_n \}$ , $w = \sum_{i=1}^n v_i$ satisfies $\displaystyle f(\varphi)^t(v) = \sum_{i=1}^n f(\varphi)^t(v_i) = \sum_{i=1}^n \prod_{j=1 \atop j \neq i}^n f_j(\varphi)^{e_j} \circ f_i(\varphi)^{e_i t}(v_i);$ as $f_i(\varphi)^{e_i t}(v_i) = 0$ since $e_i t \ge t \ge t_i$ , we get $\bigoplus_{i=1}^n \GEig(\varphi, f_i) \subseteq W$ .
For the converse, first note that $W$ is $\varphi$ -invariant. Assume $\mu_{\varphi|_W}$ has a monic prime factor $p$ distinct from $p_1, \dots, p_n$ . Then $\dim \GEig(\varphi|_W, p) > 0$ ; let $w \in \GEig(\varphi|_W, p) \setminus \{ 0 \}$ . Let $t \in \N$ be such that $p(\varphi)^t(w) = 0$ and $s \in \N$ be such that $f(\varphi)^s(w) = 0$ . As $p$ and $f$ are coprime, there exist polynomials $h, h' \in K[x]$ with $h p^t + h' f^s = 1$ . Hence, $\displaystyle 0 = h(\varphi) p(\varphi)^t(w) + h'(\varphi) f(\varphi)^s(w) = (h p^t + h' f^s)(\varphi)(w) = w,$ a contradiction. Hence, all prime factors $\mu_{\varphi|_W}$ lie in $\{ p_1, \dots, p_n \}$ . Therefore, $\displaystyle W = \bigoplus_{i=1}^n \GEig(\varphi|_W, p_i) \subseteq \bigoplus_{i=1}^n \GEig(\varphi, p_i) \subseteq W,$ which shows the claim.
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Corollary.

Assume $\varphi$ has a minimal polynomial, and let $f$ be a prime polynomial. Let $L$ be a field extension of $K$ over which $f$ splits; write $f = \prod_{i=1}^n (x - \lambda_i)^{e_i}$ with distinct elements $\lambda_1, \dots, \lambda_n \in L$ and $e_i \in \N$ . Then $\displaystyle \GEig(\varphi, f) \otimes_K L = \bigoplus_{i=1}^n \GEig(\varphi \otimes_K L, x - \lambda_i).$

Proof.

Notice that $\GEig(\varphi, f) \otimes_K L = \{ v \in V \otimes_K L \mid \exists n \in \N : f(\varphi \otimes_K L)^n(v) = 0 \}$ . Hence, the corollary follows from the previous lemma.
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Lemma.

Let $\dim_K V < \infty$ and $f$ be a prime polynomial. Then $\dim \GEig(\varphi, f) = \nu_f(c_\varphi) \deg f$ .

Proof.

We first show that “ $\le$ ” holds. Let $L$ be a splitting field of $f$ over $K$ , and write $f = \prod_{i=1}^t (x - \lambda_i)^{e_i}$ with $\lambda_1, \dots, \lambda_t \in L$ pairwise distinct and $e_i \in \N$ . We have $\GEig(\varphi, f) \otimes_K L = \bigoplus_{i=1}^t \GEig(\varphi \otimes_K L, x - \lambda_i)$ ; since $\nu_{f_i}(c_{\varphi \otimes_K L}) = e_i \nu_f(c_\varphi)$ , it suffices to know that the theorem holds in case $\deg f = 1$ , as then $\dim \GEig(\varphi \otimes_K L, x - \lambda_i) = \nu_{x - \lambda_i}(c_{\varphi \otimes_K L})$ and, therefore, $\dim_L (\GEig(\varphi, f) \otimes_K L) ={} & \sum_{i=1}^t \dim_L \GEig(\varphi \otimes_K L, x - \lambda_i) \\ {}={} & \sum_{i=1}^t e_i \nu_f(c_\varphi) = \deg f \cdot \nu_f(c_\varphi).$ Hence, assume that $f = x - \lambda$ with $\lambda \in K$ . In that case, $W := \GEig(\varphi, f) = \GEig(\varphi, \lambda)$ . Let $e = \nu_f(c_\varphi)$ and write $c_\varphi = (x - \lambda)^e g$ with $g \in K[x]$ . Note that $c_{\varphi|_W}$ divides $c_\varphi$ . But $\varphi - f(\varphi)$ is diagonalizable on $W$ with only the eigenvalue $\lambda$ , whence $c_{\varphi|_W} = c_{(\varphi - f(\varphi))|_W} = (x - \lambda)^{\dim W}$ . Therefore, $\dim W \le e$ .
The above argument shows $\dim \GEig(\varphi, f) \le \nu_f(c_\varphi) \deg f$ . If $p_1, \dots, p_n$ are all distinct prime factors of $c_\varphi$ , we get $\dim_K V ={} & \sum_{i=1}^n \dim_K \GEig(\varphi, p_i) \\ {}\le{} & \sum_{i=1}^n \nu_{p_i}(c_\varphi) \deg p_i = \deg c_\varphi = \dim_K V;$ as all summands are $\ge 0$ , the theorem follows.
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Corollary (Cayley-Hamilton over Fields).

If $\dim_K V < \infty$ and $\varphi \in \End_K(V)$ , then $c_\varphi(\varphi) = 0$ .
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Tags: Cayley-Hamliton, functional calculus, Jordan decomposition

This entry was posted on thursday, august 13th, 2009 at 06:22 utc and is filed under Analysis, Linear Algebra. You can follow any responses to this entry through the RSS 2.0 feed. You can leave a response, or trackback from your own site.

Functional Calculus in Linear Algebra, the Jordan Decomposition Reloaded and Cayley-Hamilton’s Theorem.

Abstract.

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