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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

jasmine
  • 14,457

1 Answers1

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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.