Is the fundamental representation of $SU(3)$ irreducible? I want to check if the fundamental representation of $SU(3)$ is irreducible.
The algebra is
$$\mathbb{su}(3) = \{ m \in Mat(3,\mathbb{C} )\ |\ m = -m^+,\ Tr[m] = 0 \}$$
and I've found the generators.
So, I've checked if the algebra is Abelian.
For example, taken the generators $t_1,t_2$ defined as
$$
t_1=
\left( \begin{matrix}
0& 1 & 0\\
-1 & 0 & 0 \\
0 & 0 & 0
\end{matrix} \right)\qquad
t_2=
\left( \begin{matrix}
0& 0 & 1\\
0 & 0 & 0 \\
-1 & 0 & 0
\end{matrix} \right)
$$
the commutator isn't equal to 0, so the algebra isn't Abelian.
I can restore the group elements through exponential map and taken $G,H \in SU(3)$
($G,H$ are MATRICES) $$
G = e^{\ a t_1}\\
H = e^{\ b t_2}
$$
I see that $[G,H]\neq 0$ because $[t_1,t_2]\neq 0$ and the $G,H$ doesn't admit a common basis of eigenvectors and I can't diagonalize them contemporary.
So in $\mathbb{C}^3$ the fundamental representation of $SU(3)$ is irreducible.
Is it a right proof?
 A: The Lie algebra of a non-Abelian Lie group can't be commutative, or the group itself would have to be Abelian. What you need to check is, e.g., that the only matrices that commute with any other matrix of $SU(3)$ is a multiple of the identity matrix, as required by Schur's lemma.
A: I interpret the question as follows. Let $SU(3)$ be the group of complex $3\times 3$ matrices with $U^\dagger U=I$ and $\det U=1$ acting as linear operators in $\mathbb C^3$.  Is there a subspace $M \subset \mathbb  C^3$ with $M \neq \{0\}$, $M \neq \mathbb C^3$ such that $U(M) \subset M$ for every $U \in SU(3)$?
In other words, is the natural action of $SU(3)$ on $\mathbb C^3$ irreducible?
Your proof is wrong, because the fact that  $SU(3)$ is not Abelian does not imply that it is irreducible.
Actually the irreducibility property is true and it holds for the natural action of $SU(N)$ in $\mathbb C^N$. I sketch a proof in the rest of my answer. Consider $x,y\in \mathbb C^N\setminus \{0\}$. It therefore  holds $y = cx/||x|| + dy'$, where $y' \perp x$ and $||y'||=1$ and $c,d \in \mathbb C$. 
Working in the plane generated by the orthonormal vectors $x/||x||$ and $y'$, it is easy to construct $U \in SU(N)$ such that $U x/||x|| = y/||y||$ using elementary computations (*). 
Therefore $y= s Ux$ for some $s\in \mathbb C$ and $U \in SU(N)$. 
We have established the following lemma.
If $x,y\in \mathbb C^N\setminus \{0\}$, there is $U \in SU(N)$ and $s \in \mathbb C$ such that $sUx= y$.
This fact implies that the action of $SU(N)$ on $\mathbb C^N$ is irreducible. Let us prove it. Suppose that $M \subset \mathbb C^N$ is a subspace invariant under $SU(N)$. If $M\ni x\neq 0$ and $y \in \mathbb C^N$, there is $U\in SU(N)$ such that $sUx =U(sx) =y$ for some constant $s$. As $sx \in M$ and $M$ is invariant, we have $U(sx) = y \in M$. Since $y$ is arbitrary, we finally obtain $M= \mathbb C^N$. This means that $SU(N)$ is irreducible on $\mathbb C^N$.

(*) $U$ can be costructed as follows. If $x/||x||$ and $y'$ are parallel everything is trivial. In the other case, $x/||x||, y', e_3,\ldots, e_N$ is an orthonormal basis of $\mathbb C^N$ for suitable orthogonal unit vectors $e_3,\ldots, e_N$. The unique linear map such that $$U : x/||x|| \mapsto  \frac{cx/||x|| + dy'}{\sqrt{|c|^2+|d|^2}}$$ $$U: y' \mapsto  ax/||x|| + by'$$
$$U: e_j \mapsto e_j $$
is unitary if and only if $$|a|^2+|b|^2 =1\tag{1}$$ and $$a\overline{c}+ b \overline{d}=0\:.\tag{2}$$ Here $d$ and $c$ are fixed and the system of equations (1)-(2) in $a, b \in \mathbb C$ always admit solutions. 
Notice that if $(a,b)$ is a solution, $(\lambda a, \lambda b)$ is such, for $|\lambda|=1$. 
Representing $U$ with respect to the said basis, since $U$ is represented by a unitary matrix, its determinant must be a unit complex. In other words $$|\det U| =|cb -da|/\sqrt{|c|^2+|d|^2}=1$$ so we can always fix the multiplier $\lambda$ in order to have $\det U=1$. However it does not automatically imply that $U \in SU(N)$, because the determinant is computed with respect to the wrong basis. Indeed, the relevant determinant used in the definition of $SU(N)$ is the one computed with respect to the canonical basis of $\mathbb C^N$. It is $${\det}_C U = \det M U M^{-1}$$ where $M$ is the unitary matrix relating the canonical basis of $\mathbb C^N$ to $x/||x||, y', e_3,\ldots, e_N$. Therefore $${\det}_C U = \det M \det U (\det M)^{-1} = \det U=1\:.$$ Summarizing, we have obtained that the found operator $U$, represented with respect to the canonical basis of $\mathbb  C$ is a unitary matrix with determinant $1$. In other words $U$ belongs to $SU(N)$ as wanted.
