# Show that every subspace of $\mathbb{R}^n$ is a kernel of a linear map.

Let $S$ be a subspace of $\mathbb R^n$. Show that there is an $n \times n$ matrix $A$ such that

$$S= \{x \in \mathbb R^n : Ax=0\}.$$

How to proceed?

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Could you use a more descriptive title? For instance: "Showing a subspace is always the kernel of a square matrix", if you know what the kernel of a matrix is. –  TMM Apr 11 at 14:04
Before you solve the general problem, is there any $S$ that you know how to solve the problem for? –  Hurkyl Apr 11 at 18:43

Let $(e_1,\ldots,e_p)$ a basis for $S$ and we complete this to a basis $(e_1,\ldots,e_p,e_{p+1},\ldots,e_n)$ for $\mathbb{R}^n$.

We define the endomorphism $f$ of $\mathbb{R}^n$ by $$f(e_i)=0 \quad i=1,\ldots,p\quad \mathrm{and}\quad f(e_i)=e_i\quad i=p+1,\ldots,n$$ then the matrix of $A$ on this basis verify the desired result.

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Imagine a matrix whose rows form a basis for $S$. You probably know that you can find a basis $B$ for the nullspace of this matrix. Now form a matrix whose rows are the elements of $B$. Fill this matrix out to $n\times n$ by adding rows of all zeros, if necessary. Voila! you have your matrix $A$.

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You can define a translation as $T$ from basis of $R^n$ to $S$ such that consider same base for $S$ and $R^n$ then if you define $T$ as it goes every same base to 0 then you will find $A$ matrix. you are arbitrary that your $T$ maps basis of $(R^{n}-S ) to (R^n-s)$ how you like it! then you will have alternative $T$

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Find an orthnormal basis for $S$, call it $v_1,..., v_m$ assume $1<m<n$, else take the 0 matrix or the identity marix. (assume also that $v_i$ are colum vectors

Then extend this to a basis for $\mathbb{R}^n$, say $v_1,...,v_m,u_1,...u_{n-m}$

Then the matrix given by $(u_1^T;... u_{n-m}^T; u_{n-m}^T...; u_{n-m}^T)$

; means stating a new row. so this is $n$ row vectors.

where $u_{n-m}^T$ apppears $m+1$ times. Do you see why this matrix must have kernel $S$?

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Do you know much about linear transformations? If so, you may know that $T : \mathbb{R}^n \to \mathbb{R}^n$, where $T(u)$ is the component of $u$ orthogonal to $S$, is a linear transformation. Then $\ker T = S$, so by taking $A$ to be the standard matrix of $T$, we have $$S = \ker T = \{x \in \mathbb{R}^n : T(x) = 0\} = \{x \in \mathbb{R}^n : Ax = 0\}.$$

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We have that $S\oplus S^\perp= \mathbb{R}^n$. Here $S^\perp=\{x\in\mathbb{R}^n : \langle s,x\rangle=0 \,\forall s\in S \}$ is a linear subspace of $\mathbb{R}^n$. Set the linear operator $T:\mathbb{R}^n\to \mathbb{R}^n$ whit fellow $T(s)=0\;\forall s\in S$ and $T(x)=x\;\forall x\in S^\perp$. Then $A$ is the matrix of $T$ in canonical base of $\mathbb{R}^n$.
To fix ideas we write $$e_1= \left[ \begin{array}{c} 1 \\ 0\\ \vdots\\ 0\\ \end{array} \right] \quad e_2= \left[ \begin{array}{c} 0 \\ 1\\ \vdots\\ 0\\ \end{array} \right] \; \ldots \; e_n= \left[ \begin{array}{c} 0 \\ 0\\ \vdots\\ 1\\ \end{array} \right]$$ For a construtive answer let $d=\dim(S)$ and $e_{i_1},\ldots e_{i_d}$ the vectors of canonical basis $\{e_1,\ldots e_n\}$ such that $S=\mbox{Span}\{e_{k_1},\ldots e_{k_d}\}$. Analogously, we have $n-d=\dim(S)$ and $e_{l_1},\ldots e_{l_{(n-d)}}$ the vectors of canonical basis $\{e_1,\ldots e_n\}$ such that $S=\mbox{Span}\{e_{l_1},\ldots e_{l_{(n-d)}}\}$. Define, $$\lambda_i= \left\{ \begin{array}{ll} 0 & \mbox{ if } i\in\{k_1,\ldots,k_d\}\\ 1 & \mbox{ if } i\in\{l_1,\ldots,l_{(n-d)}\} \end{array} \right.$$ Then $x\in S$ implies $x= (1-\lambda_1)\cdot x_1\cdot e_1+\ldots +(1-\lambda_n)\cdot x_n\cdot e_n$ and $$A= \left[ \begin{array}{ccccc} \lambda_1&\cdots&0&\cdots &0 \\ \vdots & \ddots &\vdots &\quad &\vdots\\ 0&\cdots&\lambda_i&\cdots &0 \\ \vdots & \quad &\vdots &\ddots &\vdots\\ 0&\cdots&0&\cdots &\lambda_n \\ \end{array} \right]$$