# Subspace basis and dimension.

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 '14 at 19:06

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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