Show that $V = G(\lambda_1,T)\oplus ... \oplus G(\lambda_m,T)$ where $G(\lambda,T)$ is the generalized eigenspace for $\lambda$ 
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!
 A: 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 $F_k = G_k$.
