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are the two vectors $a=(1 - i, 2 - 2i), b=(i - 1, 2i - 2)$ linear dependent over $\mathbb{R}$

Now over $\mathbb{C}$ its obvious(yes) but my question is can I even answer this question when looking over $\mathbb{R}$? or because $i$ is not a real number I will always say that these vectors are linear independent no matter what

Would you change your answers if now the vectors are: $a=(1 - \sqrt2, 2 - 2\sqrt2), b=(\sqrt2 - 1, 2\sqrt2 - 2)$ over $\mathbb{Q}$

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  • $\begingroup$ They are linearly dependent over $\mathbb{R}$ since $a+b=0$ has coefficients $1,1$ in $\mathbb{R}$. $\endgroup$ – KittyL Dec 27 '15 at 15:48
  • $\begingroup$ @KittyL hmm my original was "linear dependent" so this is why I said yes, but I don't understand if we can just "ignore" $i$ when looking over R? $\endgroup$ – The One Dec 27 '15 at 15:49
  • $\begingroup$ When looking over $\mathbb{R}$, it is the scalar coefficients of the vectors that we are looking at, not the element in the vector. When we say vectors over a field, it means we do multiplication of a scalar in that field and the vector with its element in its own set. $\endgroup$ – KittyL Dec 27 '15 at 15:50
  • $\begingroup$ @KittyL thanks can you look at this new part now: $a=(1 - \sqrt2, 2 - 2\sqrt2), b=(\sqrt2 - 1, 2\sqrt2 - 2)$ over $\mathbb{Q}$ $\endgroup$ – The One Dec 27 '15 at 15:54
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    $\begingroup$ When checking linear independence/dependence, yes. Of course the vectors have to satisfy certain criteria to be a vector space. So we cannot technically say "irrelevant". $\endgroup$ – KittyL Dec 27 '15 at 16:01
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$a+b=0.$ Therefore $a$ and $b$ are dependent over $\mathbb{R}.$

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  • $\begingroup$ but can we even answer this question when $i$ is not defined in $\mathbb{R}$? so both $a$ and $b$ have some undefined things inside them $\endgroup$ – The One Dec 27 '15 at 15:42
  • $\begingroup$ You mean $+$... $\endgroup$ – Tobias Kildetoft Dec 27 '15 at 15:43
  • $\begingroup$ $i$ is an element of Vector space. It has nothing to do with the field $\mathbb{R}$. $\endgroup$ – Suhail Dec 27 '15 at 15:47
  • $\begingroup$ Oh I see, and what about this $a=(1 - \sqrt2, 2 - 2\sqrt2), b=(\sqrt2 - 1, 2\sqrt2 - 2)$ over $\mathbb{Q}$ $\endgroup$ – The One Dec 27 '15 at 15:54
  • $\begingroup$ same is true here as well. $\endgroup$ – Suhail Dec 27 '15 at 15:57
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Let $V \subseteq \mathbb{C}^2$ be a vector space over the field $\mathbb{F}$. In general, two vectors $a,b \in V$ are linearly dependent, if $a$ is a scalar multiple of $b$, that is, if there is a $\lambda \in \mathbb{F}$ such that $a = \lambda \cdot b$.

Now consider your example. Let $a = \begin{pmatrix} 1 - \mathrm{i} \\ 2 - 2\mathrm{i}\end{pmatrix} \in V$ and $b = \begin{pmatrix} \mathrm{i} - 1 \\ 2\mathrm{i} - 2 \end{pmatrix} \in V$. The vectors would be linearly dependent, if $a = -1 \cdot b$ and $-1 \in \mathbb{F}$. This is true for the field $\mathbb{F} := \mathbb{C}$, but also for $\mathbb{F} := \mathbb{R}$ or any other field of which $-1$ is an element.

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