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I'm trying to prove that every subspace of $\mathbb{R}^n$ is a solution space of a homogeneous system of linear equation. I want to prove it without using orthogonal complement. So I found this question that has an answer :

Let $S$ a subspace of $\mathbb{R}^n$ and choose $(e_1,…,e_p)$ a basis of $S$ which we compete it by a basis $(e_1,…,e_p,e_{p+1},…,e_n)$ of $\mathbb{R}^n$.

Now let the endomorphism $f$ defined by $f(e_i)=0$, $ 1 \le i \le p$ and $f(e_i)=e_i, p+1 \le i \le n$ and let $A$ the matrix of $f$ in this basis then $AX=0\iff X\in S$.

My problem is finding the matrix $A$ and I also didn't unterstand the domain of the endomorphism. Is $f:\mathbb{R}^n \rightarrow \mathbb{R}^n$?

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    Do you know about dual spaces? – 2017-01-15
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    @amd No, I dont think so – 2017-01-15
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    Try using the fact that the image of an endomorphism (or any linear map, really) is spanned by the columns of its matrix. And yes, you’re looking for $f:\mathbb R^n\to\mathbb R^n$. – 2017-01-15

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The $j$th column of $A$ is $(x_1,\ldots,x_n)^T$, where $f(e_i) = \sum_{t=1}^n x_t e_t$ is the unique writing of $f(e_i)$ in the basis $\{e_1,\ldots,e_n\}$. And yes, $f: \Bbb R^n \to \Bbb R^n$. Note that to define a linear map, it suffices to define its values on the basis elements.

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    So $A=(0,0, \ldots,0)^T \quad \forall \quad 1 \le i \le p $ and $A=(0,0,\ldots,x_i,0)^T \quad \forall p+1 \le i \le n $ with $x_i=1$? And I dont see why $AX=0 \rightarrow X\in S?$ Im tryng to show that using $\ker f$ but I couldnt make any advance – 2017-01-15
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    @R.MCM yes, the $i$th rows of $A$ are as you wrote. Take $X = (x_1,\ldots,x_n)^T\in \Bbb R^n$ and compute $AX$. From $AX=0$ you'll get $x_{p+1} = \cdots = x_n = 0$, so that $x = (x_1,\ldots,x_p,0,\ldots, 0) = \sum_{i=1}^p x_i e_i \in \text{span}(\{e_1,\ldots,e_p\}) = S$. – 2017-01-15