Finding a basis for x+y+z=0
Let $V \neq \{0\}$ be a vecotspace and for $x,y,z\in V\setminus\{0\}$ is:
$x+y+z=0$
I'd like to find a basis for $span\{x,y,z\}$
$x=-y-z$ solves the above system for any $y,z$. So we could say $v_1=(2,-1,-1)$ is a basisvector. But we could also do:
$y=-x-z$ and we would get $v_2=(-1,2,-1)$. And of course we could also say, that $v_3=(-1,-1,2)$ is a basisvector.
So $L_1=span\{v_1\}$ or $L_2=span\{v_2\}$ or $L_3=span\{v_3\}$ would span the solution set for $x+y+z=0$.
But what I'm a bit confused about is, that $L_i\neq L_j$ for $i\neq j$. Should we actually get $L_i=L_j$ for any $i,j$?
What's going on here? What's the difference between the $L_i$? I mean I can see that we can basically rotate the coordinates and get the above $L_i$'s but I'm just not really satisfied with this here.
Can someone write me down the transfomration matrix for this?
$\endgroup$3 Answers
$\begingroup$If you're looking for a basis, you seem to make things more complicated than necessary.
I'd like to find a basis for this.
$x=-y-z$ solves the above system for any $y,z$.
From $\color{blue}{x=-y-z}$, you find that any vector $(\color{blue}{x},y,z) \in V$ is of the form:$$(\color{blue}{x},y,z) = (\color{blue}{-y-z},y,z) = y(-1,1,0)+z(-1,0,1)$$meaning that any element can be written as a linear combination of $(-1,1,0)$ and $(-1,0,1)$. Note that these two vectors are linearly independent, thus they form a basis.
$\endgroup$ 3 $\begingroup$Your question makes no sense. You have $x,y,z\in V\setminus\{0\}$ such that
$x+y+z=0$.
Do you want to compute a basis for $span\{x,y,z\} $ ?
If yes, then observe that $\{x,y,z\} $ is linearly dependent and $x=-y-z$, hence
$span\{x,y,z\}= span\{y,z\}$ .
If $\{y,z\}$ is linearly dependent, then $span\{x,y,z\}= span\{z\}$ and $\{z\}$ is a basis of $span\{x,y,z\} $ .
If $\{y,z\}$ is linearly independent, then $span\{x,y,z\}= span\{y,z\}$ and $\{y,z\}$ is a basis of $span\{x,y,z\} $ .
$\endgroup$ 4 $\begingroup$Since $x+y+z=0$ we can think of\begin{align}A= \begin{pmatrix} 1 & 1 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix} \end{align}and $AX=0$ with $X=(x,y,z)^T$
The basis for nullspace of $A$ is $(1,0,-1)^T$, $(0,1,-1)^T$, so its dimension is $2$.
One can find an orthogonal basis from the Gram-Schmidt process.
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