# Prove that $W$ has dimension $n$ over $\mathbb{C}$, assuming that $V$ has dimension $n$ over $\mathbb{R}$.

Let $$V$$ be a real vector space with inner product $$\langle \cdot \,,\cdot \rangle : V \times V \rightarrow \mathbb{R}$$. Let $$W = V \times V$$ with vector addition defined by

$$(v_1,w_1) + (v_2,w_2 ) = (v_1 + v_2, \, w_1 + w_2)$$

and scalar multiplication defined by

$$(a + bi)(v,w) = (av - bw, aw + bv)$$

and inner product $$\langle.\,,.\rangle' : W \times W \rightarrow \mathbb{C}$$ defined by

$$\langle(v_1 ,w_1 ),(v_2 ,w_2 )\rangle' = (\langle v_1 ,v_2 \rangle + \langle w_1 ,w_2\rangle) + i(-\langle v_1 ,w_2 \rangle + \langle w_1 ,v_2 \rangle )$$

Prove that $$W$$ has dimension $$n$$ over $$\mathbb{C}$$, assuming that $$V$$ has dimension $$n$$ over $$\mathbb{R}$$.

I know that dimension is the number of vectors in a basis for a vector space. I also know that a basis is composed of linearly independent vectors that span the space. Would I have to use a form of induction to prove this? Sorry, normally I have more to give on questions but I'm finding that as I get into more proof based math either I know how to do a problem or I don't. There doesn't seem to be a lot of in-between.

Since $$V$$ has dimension $$n$$ over $$\mathbb{R}$$, $$V$$ is simply $$\mathbb{R}^n$$. Then a pair of vectors of $$V$$ is simply $$2n$$ real numbers $$(x_i, y_i)$$ The scalar product is written as:
$$(a+bi)(x_i,y_i)=(ax_i-by_i,ay_i+bx_i)$$
(That should remind you of the formula $$(a+bi)(c+di)=(ac-bd)+(ad+bc)i$$). Can you then write an explicit isomorphism between $$W$$ and $$\mathbb{C}^n$$?
SPOILER: just send $$(x_i,y_i)\in\mathbb{R}^n\times\mathbb{R}^n=W$$ to $$(x_i+iy_i)\in\mathbb{C}^n$$. Now one needs to formally prove that this is, indeed, an isomorphism to conclude the dimension of $$W$$.
• So what I need to find is a bijection T from V to W that preserves addition and scaler multiplication. That is, for all vectors $u$, $v$ in $V$ and all scalers $c \in F$, I need to show $T(u + v) = T(u) + T(v)$ and $T(cv) = cT(v)$. – Idle Math Guy Mar 17 at 22:53
• So the isomorphism would be sending a vector that consists of real numbers to a vector that consists of complex numbers? Does the notation up above imply that all vectors consist of only two values? i.e. We would never see a vector like $(x_i,y_i,z_i)$. Trying my best. Math is hard. :( – Idle Math Guy Mar 17 at 23:02
• Ok, $V \times V$ refers to cartesian product. So if $V$ is $R^3$ then how would scalar multiplication work? i.e. what would $(a+bi)(x_i, y_i, z_i)$ be equal to?Many times once I figure out what the problem is actually asking for I can solve the problem. – Idle Math Guy Mar 18 at 1:43