How to tell if a vector set is linearly dependent in $\mathbb{C^3}$ over $\mathbb{C}$ and $\mathbb{R}$? 
Given set $A=\{(1+i,1,-2i),(1,1-i,2+i),(2+i,2,-3)\}$, is the set linearly dependent in the context of vector space:
1) $\mathbb{C^3}$ over $\mathbb C$
2) $\mathbb{C^3}$ over $\mathbb R$

As far as I understand I can present $A$ in the matrix form, then arrive to the row echelon form and check whether a row can become a row of zeroes.
So this is the matrix:
$$
\begin{bmatrix}
1+i&1&2+i\\
1&1-i&2\\
-2i&2+i&-3
\end{bmatrix}
$$
After the following elementary operations:
$$
R_1 \leftrightarrow R_2 \\
R_2 \rightarrow R_2 -(1+i)R_1 \\
R_3 \rightarrow R_3+2iR_1 \\
R_2 \rightarrow -R_2 \\
R_3 \rightarrow -(4+3i)R_2
$$
we arrive to the echelon form:
$$
\begin{bmatrix}
1&1-i&2\\
0&1&i\\
0&0&2-2i
\end{bmatrix}
$$
1) I think $A$ cannot be linearly dependent when $\mathbb{C^3}$ over $\mathbb C$ because $2-2i\neq 0$ so we can never get a row of zeroes thus we have 3 leading coefficients in $\mathbb{C^3}$ so the set is linearly independent.
2) When $\mathbb{C^3}$ over $\mathbb R$ I don't really know what happens with $i$'s in the reduced matrix.  
 A: We will use the classical definition:

Let $S=\{v_1,\dots,v_n\}$ be a subset of a linear space $V$ over a field $\mathbb{F}$. The equation
$$\sum_{i=1}^n\alpha_iv_i=0$$
in $\alpha_1,\dots,\alpha_n\in\mathbb{F}$ admits a solution $\alpha_1=\dots=\alpha_n=0$. If  it admits no other solution, then $S$ is said to be linearly independent. Otherwise, $S$ is said to be linearly dependent.

Our subset has $n=3$ and
$$v_1=(1+i,1,-2i)\\
v_2=(1,1-i,2+i)\\
v_3=(2+i,2,-3)$$
Our linear space is $\mathbb{C}^3$. You have already found an answer for the case $\mathbb{F}=\mathbb{C}$; let's tackle the case $\mathbb{F}=\mathbb{R}$. Let $\alpha_1,\alpha_2,\alpha_3\in\mathbb{R}$ and consider a linear combination of the $v_i$
\begin{align}
\sum_{i=1}^n\alpha_iv_i=\alpha_1\cdot(1+i,1,-2i)+\alpha_2\cdot(1,1-i,2+i)+\alpha_3\cdot(2+i,2,-3)=\\
\big((\alpha_1+\alpha_2+2\alpha_3)+(\alpha_1+\alpha_3)\cdot i,(\alpha_1+\alpha_2+2\alpha_3)-\alpha_2\cdot i, (2\alpha_2-3\alpha_3)+(-2\alpha_1+\alpha_2)\cdot i\big)
\end{align}
Now, if this linear combination is $0$, then we must have that each coordinate is $0$. This also means that for each coordinate, both the real part and the imaginary part must be $0$. Hence, we have a system of equations:
$$\left\{
\begin{array}{l}
\alpha_1+\alpha_2+2\alpha_3=0\\
\alpha_1+\alpha_3=0\\
\alpha_1+\alpha_2+2\alpha_3=0\\
\alpha_2=0\\
2\alpha_2-3\alpha_3=0\\
-2\alpha_1+\alpha_2=0
\end{array}
\right.$$
You can solve this however you prefer, but it's easy to see that the only solution to this system is $\alpha_1=\alpha_2=\alpha_3=0$. It follows from the definition that $\{v_1,v_2,v_3\}$ is a linearly independent set of $\mathbb{C}^3$ over $\mathbb{R}$.

Did you notice that in this solution, we didn't use any particular property that requires the $\alpha_i$ to be real (rather than complex)? Indeed, we have the following:

Corollary: Let $\mathbb{F_1}$ be a field extension of $\mathbb{F_0}$. If $S=\{v_1,\dots,v_n\}\subset V$ is linearly independent over $\mathbb{F_1}$, then it is linearly independent over $\mathbb{F_0}$.

Proof: Observe that any solution $\alpha_1,\dots,\alpha_n\in\mathbb{F_0}$ to $\sum_{i=1}^n\alpha_iv_i=0$ is also a solution in $\mathbb{F_1}$. Since $S$ is linearly independent over $\mathbb{F_1}$, the only solution in $\mathbb{F_1}$ is the trivial, zero solution. It follows that the trivial solution is also the only solution in $\mathbb{F_0}$, and hence $S$ is also linearly independent in $\mathbb{F_0}$, as claimed. $\square$
By the corollary, the fact that you proved linear independence in the case $\mathbb{F}=\mathbb{C}$ meant you automatically also proved linear independence in the case $\mathbb{F}=\mathbb{R}$.
A: Treat $A$ as a matrix whose columns are the given vectors.
Note that $A$ has an eigenvalue of multiplicity one at zero. In particular $\ker A$ is the (complex) span of the corresponding eigenvector.
Hence the columns are linearly dependent over $\mathbb{C}$.
In this case, the eigenvector is (a complex multiple of) $v=(2,-1+i,-1-i)$. Note that there is no $\lambda \in \mathbb{C}$ such that
$\lambda v $ is purely real.
Hence the columns are linearly independent over $\mathbb{R}$.
