Proving that compositions of linear operators are equal. Let $T_1$ and $T_2$ be linear operators on $\mathbb{R}^n$. Let $B = \{v_1, v_2, \dots, v_n\}$ be a basis for $\mathbb{R}^n$.
Suppose that each vector in $$ is an eigenvector of both $T_1$ and $T_2$.
Prove that $T_1 \circ  T_2 = T_2 \circ T_1$.
What I have tried:
Let $v$ be an arbitrary vector from $B$.
Then,
$T_1(v) = λ_1\cdot v$
$T_2(v) = λ_2\cdot v$
Hence,
$$T_1 \circ T_2 = T_1(T_2(v)) = T_1(λ_2\cdot v) = λ_1\cdot λ_2 \cdot(v)$$
$$T_2 \circ T_1 = T_2(T_1(v)) = T_2(λ_1\cdot v) = λ_2\cdot λ_1\cdot (v)$$
Hence, equal.
Is it sufficient proof?
Thank you in advance!
 A: Your proof is almost correct, but not quite. To figure out why, you should figure out what each step of your argument means beyond the equation (maybe it helps to explain in words), and then you can see the gaps in your reasoning. Let me rephrase your argument:

Let $v$ be an arbitrary vector in the basis $B$. Then there exist real numbers $\lambda_1,\lambda_2$ such that $T_1(v)=\lambda_1 v$ and $T_2(v)=\lambda_2 v$ by hypothesis that all vectors of $B$ are eigenvectors of $T_1$ and $T_2$. Thus, $T_1T_2(v)=T_2T_1(v)$.

But note that this only proves $T_1T_2=T_2T_1$ for all vectors $v$ in the basis $B$, but not for any of the vectors outside the basis! This is because of how you defined $v$ as a vector of the basis $B$, not an arbitrary vector of $\mathbb R^n$. If you forget how you defined your notation, they will come back to bite you, especially since $v$ is normally used to denote a general vector and it is better notation to write $v_i$ for a particular basis vector.
Note that your argument does not immediately work if you just replace "arbitrary vector in the basis $B$" by "arbitrary vector in $\mathbb R^n$". This is because you do not know $T_1(v)=\lambda_1 v$ a priori for arbitrary vectors $v$, since you are only given that vectors of $B$ are eigenvectors of $T_1,T_2$ and not for all vectors of $\mathbb R^n$.

You have shown that $T_1T_2$ and $T_2T_1$ are the same on the basis $B$. The last step: prove that if two linear operators are the same on a basis, then they are the same on the whole vector space. Can you do this?
A: You took an arbitrary $v \in \mathbb{R}^n$. You can write it as
$$
v = \alpha_1 v_1 + \cdots + \alpha_n v_n,
$$
for some $\alpha_i \in \mathbb{R}$, $i \in \{1,\dots,n\}$. That's the part that was missing on your proof, because the basis $B = \{v_1,\dots,v_n\}$ has the eigenvectors of both operators. Also, for more generality, assume that
$$
T_1 v_i = \lambda_i v_i
$$
and the same for $T_2$, but with $\lambda'_i$, for all $i \in \{1,\dots,n\}$.
