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Suppose $V$ is a finite-dimensional complex vector space and $T:V\to V$ is a linear operator on $V$. Let $\lambda_1,...\lambda_m$ be the distinct eigenvalues of $T$. Then, show that $V = G(\lambda_1,T)\oplus ... \oplus G(\lambda_m,T)$, where $G(\lambda,T)$ is the generalized eigenspace corresponding to $\lambda$.

We know that $G(\lambda, T) = \ker (T-\lambda I)^{\dim V}$ (let $\dim V$ = n for simplicity). Also, we know that generalized eigenvectors corresponding to the eigenvalues $\lambda_1,...,\lambda_m$ are linearly independent. Using this, I was able to show that $G(\lambda_i,T) \bigcap G(\lambda_j,T) = \{0\}$ for $i\neq j$.

Now, it would suffice to show that $\sum_{k=1}^m \dim G(\lambda_k,T) = n$, right? Alternatively, we could also just show that if $0 = u_1 + u_2 + ... + u_m$ for $u_k \in G(\lambda_k,T)$, then $u_k = 0$ for all $k=1,2,...,m$. I'm not sure how to proceed, and would appreciate any hints.

In addition, I was able to show that each $G(\lambda_k,T)$ is $T$-invariant, and $(T-\lambda_k I)\vert_{G(\lambda_k,T)}$ is a nilpotent operator. I don't know if these will help.

P.S. I came across an inductive proof, that does induction on the dimension of $V$ - but I'm really looking for something direct and non-inductive as above, if possible. Thanks!

stoic-santiago
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  • Are you assuming that a Jordan decomposition exists or are you trying to prove it using this claim? – NL1992 Nov 01 '20 at 08:08
  • @strawberry-sunshine If $A$ is a nonzero principal ideal domain and $M$ a torsion module, then it ($M$) decomposes as the direct sum of its primary components, which constitute a family indexed by a complete and independent system $\Pi \subseteq A$ of representatives of the set of all irreducible elements of $A$ w. rsp. to association in divisibility such that the component $M_{(\pi)}$ at index $\pi \in \Pi$ is equal to $0_M \colon {\pi^n}_{n \in \mathbb{N}}$, in other words the submodule consisting of all $x$ annihilated by a power of $\pi$. (to be cont.) – ΑΘΩ Nov 01 '20 at 08:17
  • @strawberry-sunshine (cont.) In particular, if $V$ is a vector space over a commutative field $K$ equipped with a certain endomorphism $\varphi \in \mathrm{End}_{\operatorname{\mathit{K}-\mathbf{Mod}}}(V)$, then $V$ can be naturally organised as $K[X]$-module where $f.x=f(\varphi)(x)$ for each $f \in K[X]$ and $x \in V$. If $V$ is finite dimensional, then by virtue of the Cayley-Hamilton theorem $V$ is torsion over $K[X]$, since the characteristic polynomial of $\varphi$ is a nonzero element of the annihilator. Hence, the above general decomposition applies with the mention (to be cont.) – ΑΘΩ Nov 01 '20 at 08:22
  • @strawberry-sunshine (cont.) mention that the chosen system of representatives for irreducible polynomials in $K[X]$ will consist of the monic ones and the only nonzero primary components of $V$ -- as a $K[X]$ module, of course -- are those corresponding to irreducible factors of the char. polyn. of $\varphi$. In the particular case that this characteristic polynomial splits completely over $K$, these factors are of the form $(X-\lambda)^k$ and they primary components associated to them are precisely the generalised eigenspaces associated to the eigenvalues $\lambda \in K$. – ΑΘΩ Nov 01 '20 at 08:25
  • @strawberry-sunshine Errata to a mistake I made in the first comment: the correct formal expression of the primary component of index $\pi$ is $M_{(\pi)}=\displaystyle\bigcup_{n \in \mathbb{N}}\left(0_M \colon \pi^n\right)$, where in general for $X \subseteq M$ and $\Lambda \subseteq A$ one defines the subset quotient as $X \colon \Lambda={x \in M \mid (\forall \lambda)(\lambda \in \Lambda \Rightarrow \lambda x \in X)}$. – ΑΘΩ Nov 01 '20 at 08:33
  • @NL1992 We don't know about Jordan decompositions yet. – stoic-santiago Nov 01 '20 at 08:37
  • @ΑΘΩ Your comments are probably right, but beyond me. Could you answer this as if you're answering it for an undergraduate taking a course in linear algebra? – stoic-santiago Nov 01 '20 at 08:38
  • @strawberry-sunshine You have an answer such that you desire below. – ΑΘΩ Nov 01 '20 at 09:07
  • "Now, it would suffice to show that $\sum_{k=1}^m \dim G(\lambda_k,T) = n$, right?" $S:= \ker\Big(\big(\lambda_1 I-T\big)^{k_1}\Big) \oplus \ker\Big(\big(\lambda_2 I-T\big)^{k_2}\Big)\oplus... \oplus\ker\Big(\big(\lambda_m I-T\big)^{k_m}\Big)$
    where the direct sum is justified because you've proven the intersection between these subspaces to be trivial. Then $ \dim(V)\leq \dim(S)\leq \dim(V)$ by application of Sylvester's Rank Inequality https://math.stackexchange.com/questions/3878092/prove-that-the-intersection-of-2-generalised-eigenspaces-is-the-zero-space/3878536#3878536     – user8675309 Nov 02 '20 at 06:34

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Let me denote the generalized eigenspace for $\lambda_k$ simply as $G_k$. You've shown that distinct generalized eigenspaces have trivial intersection, so to show that $V$ decomposes as a direct sum, it suffices to show that the sum of the generalized eignespaces span $V$, i.e., that every $\mathbf{v} \in V$ can be written as $$\mathbf{v} = \sum_{k=1}^m\mathbf{v}_k,\tag{1}$$ where $\mathbf{v}_k \in G_k$. One way of doing this is to use Bezout's identity.

Let $p(x)$ be the characteristic polynomial, factorized as $$p(x) = \prod_{k=1}^m (x-\lambda_k)^{a_k},$$ where $a_k$ denotes the algebraic multiplicities of each eigenvalue. For each $1 \le k \le m$, let $p_k$ denote the polynomial obtained from $p$ by dividing out the factor associated with $\lambda_k$, i.e., $$p_k(x) = \frac{p(x)}{(x-\lambda_k)^{a_k}}.$$ Let $F_k = \ker\left[(T-\lambda_k)^{a_k}\right]$. Then $F_k \subseteq G_k$. It will turn out that $F_k = G_k$, but we haven't shown this yet. We will obtain this fact automatically by showing that the $\mathbf{v}_k$ in equation $(1)$ can be taken to be elements of $F_k$.

Now, note that the collection of polynomials $\{p_k\}_{k=1}^m$ are collectively coprime. Therefore by Bezout's identity, there exists polynomials $\{f_k\}_{k=1}^m$ such that $$\sum_{k=1}^mf_kp_k = 1.$$ Evaluating at $T$, we get the operator equation $$\sum_{k=1}^mf_k(T)p_k(T) = I.$$ Now, let $\mathbf{v} \in V$ be arbitrary. Acting with the equation above on $\mathbf{v}$, we get $$\sum_{k=1}^m f_k(T)p_K(T)\mathbf{v} = \mathbf{v}.$$ Let $\mathbf{v}_k = f_k(T)p_K(T)\mathbf{v}$. I claim that $\mathbf{v}_k \in F_k$. Indeed, we have $$(T-\lambda_k)^{a_k}\mathbf{v}_k = f_k(T)p_k(T)(T-\lambda_k)^{a_k}\mathbf{v} = f_k(T)p(T)\mathbf{v} = \mathbf{0},$$ where the last equality follows from the Cayley-Hamilton theorem since $p(T) = 0$. Since $\mathbf{v}$ was arbitrary, it follows that $$F_1 + \cdots + F_m = V.$$ Moreover, since you've already shown that $G_k$ are independent, it follows that the $F_k$ are also independent, and we can upgrade the sum above to a direct sum $$F_1 \oplus \cdots \oplus F_m = V.$$ Finally, this also allows us to conclude that we must have $F_k = G_k$.

EuYu
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