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I have a homework assignment in linear algebra and my answer seems kind of fuzzy. Is this proof valid?

Let $\{a,b,c\}$ be linearly independent set in real vector space $\mathbb{R}^3$. Show that the same set is also linearly independent in complex vector space $\mathbb{C}^3$.

$a,b,c\in\mathbb{R}^3\Rightarrow a,b,c\in\mathbb{C}^3$

The only transformation which does not require another vector is multiplication by scalar value:

$\forall\lambda\in\mathbb{C}\exists a':a'=\lambda a$

Let's assume that $\exists\lambda\neq0:\exists\beta, \gamma\in\mathbb{C}^3:a'=\beta b+\gamma c$.

$a'=\beta b+\gamma c$

$\lambda a=\beta b+\gamma c$

$\lambda(a_x\vec{i}+a_y\vec{j}+a_z\vec{k})=\beta(b_x\vec{i}+b_y\vec{j}+b_z\vec{k})+\gamma(c_x\vec{i}+c_y\vec{j}+c_z\vec{k})$

For new set to be linearly independent, there must be a unique solution to the following system:

$(\lambda a_x-\beta b_x-\gamma c_x)\vec{i}=0$

$(\lambda a_y-\beta b_y-\gamma c_y)\vec{j}=0$

$(\lambda a_z-\beta b_z-\gamma c_z)\vec{k}=0$

Since $\vec{i},\vec{j},\vec{k}\neq0$:

$\lambda a_x=\beta b_x+\gamma c_x$

$\lambda a_y=\beta b_y+\gamma c_y$

$\lambda a_z=\beta b_z+\gamma c_z$

Let $\beta' = \frac{\beta}{\lambda}$ and $\gamma'=\frac{\gamma}{\lambda}$. This is allowed since $\lambda\neq0$ from our assumption.

$a_x=\beta'b_x+\gamma'c_x$

$a_y=\beta'b_y+\gamma'c_y$

$a_z=\beta'b_z+\gamma'c_z$

Since $Im(a)=Im(b)=Im(c)=0$, $\beta'b_x=\beta'Re(b_x)+\beta'Im(b_x)=\beta'Re(b_x)+\beta'\bullet0=\beta'Re(b_x)$

This is also true for other multiplication operations within the system and shows that the ratio of real and imaginary parts does not change. This shows that $\lambda a=\beta b+\gamma c$ is only true for $\lambda=\beta=\gamma=0$ which proves that $\{a,b,c\}$ is linearly independent in $\mathbb{C^3}$.

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  • $\begingroup$ Admittedly I did not read the whole 'proof' carefully but it looks dodgy at best. When you are proving linear independence, make sure you use the basic definition of linear independence rather than its equivalent intuitive definition. (ie if $\sum_i a_iv_i=0$ then for all $i$ $a_i=0.$ $\endgroup$ – Jack Yoon Nov 9 '15 at 12:07
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    $\begingroup$ This proof is correct, but it is quite long. If you know something about determinants, then the proof could be writte in one line. $\endgroup$ – Crostul Nov 9 '15 at 12:08
  • $\begingroup$ Suppose it were not true. You could write down a linear combination of the (real) vectors which summed to zero using complex coefficients. Examine the real part. $\endgroup$ – TheMathemagician Nov 9 '15 at 12:17
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Your proof is over complicated and it looks like you're assuming that $\beta$ and $\gamma$ has to be real (which might invalidate the proof).

Instead just assume that you have a linear combination $\alpha a + \beta b + \gamma c=0$ and since $a$, $b$ and $c$ are real you can separate the equation into real and imaginary parts:

$$\Re\alpha a + \Re\beta b + \Re\gamma c = 0$$ $$\Im\alpha a + \Im\beta b + \Im\gamma c = 0$$

Which we know that neither has non-zero solutions.

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