Question

Show that the vectors (1-I,i) and (2,-I+i) in R² are Linearly Dependent over Field but Linearly Independent over R , where .i = √-1

Answer 1

Answer on Question #43816 – Math – Linear Algebra

Show that the vectors $(1-i,i)$ and $(2,-l+i)$ in $R^2$ are Linearly Dependent over Field but Linearly Independent over $R$, where $i = \nu-1$

Solution.

There are several inaccuracies in the condition. The first one vectors $(1-i,i)$ and $(2,-l+i)$ do not belong to $R^2$ because their coordinates do not belong to $R$, so $R^2$ should be changed on $C^2$ and there missed a field over which these vectors are linearly dependent, we guess there would have to be $C$.

Two vectors $v_{1}, v_{2}$ are Linearly Dependent over field $F$ if there exist scalar $a$ and $b$ () in $F$ such that

$$a v _ {1} + b v _ {2} = 0$$

Let's try to find $d$ these scalars

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

We get a system:

$$\left\{ \begin{array}{l} a (1 - i) + 2 b = 0 \\ a i + b (- 1 + i) = 0 \end{array} \right. \quad \left\{ \begin{array}{c} b = \frac {a}{2} (- 1 + i) \\ a i + \frac {a}{2} (- 1 + i) (- 1 + i) = 0 \end{array} \right.$$ $$a i + \frac {a}{2} (- 1 + i) (- 1 + i) = 0$$ $$a i + \frac {a}{2} (1 - 2 i - 1) = 0$$ $$a - a = 0$$

So, we get that $a$ can be arbitrary and $b = \frac{a}{2} (-1 + i)$. Indeed, if take $a = 2$ then $b = (-1 + i)$.

And it is easy to check that $2(1 - i, i) + (-1 + i)(2, -1 + i) = 0$, since $(-1 + i) \in C$, we get that these vectors are linearly dependent over $C$. And as $b = \frac{a}{2} (-1 + i)$ we see that $a$ and $b$ can not be real simultaneously, so these vectors are not linearly dependent over $R$, hence they are linearly independent over $R$.

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