1
$\begingroup$

Consider the set of (column) vectors defined by $$X = \{x \in \mathbb{R}^3 | x_{1} + x_{2} + x_{3} = 0\},$$ where $X^{T} = [x_{1}, x_{2}, x_{3}]^{T}$, I need to prove whether(or not) given vectors, $[1, -1, 0]^{T}$, $[1, 0, -1]^{T}$ span $X$?

I find examples to prove the same for $\mathbb{R}^3$ but not for a subspace.

$\endgroup$
  • 1
    $\begingroup$ Any $[a,b,c]$ in $X$ can be expressed as $-b[1,-1,0]-c[1,0,-1]$. $\endgroup$ – Kenny Lau Jun 29 '16 at 8:14
  • $\begingroup$ can you suggest a proof $\endgroup$ – Pranav Jun 29 '16 at 8:16
  • 1
    $\begingroup$ Well, because they are. $a=-b-c$, so you only have two parameters. $\endgroup$ – Kenny Lau Jun 29 '16 at 8:17
1
$\begingroup$

Note that $$ \begin{array}{rcl} \mathrm{X}=\{(x_1,x_2,x_3):x_1+x_2+x_3=0\} &=& \{(x_1,x_2,x_3):x_1=-x_2-x_3\} \\ &=& \{(-x_2-x_3,x_2,x_3):x_2,x_3\in\mathbb{R}\}\\ &=& \{x_2(-1,1,0)+x_3(-1,0,1):x_2,x_3\in\mathbb{R}\}\\ &=&\mathrm{span}\{(-1,1,0),(-1,0,1)\}\\ &=&\mathrm{span}\{-(-1,1,0),-(-1,0,1)\}\\ &=&\mathrm{span}\{(1,-1,0),(1,0,-1)\}, \end{array} $$ Aa requared...

$\endgroup$
4
$\begingroup$

Let $\textbf{y} = (y_1,y_2,y_3)^T$ be any vector of $X$. We know that $y_1+y_2+y_3=0$ since $\textbf{y} \in X$.

We need to know if it is always possible to find $a,b\in \mathbb{R}$ such that : $$ a\cdot(1,-1,0)^T + b\cdot (1,0,-1)^T = (y_1,y_2,y_3)^T $$ Developing we get : $$ (1)\ \ \ a+b = y_1 \\ (2)\ \ \ -a = y_2 \\ (3)\ \ \ -b = y_3 $$ Doing $(1) + (2) + (3)$ tells us that $0 = y_1 + y_2 + y_3$ which is always true.

$\endgroup$
1
$\begingroup$

There is one equation and $3$ variables. So clearly there are 2 Free variables. Let $y$ & $z$ be the free variables. Now for first vector, put $y=0$ and $z=1$, you will get $v_1=[-1,0,1]^T$ and after that put $y=1$ and $z=0$, you will get $v_2=[-1,1,0]^T$. So Subspace v can be written as $-:$

$$v=spam([-1,0,1]^T,[-1,1,0]^T)$$ Note that magnitude of these vectors are highly irrelavant. So you can multiply $v_1$ & $v_2$ with any non-zero number.

Hope this will be helpful !

$\endgroup$
1
$\begingroup$

The vectors $\begin{bmatrix} 1 \\ -1 \\ 0\end{bmatrix}$ and $\begin{bmatrix} 1 \\ 0 \\ -1\end{bmatrix}$ are in the subspace of $\mathbb{R}^3$ where $x_1+x_2+x_3=0$ for vectors of the form $\begin{bmatrix} x_1 \\ x_2 \\ x_3\end{bmatrix}$.

Note that this subspace has dimension $2$ (since $x_3=-x_2-x_1$, it has only two vectors in it's basis). This could also be inferred by noting that in the standard basis for $\mathbb{R}^3$, $f(x)=x_1+x_2+x_3$ is a linear functional and $x_1+x_2+x_3=0$ is the null space annihilated by this functional and it is a hyperspace, that is of dimension $3-1=2$.

Since the vectors $\begin{bmatrix} 1 \\ -1 \\ 0\end{bmatrix}$ and $\begin{bmatrix} 1 \\ 0 \\ -1\end{bmatrix}$ are also linearly independent (their linear combination is zero only in the trivial case), they necessarily form a basis and hence span the space.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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