This problem lets $V$ be a finite-dimensional vector space over $\cal{C}$ (complex field), and lets $T \in \cal{L}$$(V)$. Suppose we have an invariant subspace of $T$ called $U$, and we want to show that $$ U = \bigoplus_{i=1}^r (U \cap G(\lambda_i, T)) $$ where each $\lambda_i$ are distinct eigenvalues of $T$ and $G(\lambda_i, T)$ is the generalized eigenspace for each $\lambda_i$.
I've looked up similar questions and gathered that Jordan form could be relevant, but I have not covered Jordan form yet, and this problem is meant to be approached without it.
My attempt at a solution
Since $V$ is finite-dimensional and the field is over complex numbers, we can write $$G(\lambda_1,T) \oplus \dots \oplus G(\lambda_r, T) = V$$ I was hinted that we should try to decompose $U$ into a direct sum of generalized eigenspaces, or apply the same theorem to $U$ the same way we did to $V$, so since $U$ is an invariant subspace of $T$, we can also define a restriction operator $T_U$ where $T_U \in \cal{L}$$(U)$, $T_U(u) = Tu$ for $u \in U$, and so we can decompose $U$ into $$U = G(\lambda_i, T_U) \oplus \dots \oplus G(\lambda_m, T_U)$$ where $\lambda_i$ through $\lambda_m$ are eigenvalues of $T_U$. Thus, each element $u \in U$ is $u = u_i \oplus \dots \oplus u_m$ where each $u_j \in G(\lambda_j, T_U)$ and $u_j \in U$ by restriction of $T_U$.
Now, we have to show that each $u_j \in G(\lambda_j,T)$ to complete the proof. We know that $u_j \in G(\lambda_j, T_U)$, or by definition: for some $m_j$ we have $$(T_U - \lambda_jI)^{m_j}u_j = 0 = T_U^{m_j}u_j + ... + (\text{cross terms of the form } \lambda_j^xT_U^yu_j \text{ for some x and y}) + ... + \lambda_j^{m_j}u_j$$ and since $T_U(u) = Tu$ for $u \in U$ we can rewrite the above as $$T^{m_j}u_j + ... + (\text{cross terms of the form } \lambda_j^xT^yu_j \text{ for some x and y}) + ... + \lambda_j^{m_j}u_j$$ which I assert is equal to $$(T - \lambda_jI)^{m_j}u_j$$ Now this is where I'm a bit shaky: I claim that if $\lambda_j$ is an eigenvalue of $T_U$, then we know that $\lambda_j$ is also an eigenvalue of $T$. I also claim that from the above, $u_j \in null(T-\lambda_jI)^N$ for some $N$, which is the definition of $u_j \in G(\lambda_j, T)$ which is what we set out to prove.
Now, I don't know if this is necessary or true, but I also claim that $m_j \leq n_j$ where $m_j$ is the exponent associated with the generalized eigenspace $G(\lambda_j, T_U)$ and $n_j$ is the exponent associated with the generalized eigenspace $G(\lambda_j, T)$. I tried to say something along the lines of if $u_j \in G(\lambda_j, T_U)$ then $u_j \in G(\lambda_j, T)$ but not the other way around (just by the definition of generalized eigenspaces and the restriction operator). Again, not sure if this true nor if it is necessary, and I'm not sure if there are repercussions should $m_j > n_j$.
So now we see that each $u \in U$ is $u = u_1 \oplus \dots \oplus u_m$ where each $u_j \in U \cap G(\lambda_j, T)$ for $i = 1, \dots, m$ and also $u = u_1 \oplus \dots \oplus u_m \oplus u_{m+1} \oplus \dots \oplus u_r$ where each $u_{m+1} = \dots = u_r = 0$ (this came from the fact that the number of eigenvalues of $T_U$ are less than or equal to the number of eigenvalues of $T$). Thus, we have shown $$ U = \bigoplus_{i=1}^r (U \cap G(\lambda_i, T)) $$
My questions:
Is this a valid proof of the problem? If not, is this on the right track (what would be the correct approach)? Which parts of my logic are unneccessary and/or untrue? Are there parts that need to be better formalized?