Linear independence of certain vectors of $\mathbf{C}^2$ over $\mathbf{R}$

Suppose that $\{e_1,e_2\}$ where $e_1=(1,0)$ and $e_2=(0,1)$ is the standard basis of $\Bbb C^2$ as a vector space over $\Bbb C$. Show that $\{e_1,ie_1,e_2,ie_2\}$ is a basis of $\Bbb C^2$ as a vector space over $\Bbb R$ and conclude that $\dim_\Bbb R \Bbb C^2=2\dim_\Bbb C \Bbb C^2$.

I know that in general in order to show that a set of vectors form a basis you must show that they are linearly independent and are a spanning set. So normally I would solve $Ax=0$ and if $x$ was $0$ I would know they were linearly independent. However, I'm not sure how to use that here.

• Do you mean to have $\{e_1, ie_1, e_2, ie_2\}$? – Eli Rose Feb 17 '16 at 14:54

As another user has described linear independence (and hopefully made it clear that $dim_{\mathbb{R}} \mathbb{C}^2 = 4 = 2\dim_{\mathbb{C}}\mathbb{C}^2$ ), you need now to show that $\{e_1,e_2,ie_1,ie_2\}$ spans $\mathbb{C}^2$. Let $\xi \in \mathbb{C}^2$ be arbitrary. Then, by our assumptions, there exists $x_1,x_2 \in \mathbb{C}$ so that $\xi = x_1e_1+x_2e_2$. Since each $x \in \mathbb{C}$ we can write $x = a + bi$ for $a,b$ real. Hence, \begin{equation} \xi = x_1e_1+x_2e_2 \\ =(a_1+b_1i)e_1+(a_2+b_2i)e_2 \\ =a_1e_1+b_1ie_1+a_2e_2+b_2ie_2. \end{equation} This should complete your proof.

Let $a,b,c,d \in \mathbb R$ and $ae_1+be_2+ce_1i+de_2i=0$. Then we have:

$$ae_1+be_2+ce_1i+de_2i=0 \iff (a+ci)e_1+(b+di)e_2=0 \iff a,b,c,d=0$$ since we have $$(a+ci)\left(\begin{array}{c}1 \\0 \end{array}\right)+(b+di)\left(\begin{array}{c}0 \\1 \end{array}\right)=\left(\begin{array}{c}0 \\0 \end{array}\right) \iff a+ci=0,b+di=0$$

and $a+ci,b+di$ are just complex numbers since $a,b,c,d$ are reals and a complex number $z=x+yi$ with $x,y \in \mathbb R$ is equal $0$ if and only if imaginary and realpart ($Re(z)=x, Im(z)=y$) are equal 0.

So we have that $\mathbb C^2$ over $\mathbb R$ is a $4$ dimensional vector space.

Using the fact $\mathbb C \cong \mathbb R^2$ you can conclude the claim.

Edit:

Even though there is a seperate answer above how to deal with the span I'll give a proof too for the sake of completeness and because I know that sometimes it's good to see the same story twice, so here it is:

Let $z \in \mathbb C^2$ be arbitrary and of the form $z=(z_1,z_2)$ with $z_1,z_2 \in \mathbb C$. Since elements in the complex plane consist of a real- and imaginary part we can 'decompose' our $z$ even further by

$$z_1=x_1+y_1i; \space x_1,y_1 \in \mathbb R$$ $$z_2=x_2+y_2i, \space x_2,y_2 \in \mathbb R$$

So we can express our arbitrary $z \in \mathbb C^2$ as follows:

$$z=(z_1,z_2)=(z_1,0)+(0,z_2)=z_1e_1+z_2e_2=(x_1+y_1i)e_1+(x_2+y_2i)e_2=x_1e_1+x_2e_2+y_1e_1i+y_2e_2i$$

And as you can see, we were able to find a way to express an arbitrary (and hence any!) element in $\mathbb C^2$ in terms of real valued coefficients and the linear indepentent vectors $\{e_1,e_2,e_1i,e_2i\}$.

• Could you explain how to do the spanning? I've attempted it but I keep ending up with 4 unknowns and 2 equations so I guess I'm setting it up wrong. – user313163 Feb 17 '16 at 15:25
• Is everything clear now? Don't be afraid to ask if something is an unclear!:) Note though that Sloan's proof and mine are identical, I just wanted to give you a second view of the same thing to make things more clear. Hope this helps. – noctusraid Feb 17 '16 at 17:29
• Yes that has answered my question. Thank you very much :) – user313163 Feb 17 '16 at 17:32
• Then you may accept my answer so that we have less open questions on this site. ;) This applies to your other questions too. Accept an answer if you think the question you posed has been well recieved and answered. I saw that you are rather new on math.stackexchange that's why I kindly ask you to.:) – noctusraid Feb 17 '16 at 17:47
• Oh okay, I didn't realise you could do this, I'll do it now. – user313163 Feb 17 '16 at 17:58