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Let the vector space be defined as $ \{(x, y, z) \in \mathbb{R}^3 \mid x + y + z= 0\}$. Find its basis and its dimension.

Basis is $\{(1,1,1)\}$ and its dimension is $1$, or basis (of a null space) is $\{(-1,1,0), (-1,0,1)\}$ and its dimension is $2$. Not sure.

Rank equals to $1$, and the nullity is $2$, that satisfies: $$r(A) + n(A) = n.$$

Dimension is defined as the rank of a row space (then $\dim V = 1$).

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    You're confusing the dimension of the kernel of a linear map and the rank of this linear map. – 2017-01-29
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    You're mixing stuff here: what "null space" are you talking about? You don't even have a linear map here (though you could easily construct one) ... – 2017-01-29
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    I did a mistake, just as I wrote it I realised it, that dimension is 2 – 2017-01-29
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    $(1, 1, 1)$ doesn't belong to your vector space. – 2017-01-29
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    That is true. It would never give me that zero on the right side. – 2017-01-29

4 Answers 4

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The vector space is defined by 1 linear equation. As we're in $\mathbf R^3$, this implies, by the Rank-nullity theorem that its dimension is $3-1=2$.

The equation shows this space is isomorphic to $\mathbf R^2$, by the isomorphism \begin{align}\varphi\colon\mathbf R^2&\longrightarrow V,\\(x,y)&\longmapsto (x,y,-x-y).\end{align} A basis of $V$ can be obtained as the image by $\varphi$ of the canonical basis of $\mathbf R^2$, for instance, i.e. $$v_1=\begin{bmatrix}1\\0\\-1\end{bmatrix},\quad v_2=\begin{bmatrix}0\\1\\-1\end{bmatrix}.$$

Note: In general, , for a, $r$-dimensional subspace $V$ of $\mathbf R^n$, we call the number $n-r$ the codimension of $V$. Now, there results from the Rank-nullity theorem that, if a subspace is defined by a system of linear equations, represented by a matrix of coefficients $A$, $$\operatorname{rank}A=\operatorname{codim}V.$$

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$$x+y+z=0\Leftrightarrow \begin{cases}x=-\lambda-\mu\\y=\lambda \\z=\mu\end{cases}\; (\lambda,\mu\in\mathbb{R})\Leftrightarrow \begin{bmatrix}{x}\\{y}\\{z}\end{bmatrix}=\lambda \underbrace{\begin{bmatrix}{-1}\\{1}\\{0}\end{bmatrix}}_{u}+\mu\underbrace{\begin{bmatrix}{-1}\\{0}\\{1}\end{bmatrix}}_{v}.$$ The set $B=\{u,v\}$ spans the subspace and is linearly independent so, is basis of the given subspace.

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Note that your Vector Space is nothing but a plane in $\Bbb R^3$

So, any $2$ independent solutions of the equation $x+y+z=0$ will serve as a basis for your Vector Space.

$\{(1,1,1)\}$ cannot be a basis, since it does not even lie on the plane.

Whereas $\{(1,-1,0),(1,0,-1)\}$ given by you works well.

Now, comming to your confusion, you can talk about a Null Space with respect to a particular Linear Map only. Here in this case we dont even have a Linear Map, so it dosent make sense to talk about a Null Space.

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The given subset W={(x,y,z)∈R^3|x+y+z=0} of R^3 = solution space of the system AX=0 = the null space of A=[1 1 1]. Rank(A)=1, then nullity(A)=dimension of W=2. Basis for W={(1,-1,0),(1,0,-1)}= basis for the solution space. Also, (a,b,c)∈W iff a+b+c=0 iff a=-b-c. Thus W={(-b-c,b,c): b,c∈R}. Since, (-b-c,b,c)=b(-1,1,0)+c(-1,0,1) Thus, W=span{(-1,1,0),(-1,0,1)}. Then basis for W={(-1,1,0),(-1,0,1)}, because the two vectors (-1,1,0) and (-1,0,1) are also, linearly independent.

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    Use Proper formatting so that users can read your solution effectively and can help better ! – 2017-01-29