Prove Φ is a linear transformation Suppose $\dim U_1 + \dim U_2 + \dots + \dim U_m = \dim V$. Show that there exist an isomorphism $\Phi : V \to W$ such that $\Phi (T(v)) = \Psi (\Phi (v))$ for every $v \in V$.
$T:V \rightarrow V$ is a linear operator on a finite-dimensional vector space V. $U_i$ is the eigenspace of T. $Ψ:W\rightarrow W$ is a linear operator on W. W is the vector space $U_1 ⊕ U_2 ⊕ ... ⊕ U_m$. 
I am solving this question, and I am up to this point where I need to first show 
$Φ : V\rightarrow W$ is a linear transformation. 
To prove vector addition, I set up $Φ(T(v_1+v_2)$, and apply the definition of Φ so that I have $Ψ(Φ(v_1+v_2)$, but then at this point I realize I need to assume it is a linear transformation so I can say $Φ(v_1+v_2)=Φ(v_1)+Φ(v_2)$. After this assumption, I can say the equation $=Ψ(Φ(v_1)+Ψ(Φ(v_2)=Φ(T(v_1))+Φ(T(v_2))$ 
Then how can I prove this is a linear transformation? 
 A: From what I can tell, you aren't even given a defintion of $\Phi: V \rightarrow W$.  I believe what you are supposed to do is define the function yourself, then show it is a linear transformation.
I am guessing $\Psi: W \rightarrow W$ is the linear transformation $(u_1, ... , u_m) \mapsto (\lambda_1 u_1, ... , \lambda_m u_m)$.
In that case, it is easier to define a linear transformation of $W = U_1 \oplus \cdots \oplus U_m$ into $V$, not the other way around.  By a standard result, namely the linear independence of distinct eigenspaces, the natural map $(u_1, ... , u_m) \mapsto u_1 + \cdots + u_m$ is an injective linear transformation $S: W \rightarrow V$.  
The image of $S$ in $V$ is a subspace whose dimension is $\textrm{Dim } W = \sum\limits_{i=1}^m \textrm{Dim } U_i = \textrm{Dim } V$.  Since the image of $S$ is contained in $V$, and they have the same dimension, they must be equal.  Hence $S$ is a linear transformation which is a vector space isomorphism (linear transformation and a bijection).
You can then define $\Phi$ to be the inverse function of $S$.  It is a linear transformation, because $S$ is.
Now if $v$ is in $V$, let $\Phi(v) = (u_1, ... , u_m)$ for some $u_i \in U_i$, so that $v = u_1 + \cdots + u_m$.
Then $$\Phi \circ T(v)= \Phi(\lambda_1u_1 + \cdots + \lambda_m u_m) = (\lambda_1u_1, ... , \lambda_m u_m) = \Psi(u_1, ... , u_m) = \Psi \circ \Phi(v)$$
A: Lemma: Let $T$ be a linear operator on $V$.  Let $u_1, ... , u_t$ be nonzero eigenvectors of distinct eigenvalues $\lambda_1, ... , \lambda_t$ for $T$.  Then $u_1, ... , u_t$ are linearly independent.
Proof: By induction on $t$.  If $t = 1$, this is immediate.  More generally, assume $c_1u_1 + \cdots + c_tu_t = 0$ for scalars $c_i$.  One of these scalars, say $c_1$, is not zero.  Multiplying this equation by the scalar $\lambda_1$, we get 
$$\lambda_1c_1 u_1 + \cdots + \lambda_1 c_t u_t = 0$$
Alternatively, we can apply $T$ to get 
$$\lambda_1c_1u_1 + \cdots + \lambda_t c_tu_t = 0$$
Subtracting these two equations gives
$$(\lambda_2 - \lambda_1)c_2u_2 + \cdots + (\lambda_t - \lambda_t)c_tu_t = 0$$
By induction, $u_2, ... , u_t$ are linearly independent, so all the scalars $(\lambda_i - \lambda_1)c_i$ ($i =2, ... , t$) are zero.  But $\lambda_i - \lambda_1 \neq 0$, so $c_i = 0$.  $\blacksquare$
Proposition: Let $T$ be a linear operator on $V$, with eigenvalues $\lambda_1, ... , \lambda_m$.  Let $U_i$ be the eigenspace of $\lambda_i$, and let $W$ be the (external) direct sum $U_1 \oplus \cdots \oplus U_m$.  Define a function $S: W \rightarrow V$ by 
$$S(u_1, ... , u_m) = u_1 + \cdots + u_m$$
Then $S$ is a linear transformation, and it is injective.
Proof: Let $w = (u_1, ... , u_m), w' = (u_1', ... , u_m')$ be elements of $W$, and let $c$ be a scalar.  Then
$$S(w+cw') = S(u_1 + cu_1', ... , u_m + cu_m')= u_1 + cu_1' + \cdots + u_m + cu_m'$$
$$ = u_1 + \cdots + u_m + cu_1' + \cdots + cu_m' = S(w) + S(cw')$$
so $S$ is a linear transformation.
To show that $S$ is injective, it suffices to show that if $w = (u_1, ... , u_m) \in W$, and $S(w) = 0$, then $w = (0, ... , 0)$.  So by hypothesis,
$$u_1 + \cdots + u_m = 0$$
But nonzero eigenvectors of distinct eigenvalues are linearly independent by the lemma, so we must have $u_1 = \cdots = u_m = 0$.
