Is a set of complex numbers linear dependent?

Question

Is the set $\{(1-i,i),(2, -1 +i)\} \subset \mathbb{C}^2$ linear independent when:

1) we consider $\mathbb{C^2}$ as complex vector space (complex scalars)

2) we consider $\mathbb{C^2}$ as real vector space (real scalars)

I started with setting:

$a(1-i,i) + b(2, -1 + i) = (0,0)$

But then I always find that $a = b = 0$, although my solution tells me that:

$(1-i,i) = (1/2 - 1/2i)(2, -1 + i)$

Thus the set is linear dependent when we consider it as complex vector space. How do I solve this complex system of equations?

Answer 1

Perform elimination on \begin{align} \begin{bmatrix} 1-i & 2 \\ i & -1+i \end{bmatrix} &\to \begin{bmatrix} 1 & 1+i \\ i & -1+i \end{bmatrix} && R_1\gets \tfrac{1}{1-i}R_1 \\[6px]&\to \begin{bmatrix} 1 & 1+i \\ 0 & 0 \end{bmatrix} && R_2\gets R_2-iR_1 \end{align} This means that $$ (2,-1+i)=(1+i)(1-i,i) $$ In particular, there is no real number $r$ such that $$ (2,-1+i)=r(1-i,i) $$ so the two vectors are linearly independent over $\mathbb{R}$.

Answer 2

You get the system

$$ a(1 - i) + 2b = 0, \\ ai + b(-1 + i) = 0. $$

Let us multiply the first equation by $\frac{i}{1 - i}$ so that the coefficients of $a$ in the first equation will be the same as the coefficient of $a$ in the second equation. We have

$$ \frac{i}{1 - i} = \frac{i(1 + i)}{(1 - i)(1 + i)} = \frac{-1 + i}{2} $$

and so the first equation becomes

$$ ai + (-1 + i)b = 0 $$

which is precisely the second equation. Hence, the two equations are linearly dependent and we have (infinitely many) non-zero solutions. If we choose $a = 2$, we get

$$ b = -\frac{a(1-i)}{2} = -(1-i) = -1 + i $$

and

$$ 2(1 - i, i) + (-1 + i)(2, -1 + i) = 0. $$

Answer 3

Let $v_1 = (1-i, i)^T, v_2 = (2, -1+i)^T$.

Note that ${2 \over 1-i} v_1 = v_2$, so we see that $v_1,v_2$ are linearly dependent over $\mathbb{C}$.

Now suppose $a v_1+ b v_2 = 0$ where $a,b$ are real.

This gives the equations $a+2b -i a = 0$ and $-b + i (a+b) = 0$.

If $x+iy = 0$, with $x,y$ real, then we have $x=y=0$.

hence $a=0, b=0$ and so $v_1,v_2$ are linearly independent over $\mathbb{R}$.