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

Question

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.

Answer 1

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 $W$.