A family of commuting endomorphisms is semisimple if each element is semisimple If $\phi : V \rightarrow V$ is an endomorphism of a finite-dimensional (say real) vector space, $\phi$ is called "semisimple" if any $\phi$-invariant subspace of $V$ has a complimentary $\phi$-invariant subspace ($W \subset V$ is $\phi$-invariant if $\phi (W) \subset W$). We say that a collection $\Phi= \{ \phi_i\}_{i \in I}$ of such endomorphisms is semisimple if any subspace which is invariant under every element of $\Phi$ has a complimentary subspace which is invariant under every element of $\Phi$. 
How can I prove that
"If $\Phi$ is a commuting family of endomorphisms then $\Phi$ is semisimple if each $\phi_i$ is semisimple"?
Thank you for your help.
 A: Short answer : this is because “semisimple” equals “diagonalizable over $\mathbb C$”.
The details : 
Working on $\mathbb C$ instead of $\mathbb R$, our initial space
$V$ become another vector space which we denote by $T(V)$, 
(others denote it by $V \otimes {\mathbb C}$) and we have
a $\mathbb R$-linear bijection $T : V\to T(V)$. Similarly, any 
$\phi : V \to V$ induces a map $T(\phi) : T(V) \to T(V)$
defined in the obvious way (if $B$ is a $\mathbb R$-basis of $V$,
then the matrix of $T(\phi)$ relatively to $T(B)$ is the same as
the matrix of $\phi$ relatively to $B$). Also, let us
put $T\Phi=\lbrace T\phi | \phi \in \Phi \rbrace$, for any
$\Phi \subseteq {\sf End}(V)$.
Lemma 1 (Engel’s Lemma) Let $\Phi$ be a commuting family in
${\sf End}(V)$. Then there is a nonzero vector $v\in T(V)$ which
is an eigenvector of every $\phi\in T(\Phi)$.
Proof of lemma 1 We argue by induction on $d={\sf dim}(V)$. If
$d\leq 1$ the result is obvious. So assume that $d\geq 2$ and that the
result holds for any $d'<d$. Let $\Phi$ be a commuting family in
${\sf End}(V)$. If $\Phi$ consists only of homotheties (i.e. multiples
of the identity), then any $v$ will do the job. Otherwise there is
a $\phi\in V$ which is not a homothety. By the fundamental theorem of
algebra, the characteristic polynomial $\chi_{\phi}$of $\phi$ can be factorized
as $\chi_{\phi}(x)=\displaystyle\prod_{k=1}^r (x-\lambda_k)^{m_k}$ where
the $\lambda_k$ are distinct complex numbers and the $m_k$ are integers
summing to $n$. Since $\phi$ is not a homothety, we must have $r>1$. Consider 
the characteristic susbpace $W={\sf Ker}((T(\phi)-\lambda_1{\sf id})^{m_1})$. Since any 
$\psi \in T(\Phi)$ commutes with $\phi$ and hence with $(T(\phi)-\lambda_1{\sf id})^{m_1}$ also, we see that $W$ is invariant by every $\psi \in T(\Phi)$. Since
${\sf dim}(W)<{\sf dim}(V)$, by the induction hypothesis there is a
$w\in W$ which is an eigenvector of every $\phi\in T(\Phi)$, and this finishes 
the proof.
Lemma 2(semisimple equals diagonalizable) Let $\phi \in {\sf End}(V)$. Then
$\phi$ is semisimple iff $T(\phi)$ is diagonalizable, i.e. iff there is a basis
$B'$ of $V$ such that the matrix of $T(\phi)$ relatively to $B'$
is diagonal.
Proof of lemma 2 Let $\phi \in {\sf End}(V)$, and  let $\chi_{\phi}$ be the 
characteristic polynomial of $\phi$. By the fundamental theorem of
algebra, the characteristic polynomial $\chi_{\phi}$of $\phi$ can be factorized
as $\chi_{\phi}(x)=\displaystyle\prod_{k=1}^r (x-\lambda_k)^{m_k}$ where
the $\lambda_k$ are distinct complex numbers and the $m_k$ are integers
summing to $n$. Let $C_j={\sf Ker}((T(\phi)-\lambda_j{\sf id})^{m_j})$ be
the $j$-th characteristic space. Then $T(V)$ is the direct sum of the $C_j$, and in fact any $\phi$-invariant subspace $W$ of $T(V)$ will be the direct sum of the  $W\cap W_j$. 
Suppose $\phi$ is semisimple. The eigenspace 
 $E_j={\sf Ker}((T(\phi)-\lambda_j{\sf id}))$ is a subspace of $C_j$, and is 
 invariant by $\phi$. By semisimplicity, there is a $\phi$-invariant $F_j$ with
 $T(V)=E_j\oplus F_j$. Now $F_j$ is the direct sum of the $F_j\cap C_k$ as noted 
 above, and hence $C_j=E_j \oplus (F_j \cap C_j)$. By definition of $E_j$, the 
 linear map $T(\phi)-\lambda_j{\sf id}$ is injective on $F_j \cap C_j$, but at 
 the same time $(T(\phi)-\lambda_j{\sf id})^{m_j}$ is zero on $F_j \cap C_j$. This 
 imples $F_j\cap C_j=\lbrace 0 \rbrace$, so $C_j=E_j$. Then $T(V)$ is the direct sum
 of the $E_j$ and $T(\phi)$ is diagonalizable.
Conversely, suppose $T(\phi)$ is diagonalizable. The eigenvalues of $T(\phi)$
 are either real (call those $\rho_1,\rho_2,\ldots ,\rho_s$), or they come
 in conjugate pairs (call those $\alpha_1\pm \beta_1i, \ldots, \alpha_s\pm \beta_si$). If we define in $V$ the characteristic spaces
 $C_k={\sf Ker}(\phi-\rho_k {\sf id})$ and $D_k={\sf Ker}((\phi-\alpha_k{\sf id})^2+
 (\beta_k)^2{\sf id})$,  any $\phi$-invariant subspace of $C_k$ or $D_k$
 has an invariant complement in $C_k$ or $D_k$. So $\phi$ is semisimple.
Lemma 3(semisimple equals diagonalizable again) Let $\Phi \subseteq {\sf End}(V)$. 
We say that   $T(\Phi)$ is diagonalizable when  there is a basis
$B'$ of $V$ such that the matrix of $T(\phi)$ relatively to $B'$
is diagonal, for every $\phi\in \Phi$.
If $T(\Phi)$ is diagonalizable then $\Phi$ is semisimple.
The converse holds for a commuting family : 
If $\Phi$ is commuting and semisimple, then  $T(\Phi)$ is diagonalizable.
Proof of lemma 3  If $T(\Phi)$ is 
diagonalizable, then we may (as in the final part of the proof
of lemma 2) decompose $T(V)$ as a sum of characteristic spaces,
in which every invariant subspace has an “internal” invariant complement.
So $\Phi$ is semisimple.
Conversely, suppose that $\Phi$ is semisimple and commuting. By Engel's lemma, there
is a vector $v_1$ which is an eigenvector for every $\phi\in V$. Since $\Phi$
is semisimple, there is a $\Phi$-invariant subspace $G$ with
$V={\sf span}(v_1)\oplus G$. By Engel's lemma, there
is a vector $v_2\in G$ which is an eigenvector for every $\phi\in V$. Iterating this
procedure (which must terminate since $V$ has finite dimension), we eventually
obtain a basis in which the matrix of every $\phi\in \Phi$ is diagonal as wished.
Now we can proceed to the proof of the properties you're asking.
Theorem 1. Let $\Phi$ be a commuting family in ${\sf End}(V)$. If every
$\phi \in \Phi$ is semisimple, then $\Phi$ is semisimple.
Proof of theorem 1. By Engel's lemma, there
is a vector $v_1$ which is an eigenvector for every $\phi\in V$. Since $\Phi$
is semisimple, there is a $\Phi$-invariant subspace $G$ with
$V={\sf span}(v_1)\oplus G$. By Engel's lemma, there
is a vector $v_2\in G$ which is an eigenvector for every $\phi\in V$. Iterating this
procedure (which must terminate since $V$ has finite dimension), we eventually
obtain a basis in which the matrix of every $\phi\in \Phi$ is diagonal.
So $T\Phi$ is diagonalizable, and we are now done by lemma 3.
Theorem 2. Let $\Phi$ be a commuting family in ${\sf End}(V)$. If 
$\Phi$ is semisimple, then every $\phi \in \Phi$ is semisimple.  
Proof of theorem 2. If $\Phi$ is semisimple, then by lemma 3 $T\Phi$
is diagonalizable. So  $T\phi$ is diagonalizable for every $\phi\in\Phi$,
which means that every $\phi\in\Phi$ is semisimple by lemma 2.
