# Primary Decomposition Theorem - Questions

Let $T$ be a linear operator on a finite dimensional vector space $V$ with characteristic polynomial $f(x)=(x-c_1)^{d_1} \cdots (x-c_k)^{d_k}$ and minimal polynomial is $p(x)=(x- c_1)^{r_1} \cdots (x-c_k)^{r_k}$.

If $W _i = \ker ( T -c_i)^{r_i}$ then, I have to show that the dimension of $W_i$ is $d_i$.

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I intended 1, 2, ...k as subscripts of the power r and d. – preeti Apr 19 '12 at 14:47
You enclose them in such cases by {}. Please see my edit... – user21436 Apr 19 '12 at 14:51

This is true even if we replace $x-c_i$ with an arbitrary irreducible factor, though in that case we must multiply $d_i$ by the degree of the factor. But I'll keep it to the linear case:

Let $K_{i} = \{v\in V\mid (T-c_iI)^m(v)=0\text{ for some }m\geq 0\}$. Then $W_i\subseteq K_i$.

In fact, $K_i=W_i$. The inclusion $W_i\subseteq K_i$ is immediate. For the converse, restrict $T$ to $K_i$. The only eigenvalue is $c_i$, so the characteristic polynomial of $T\bigm|_{K_i}$ is $(x-c_iI)^p$. But the characteristic polynomial of $T\bigm|_{K_i}$ divides the characteristic polynomial of $T$, so $p\leq d_i$. Since $f(T)$ is zero on $V$, it is zero on $K_i$, so the minimal polynomial of $T|_{K_i}$ divides both $(x-c_i)^{d_i}$ and $f(x)$. Therefore, the minimal polynomial of $T|_{K_i}$ is $(x-c_i)^{r_i}$, so $(T-c_iI)^{r_i}$ is identically zero on $K_i$. That is, $K_i\subseteq\mathrm{ker}(T-c_iI)^{r_i}=W_i \subseteq K_i$.

Thus, $K_i=W_i$. Note also that the $p$ in the above argument is the dimension of $K_i$, so $\dim(W_i) = \dim(K_i) = p \leq d_i$.

Lemma. If $c\neq c_i$, then $T-cI$ is one-to-one on $K_i$.

Proof. Let $x\in K_i$ be such that $(T-cI)x=0$; let $p$ be the smallest nonnegative integer such that $(T-c_iI)^p = 0$. If $p\gt 0$, let $y=(T-c_i)^{p-1}x$. Then $y\neq 0$, $(T-c_iI)y = 0$, so $y$ is an eigenvector of $c_i$. But $(T-cI)y =(T-cI)(T-c_iI)^{p-1}x = (T-c_iI)^{p-1}(T-cI)x = (T-c_iI)^{p-1}0 = 0$, so $y$ is also an eigenvector of $c\neq c_i$, which is impossible. So $p=0$, hence $x=0$. Thus, $T-cI$ is one-to-one on $K_i$. $\Box$

Note also that $\dim(K_i) = That means that $$K_i\bigcap \sum_{j\neq i}K_j = \{0\}.$$ (If something is in the intersection, then it lies in some$K_j$with$j\neq i$; since$T-c_iI$is one-to-one on every$K_j$,$j\neq i$, then$(T-c_iI)^mx=0$implies$x=0$). Thus,$\dim(W_1)+\cdots+\dim(W_k) \leq d_1+d_2+\cdots+d_k = \dim V$. So it suffices to show that$W_1+\cdots+W_k = V$. Induction on$k$. If$k=1$, then the characteristic polynomial of$V$is$(x-c_1)^d_1$, and by the Cayley-Hamilton Theorem we have$V=K_1=W_1$, and we are done. Assume the result holds for linear transformations with fewer than$k\gt 1$eigenvalues. Let$g(t) = (x-c_1)^{d_1}\cdots(x-c_{k-1})^{d_{k-1}}$. If$W=\mathrm{Range}(T-c_kI)^{d_k}$. Since$T-c_kI$maps$K_i$into itself if$i\neq k$, is one-to-one, and so onto. So$K_i\subseteq W$for all$i\neq k$. So$c_i$is an eigenvalue of$T|_W$for all$i\lt k$. On the other hand,$c_k$is not an eigenvalue of$T|_W$, because$g(T)$is the zero linear transformation on$W$by the Cayley-Hamilton Theorem; so the minimal polynomial of$T|_W$divides$g(t)$, and so the only possible eigenvalues are$c_1,\ldots,c_{k-1}$. So$T|_W$has fewer than$k$eigenvalues. By induction, we have$W=K_1+\cdots+K_{k-1}$. On the other hand,$\mathrm{ker}(T-c_kI)^{d_k}\cap W=\{0\}$, so by the Rank-Nullity Theorem we have that $$V=\mathrm{Range}(T-c_kI)^{d_k}+\mathrm{ker}(T-c_kI)^{d_k} = K_1+\cdots+K_{k-1} + K_k.$$ Therefore, $$n = r_1+\cdots+r_k \geq \dim(K_1)+\cdots+\dim(K_n)\geq \dim(K_1+\cdots+K_n) = \dim(V) = n,$$ so we have equality throughout. In particular,$\dim(K_i)=r_i$. - For any$i$write$p_i(x)=(x-c_i)^{r_i}$and$p(x)=q_i(x)p_i(x)$for the minimal polynomial, so that$W_i=\ker(p_i(T))$, and$q_i(x)$groups all factors of$p(x)$without$c_i$as root. Then one has$0=p(T)=q_i(T)\circ p_i(T)$, and it follows easily that$q_i(x)$is the minimal polynomial of$T$restricted to the image of$p_i(T)$, which image I shall call$Q_i$: indeed$q_i(T|_{Q_i})=0$and a nonzero polynomial of lower degree than$q_i$that annihilates$T|_{Q_i}$would contradict the minimality of$p(x)$. As a consequence the characteristic polynomial of this restriction$T|_{Q_i}$does not have$c_i$as root. Now$V=W_i\oplus Q_i$, since$\dim W_i+\dim Q_i=\dim V$by the rank-nullity theorem, and one has$W_i\cap Q_i=\{0\}$, since this is a$T$-stable subspace for which the eigenvalues of the restriction of$T$to it can neither be distinct from$c_i$nor equal to$c_i$. Therefore$f(x)$is the product of the characteristic polynomials of the restrictions of$T$to$W_i$and to$Q_i$. The former polynomial can contain only irredicible factors$x-c_i$(since a power of$T-c_iI$annihilates$W_i$), while the latter polynomial factor can contain no such factors, so the characteristic polynomial of the restriction of$T$to$W_i$contains exacly all such factors of$f(x)$: it equals$(x-c_i)^{d_i}$. Then$d_i=\dim W_i$, QED. The essential point in the above argument is that$Q_i$contains no eigenvectors for$c_i$. The direct sum decomposition is useful to know (it shows that the generalised eigenspace$W_i$is a$T$-stable direct summand of$V$), but is not crucial to the argument: it can be shown that for any linear operator$S$commuting with$T$(in particular for any polynomial in$T$), the characteristic polynomial$f(x)$decomposes as a product of the characteristic polynomials of the restrictions of$T$to the kernel and to the image of$S$, whether or not these two spaces form a direct sum (by an isomorphism theorem the image of$S$is isomorphic to the quotient of$V$by the kernel of$S$, via an isomorphism compatible with the action of$T\$).

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