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Sorry if this is really basic but for some reason it's giving me trouble. If $z,w,t \in \mathbb{C}$, do there exist real numbers $a,b,c \in \mathbb{R},$ not all zero, such that $$ az+bw+ct = 0 $$ I would imagine the answer is no generally because you can't necessarily eliminate the imaginary parts of the numbers. Is there any way to write this out formally or by some argument I'm not seeing? Thanks in advance for any hints!

Adam Rubinson
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Irving Rabin
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2 Answers2

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Here, $\mathbb{C}$ can be thought of as the vector space $\mathbb{R}^2,$ and since $a,b,c$ are in $\mathbb{R},$ we can say that our field for our vector space is $\mathbb{R}$. So the question is essentially, "prove that $n+1$ vectors in the vector space $\mathbb{R}^n$ over $\mathbb{R}$ are linearly dependent", which is just a famous/standard result from linear algebra.

Adam Rubinson
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  • If you're going to use results from linear algebra, then just say "since $z$, $w$ and $t$ are tree vectors in $\Bbb R^2$ they're linearly dependent and we are done". – jjagmath Feb 13 '22 at 23:52
  • @jjagmath yes that's a very good point. – Adam Rubinson Feb 13 '22 at 23:57
  • Oh so because $\dim(\mathbb{C}) = 2$ and there are three vectors they are necessarily linearly dependent? I guess the issue was I was thinking of it as $\mathbb{C}$ over $\mathbb{R}$ but I suppose everything is going to be fine because the reals are a subring of $\mathbb{C}$. – Irving Rabin Feb 14 '22 at 00:31
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    @AndreyYanyuk Careful. The dimension of $\mathbb{C}$ depends on the base field. Over $\mathbb{R}$ it's two, over $\mathbb{C}$ it's one. – CyclotomicField Feb 14 '22 at 00:40
  • Ohhh I hadn't realized that! Thanks very much! @CyclotomicField – Irving Rabin Feb 14 '22 at 00:55
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    @AndreyYanyuk no problem. This will be important later when you study Galois theory and field extensions over $\mathbb{Q}$. Say for example we add two algebraic numbers $\sqrt{2}$ and $\sqrt{3}$ and all linear combinations of them to make a subfield of $\mathbb{R}$, which includes $\sqrt{6}$ because $\sqrt{2} \sqrt{3}=\sqrt{6}$. So all elements in this field will be of the form $a\sqrt{2}+b\sqrt{3}+c\sqrt{6}+d$, which is a four dimensional space over $\mathbb{Q}$. – CyclotomicField Feb 14 '22 at 16:34
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Let $z = x_1 + iy_1, w = x_2 + iy_2, t = x_3 + iy_3$. Thus the equation $az+bw+ct = 0 \implies ax_1+bx_2+cx_3=0, ay_1+by_2+cy_3 = 0$. Thus $(a,b,c)$ is the solution of the system of linear equations $AX = 0$ where $A = \begin{pmatrix} x_1 & x_2 & x_3 \\ y_1 & y_2 & y_3 \end{pmatrix}, X = \begin{pmatrix} a \\ b \\ c \end{pmatrix}, 0 = \begin{pmatrix} 0 \\ 0 \end{pmatrix}$. The columns of $A$ are linearly dependent because there are $3$ vectors and there are $2$ entries in each vector. This means that the equation $AX = 0$ must have a non-trivial solution, and this means that the answer is "yes" to the original question.

Wang YeFei
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