Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
    
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 '13 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 '13 at 18:43
add comment

6 Answers

up vote 6 down vote accepted

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.

share|improve this answer
add comment

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 $A$.

share|improve this answer
add comment

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 $T$

share|improve this answer
add comment

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 $S$?

share|improve this answer
add comment

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

share|improve this answer
add comment

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 $\mathbb{R}^n$.

Update.

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 $$ A= \left[ \begin{array}{ccccc} \lambda_1&\cdots&0&\cdots &0 \\ \vdots & \ddots &\vdots &\quad &\vdots\\ 0&\cdots&\lambda_i&\cdots &0 \\ \vdots & \quad &\vdots &\ddots &\vdots\\ 0&\cdots&0&\cdots &\lambda_n \\ \end{array} \right] $$

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.