On proof that for $U$-invariant subspace of diagonalisable map $A$, the restriction $A_{|U}$ is diagonalisable too A linear map $A \in \mbox{hom}(V,V)$ is called diagonalisable iff 
$$
 V = \oplus_{i=1}^m \mbox{Eig}(a_i)
$$
i.e. $V$ is a direct sum of eigenspaces of $A$. I have some questions on the following proof.

Let $\mathcal V$ be a finite dimensional vector space and $A \in \mbox{hom}(\mathcal V,\mathcal V)$, $A$ diagonalisable. Let $U \le \mathcal V$ be a subspace such that $AU \subseteq U$, then $A_{|U}$ is diagonalisable too.

Proof: Because $A$ is diagonalisable, we have
$$
 \mathcal V = \bigoplus_{i=1}^m \mbox{Eig}(a_i) \quad \mbox{ with } a_i \ne a_j \mbox{ for } i \ne j.
$$
Set $U_1 = \sum_{i=1}^m (U \cap \mathcal V(a_i))$. It is enough to show $U = U_1$. Let $u \in U \setminus U_1$. We have $u = \sum_{i=1}^m v_i$ with $v_i \in \mbox{Eig}(a_i)$. Choose $u$ such that in its representation the number of $v_i$ which are zero is maximal. Let $\mathcal v_k \ne 0$. We have
$$
 Au = \sum_{i=1}^m A v_i = \sum_{i=1}^m a_i v_i \in U
$$
and
$$
 a_k \sum_{i=1}^m v_i \in U.
$$
Therefore
$$
 \sum_{i=1}^m a_i v_i - a_k \sum_{i=1}^m v_i \in U
$$
and so
$$
 \sum_{i=1,~ i \ne k}^m (a_i - a_k) v_i \in U.
$$
By the choice of $u$ we have
$$
  \sum_{i=1,~ i \ne k}^m (a_i - a_k) v_i \in U_1.
$$
So that
$$
  \sum_{i=1,~ i \ne k}^m (a_i - a_k) v_i = \sum_{j=1}^m u_j 
$$
with $u_j \in U \cap \mbox{Eig}(a_j)$. Comparing coefficients yields $u_i = (a_i - a_k) v_i \in U_1$ for all $i \ne k$.
By $a_i - a_k \ne 0$ we have $v_i \in U_1$ for all $i \ne k$. Therefore
$$
 v_k = u - \sum_{i\ne k} v_i \in U \cap \mbox{Eig}(a_k) \subseteq U_1.
$$
And so $u \in U_1$, a contradiction. $\square$
I have some questions on this proof, the part where it goes

By the choice of $u$ we have
  $$
  \sum_{i=1,~ i \ne k}^m (a_i - a_k) v_i \in U_1.
$$

is not clear to me, why does the choice of $u$ implies the containment in $U_1$?
Also why:

By $a_i - a_k \ne 0$ we have $v_i \in U_1$ for all $i \ne k$. Therefore
  $$
 v_k = u - \sum_{i\ne k} v_i \in U \cap \mbox{Eig}(a_k) \subseteq U_1.
$$

Why is $u - \sum_{i\ne k} v_i \in U \cap \mbox{Eig}(a_k)$?
And one question left, here

Comparing coefficients yields $u_i = (a_i - a_k) v_i \in U_1$ for all $i \ne k$.

Does this comparison of the coefficients also yields $u_k = 0$?
Thanks!!
 A: First question: 


By the choice of $u$ we have
    $$
  \sum_{i=1,~ i \ne k}^m (a_i - a_k) v_i \in U_1.
$$

is not clear to me, why does the choice of $u$ implies the containment in $U_1$?

The choice in the proof goes this: First, assume $U\setminus U_1\ne \{0\}$. Now one chooses an element $u$ in $U\setminus U_1$ that is a linear combination of a minimal number of eigenvectors. Every vector in $\tilde u\in U$ that is the linear combiination of less eigenvectors than $u$ must be in $U_1$.
This argument is used here: $\sum_{i\ne k}(a_i-a_k)v_i$ has one more zero vector in the linear combination than $u$, hence it must belong to $U_1$.


By $a_i - a_k \ne 0$ we have $v_i \in U_1$ for all $i \ne k$. Therefore
    $$
 v_k = u - \sum_{i\ne k} v_i \in U \cap \mbox{Eig}(a_k) \subseteq U_1.
$$

Why is $u - \sum_{i\ne k} v_i \in U \cap \mbox{Eig}(a_k)$?

This sum expresses $v_k$ as a difference of elements in $U$. Moreover, $v_k \in \text{Eig}(a_k)$. Hence $v_k\in U_1$ by the construction of $U_1$.


Comparing coefficients yields $u_i = (a_i - a_k) v_i \in U_1$ for all $i \ne k$.

Does this comparison of the coefficients also yields $u_k = 0$?

Yes. However, this is irrelevant for the proof. The coefficient comparison was made to conclude $v_i\in U_1$ for $i\ne k$.
