Consider the $2 \times 2$ Hermitian matrix $$A = \begin {bmatrix} 1& i \\ -i& 1 \end{bmatrix}.$$
Let $$W_{\mathbb{R}} = \{ Ax | x \in \mathbb{R^2}\}$$ and $$ W_{\mathbb{C}} = \{ Ax |x \in \mathbb{C^2}\}$$
a) Find $\dim_{\mathbb R}W_{\mathbb{R}}$
b) Find $\dim_{\mathbb R}W_{\mathbb{C}}$
c) Find $\dim_{\mathbb C}W_{\mathbb{C}}$
My answer: For a) $\dim_{\mathbb R}W_{\mathbb{R}} = 2$ because
$A(x,y)^t =(x+iy,-ix +y)^t =x(1,-i)^t +y(i,1)^t$
I'm confused about b) and c)
Any hints/solution will be appreciated
Both $W_\mathbb R$ and $W_\mathbb C$ are subsets of $\mathbb C^2$ by definition. Let $v_1=(1,-i)^t$, $v_2=(i,1)^t$ be the columns of $A$, then \begin{align*} W_\mathbb C &= \langle v_1, v_2\rangle_\mathbb C = \left\{\,\lambda v_1 + \mu v_2 \,\middle|\, \lambda,\mu\in \mathbb C\,\right\}, \\ W_\mathbb R &= \langle v_1, v_2\rangle_\mathbb R = \left\{\,\lambda v_1 + \mu v_2 \,\middle|\, \lambda,\mu\in \mathbb R\,\right\}. \end{align*} Note that $v_1, v_2$ are $\mathbb R$-linearly independent but $\mathbb C$-linearly dependent, since $v_2= i v_1$. Hence $W_\mathbb R$ is a $2$-dimensional $\mathbb R$-vector space and $W_\mathbb C$ is a $1$-dimensional $\mathbb C$-vector space. Finally, whenever $V$ is a complex vector space with $\dim_\mathbb C V=n$, we have $\dim_\mathbb R V=2n$, since a $\mathbb C$-basis $w_1,\dots,w_n$ gives rise to as $\mathbb R$-basis $w_1, iw_1, \dots, w_n, iw_n$. In particular $$ \dim_\mathbb R W_\mathbb C = 2\dim_\mathbb C W_\mathbb C = 2. $$
The example is a bit confusing, since we actually have $W_\mathbb R=W_\mathbb C$ in this case, since $v_2 = iv_1$.