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

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

Let us make this more precise. For that, let be a subalgebra of the -algebra $\End_K(V)$ of endomorphisms of , containing the identity $\id_V$ , and let be a -subalgebra of the -algebra 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 -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 -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 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 , 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 -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, 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. 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 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 , for continuous functions and even for certain Borel-measureable functions . But for today, we want to stick to the situation of an arbitrary . 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 ; 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 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 $\mu_\varphi(\lambda) = 0$ , . 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 is a proper divisor of $\mu_\varphi$ , this cannot be.

The minimal polynomial is a rather powerful tool. In case it exists, one gets an $\varphi$ -invariant decomposition of as follows:

Lemma.

Let be a prime divisor of $\mu_\varphi$ . Then

$\GEig(\varphi, f) := \{ v \in V \mid \exists n : f(\varphi)^n(v) = 0 \}$

is an $\varphi$ -invariant subspace of . If is another prime divisor of $\mu_\varphi$ coprime to , then $g(\varphi)$ is $\GEig(\varphi, f)$ -invariant and $g(\varphi)|_{\GEig(\varphi, f)}$ is an monomorphism.

If 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 . 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 are coprime, there exist $h, h' \in K[x]$ with . 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.

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 is minimal under this condition. Let 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 -th power. But this is only possible by the minimality assumption of for all $j \neq i$ , i.e. , contradicting the choice of . Therefore, the sum is a direct sum.

Lemma.

Let be a prime factor of $\mu_\varphi$ and let $e \in \N$ be the maximal exponent of 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 and 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 and are coprime, there exist $h, h' \in K[x]$ with . Therefore,

$0 ={} & h(\varphi) f(\varphi)^n(w) + h'(\varphi) g(\varphi)(w) \\ {}={} & (h f^n + h' g)(\varphi)(w) = w = f(\varphi)^e(v).$

Note that one can in fact show that $\image f(\varphi)^{e+1} = \image f(\varphi)^e$ .

Lemma.

Let 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 . 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.

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 . In case , we have $\mu_\varphi = 1$ , which is only possible if . In that case, the statement is obvious. Hence, assume .

Let $W_1 := \GEig(\varphi, f_n)$ and be an $\varphi$ -invariant direct complement of . Clearly $f_n(\varphi)^{e_n}$ is injective on , 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 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.

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 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 , where 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 . Moreover, $\varphi_d$ is diagonalizable if, and only if $\deg f = 1$ . Finally, $\varphi$ is diagonalizable if, and only if, $\deg f = 1$ and .

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 .

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 . 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 and $\varphi_d = \varphi$ is diagonalizable, i.e. $\deg f = 1$ . Conversely, assume $\deg f = 1$ and ; then $\varphi_n = 0$ and $\varphi = \varphi_d$ is diagonalizable.

Corollary.

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

Proof.

Write $\mu_\varphi = \prod_{i=1}^n f_i^{e_i}$ with $e_i \in \N$ and pairwise distinct, monic prime polynomials . 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 . For a fixed , 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 .

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 . Write $\mu_\varphi = \prod_{i=1}^n f_i^{e_i}$ with $e_i \in \N$ and pairwise distinct monic primes , 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 $f(\varphi)|_{V_i} = (f \mymod f_i^{e_i})(\varphi)|_{V_i}$ , 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 exists by the Chinese Remainder Theorem.

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 is a splitting field of $\mu_\varphi$ over , $\varphi_n \otimes_K L \in \End_L(V \otimes_K L)$ 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 is separable, $\prod_{i=1}^n f_i$ is squarefree and splits into linear factors over . 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 $\varphi_n = \varphi'_n$ and $\varphi_d = \varphi'_d$ .

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 be the nilpotence index of $\varphi_n$ , i.e. let satisfy $\varphi_n^e = 0$ . Then

$f(\varphi) = \sum_{i=0}^{e-1} \frac{f^{(i)}}{i!}(\varphi_d) \varphi_n^i.$

Proof.

Consider , 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

$f(\varphi) = \sum_{i=0}^{\deg f} \frac{f^{(i)}}{i!}(\varphi_d) (\varphi - \varphi_d)^i = \sum_{i=0}^{\deg f} \frac{f^{(i)}}{i!}(\varphi_d) \varphi_n^i.$

Now $\varphi_n^i = 0$ for $i \ge e$ gives the formula.

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 , and in that case,

$\exp(\varphi) = \sum_{i=0}^{e-1} \frac{\exp(\varphi_d)}{i!} \varphi_n^i = \exp(\varphi_d) \sum_{i=0}^{e-1} \frac{\varphi_n^i}{i!}.$

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

Proof.

Let be a splitting field of over , 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. $c_{\varphi_L}(\lambda) = 0$ .

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 , where $\nu_f : K[x] \setminus \{ 0 \} \to \N$ gives the exponent of in the prime factor decomposition of a non-zero element of .

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 be a splitting field of $c_\varphi$ over , 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 -endomorphisms of $V \otimes_K L$ . Hence, we can assume that $c_\varphi$ splits over . In that case, there exists a basis of such that the representation matrix $M_B(\varphi)$ of $\varphi$ with respect to 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 $c_{\varphi_d} = c_\varphi$ .

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

$W := \{ v \in V \mid \exists n \in \N : f(\varphi)^n(v) = 0 \}.$

Then

$W = \bigoplus_{i=1}^n \GEig(\varphi, f_i).$

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

$f(\varphi)^t(v) = \sum_{i=1}^n f(\varphi)^t(v_i) = \sum_{i=1}^n \prod_{j=1 \atop j \neq i}^n f_j(\varphi)^{e_j} \circ f_i(\varphi)^{e_i t}(v_i);$

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 is $\varphi$ -invariant. Assume $\mu_{\varphi|_W}$ has a monic prime factor 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 and are coprime, there exist polynomials $h, h' \in K[x]$ with . Hence,

$0 = h(\varphi) p(\varphi)^t(w) + h'(\varphi) f(\varphi)^s(w) = (h p^t + h' f^s)(\varphi)(w) = w,$

a contradiction. Hence, all prime factors $\mu_{\varphi|_W}$ lie in $\{ p_1, \dots, p_n \}$ . Therefore,

$W = \bigoplus_{i=1}^n \GEig(\varphi|_W, p_i) \subseteq \bigoplus_{i=1}^n \GEig(\varphi, p_i) \subseteq W,$

which shows the claim.

Corollary.

Assume $\varphi$ has a minimal polynomial, and let be a prime polynomial. Let be a field extension of over which 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

$\GEig(\varphi, f) \otimes_K L = \bigoplus_{i=1}^n \GEig(\varphi \otimes_K L, x - \lambda_i).$

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.

Lemma.

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

Proof.

We first show that “ $\le$ ” holds. Let be a splitting field of over , 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 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.

Corollary (Cayley-Hamilton over Fields).

If $\dim_K V < \infty$ and $\varphi \in \End_K(V)$ , then $c_\varphi(\varphi) = 0$ .

Felix' Math Place

Focussed on, but not limited to Computational Number Theory

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

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Felix' Math Place

Focussed on, but not limited to Computational Number Theory

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

Comments.

About Me.

About This Blog.

Overview Pages.

Categories.

Recent Posts.

Archives.

Tags.

Impressum.

Data Protection.