Linear Independence Condition

Question

Let $\mathbb{C}^2(\mathbb{R})$ be the vector space of all complex two-tuples over $\mathbb{R}$. I know that the basis is $\{(1,0),(i,0),(0,1),(0,i)\}$. I also know that showing LINEAR DEPENDENCE of vectors in $\mathbb{R}^n$ over $\mathbb{R}$ amounts to showing the system $Ax=0$ has a non-pivot column( free variables). But why does this theory does not work here $$\begin{bmatrix} 1 & i& 0& 0\\ 0& 0 & 1 & i\end{bmatrix}\begin{bmatrix} a\\b\\c\\d\end{bmatrix}=\begin{bmatrix}0\\0\end{bmatrix}$$ There are NON-Pivot Columns here but the vectors are Linearly Independent. I do not want to equate real and imaginary parts to get the answer.

Answer 1

The issue is that you're considering a real vector space. Hence all coordinates have to belong to $\mathbb R$. Which is not the case of $i$.

The linear system you wrote

$$\begin{bmatrix} 1 & i& 0& 0\\ 0& 0 & 1 & i\end{bmatrix}\begin{bmatrix} a\\b\\c\\d\end{bmatrix}=\begin{bmatrix}0\\0\end{bmatrix}$$ has no meaning in a real space.

$(i,0)$ are not the coordinates of a real vector.

$$\mathcal B = \{e_1=(1,0),e_2=(i,0),e_3=(0,1),e_4=(0,i)\}$$ is indeed a basis of $\mathbb C^2(\mathbb R)$. And for example

$$e_2 = 0 \cdot e_1 + 1 \cdot e_2 + 0 \cdot e_3 + 0 \cdot e_4$$ which means that the coordinates of $e_2 = (i,0)$ in the basis $\mathcal B$ are not $(i,0)$ in $\mathbb C^2(\mathbb R)$ but $(0,1,0,0)$.