For an $m \times n$ real matrices A, let $W_{\mathbb{R}}$ = { $Ax |x \in \mathbb{R^n}$} and $ W_{\mathbb{C}}$ = { $ Ax |x \in \mathbb{C^n}$}.
Is this statement is true/false ?
The dimension of $W_\mathbb{R}$ as a subspace of $\mathbb{R^m}$ over $\mathbb{R}$ is the same dimension of $W_{\mathbb{C}}$ as a subspace of $\mathbb{C^m}$ over $\mathbb{C}$
My answer :i thinks this statement is false because $W_{\mathbb{R}}$ and $W_{\mathbb{C}}$ are not same so their dimension will be different
is its correct ??
Pliz help me
Any hints/solution will be appreciated
thanks in advance
The dimensions are the same; the key of the proof is:
Let $k\leq n$ and $S=\{e_1,\cdots,e_k\}$ (where $(e_i)_{i\leq n}$ is the canonical basis of $\mathbb{R}^n$). Then $S$ is a free system over $\mathbb{R}$ IFF $S$ is a free system over $\mathbb{C}$.
Proof. $(\Rightarrow)$ Assume that $\sum_{j\leq k} \lambda_j Ae_j=0$ where the $\lambda_j=a_j+ib_j$ are complex numbers. Then $\sum_{j\leq k} a_jAe_j+i(\sum_{j\leq k}b_jAe_j)=0$ implies that $\sum_{j\leq k}a_jAe_j=0,\sum_{j\leq k}b_jAe_j=0$...
$(\Leftarrow)$ Assume that $\sum_{j\leq k} a_jAe_j=0$ where the $(a_j)$ are real numbers that are not all zero. Then $\sum_{j\leq k} a_jAe_j=0$ where the $(a_j)$ are complex numbers that are not all zero.