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Find a basis for the subspace $W=\{[x,y,z]\mid 3x+5y+7z=0\}$ of $\mathbb{R}^3$. What is $dim(W)$?

This question is really throwing me off. I am interpreting the subspace as the set of all linear combinations $cX$, where $X=[x,y,z]$ and solves $3x+5y+7z=0$, for all $c$ in $\mathbb{R}$. So it would be a line in $\mathbb{R}^3$ geometrically I believe. So the basis would be any $X=[x,y,z]$ which solves $3x+5y+7z=0$, correct? Also, $dim(W)=1$, correct?

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It's a plane. Being the null space of a $1\times 3$ matrix, we can look at the rank theorem. The matrix is rank $1$, and has $3$ columns, so the dimension of its null space is $2$. –  G Tony Jacobs Mar 30 at 19:06

3 Answers 3

up vote 3 down vote accepted

We have $$3x+5y+7z=0\iff x=-\frac53y-\frac73z$$ hence $$u=(x,y,z)\in W\iff u=\left(-\frac53y-\frac73z,y,z\right)\\\iff u=y\underbrace{\left(-\frac53,1,0\right)}_{=v_1}+z\underbrace{\left(-\frac73,0,1\right)}_{=v_2}=yv_1+zv_2$$ hence $$W=\operatorname{span}(v_1,v_2)$$ and since $v_1$ and $v_2$ are linearly independent then $\dim W=2$.

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Here's a quick solution:

Set $x = 0$.
$5y + 7z = 0$ defines a line of solutions in $\textbf{R}^3$.

Now set $z = 0$.
$3x + 5y = 0$ defines a different line of solutions in $\textbf{R}^3$.

We can conclude that $\dim(W) > 1$ since our subspace cannot be a single line. However, $\dim(W)<3$, otherwise it would be spanned by $3$ linearly independent vectors (which would also span $\textbf{R}^3$). However, note that not every vector in $\textbf{R}^3$ is a solution to $3x + 5y + 7z = 0.$ Thus, $\dim(W) = 2$ by a quick process of elimination.

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Since it seems that it is a homework, I suggested to add the tag. Also, it would be good for you to check the definition of subspace (closed under addition and scalar mutliplication, and contains $0$).

Then, you should see $W$ as a set of points $(x,y,z)$ that satisfy the equation. You could for example try to find a few examples and then try to get all solutions (for example expressing one coordinates from others). This could then help you to find how many parameters you need, and then to find a basis, and then prove that it is a basis of $W$.

Hint: $Dim(W)\not=1$, and your set is not a line.

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I marked it as invalid. Read the tag description: "Please do not add this tag to questions by other people unless they explicitly say that their question is part of their homework." –  user2345215 Mar 30 at 19:09
    
Ok. I thought this was obvious, but maybe not. Sorry for that. –  Jérémy Blanc Mar 30 at 20:19
    
Maybe, maybe not. But the description specifically says not to add. But you can always ask the OP to add it himself in a comment :) –  user2345215 Mar 30 at 20:23
    
Ok, I will do this next time. –  Jérémy Blanc Mar 30 at 20:23

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