Every subspace of $\mathbb{R}^m$ is the range of some linear function

linear algebravector-spaces

We have the following statement (Matrix analysis and applied linear algebra, Mayer)

The range of every linear function $f:\mathbb{R}^n\rightarrow\mathbb{R}^m$ is a subspace of $\mathbb{R}^m$, and every subspace of $\mathbb{R}^m$ is the range of some linear function.

I understand the proof of the first part, but I need to understand the argument in the second part of statement.

He starts as follows.

Let $V$ be a subspace of $\mathbb{R}^m$. Suppose that $\{v_1, v_2, \dots, v_n\}$ is spanning set for $V$ so that $$V = \{\alpha_1v_1+ \dots + \alpha_nv_n \mid \alpha_i \in \mathbb{R}\}.$$

I know that every vector space has spanning set. Clearly, if the spanning set of $V$ has $k$ vectors, then there is also a spanning set of $V$ which has $k+1$ vector.

But I do not know why he is sure that the spanning set of $V$ has $n$ vectors. What am I missing? Thank you.

Best Answer

The statement is not very clear in this way. There are two true things.

  1. The range of any linear function $f : \mathbb{R}^n \to \mathbb{R}^m$ is a subspace of $\mathbb{R}^m$.
  2. Any subspace of $\mathbb{R}^m$ is the range of some linear function.

The second statement does not specify the domain of the function, and, for reasons of dimension, it has to be $\mathbb{R}^n$ with $n \geq m$. A better statement would be the following

For any $n \geq m$ and any subspace $V$ of $\mathbb{R}^m$, there is a linear function $f : \mathbb{R}^n \to \mathbb{R}^m$ such that $V$ is the range of $f$.

Then the proof is essentially as you say: take a spanning set $\{ v_1, \cdots, v_m \}$ of $V$, and define $$ f(e_i) = v_i, \quad i = 1, 2, \dots, m, $$ and $$ f(e_i) = v_m, \quad i = m+1, \dots, n. $$

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