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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.

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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 $W$.

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  • $\begingroup$ 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 $T(cv) = cT(v)$. $\endgroup$ – Idle Math Guy Mar 17 at 22:53
  • $\begingroup$ 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. :( $\endgroup$ – Idle Math Guy Mar 17 at 23:02
  • $\begingroup$ 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. $\endgroup$ – Idle Math Guy Mar 18 at 1:43

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