0
$\begingroup$

I want to prove that two vector-subspace are equal by showing that there dimensions are equal where $W = \{(x,y,z)\mid (z - 3y +3x = 0)\}$ and $U = \operatorname{Span}\{((1,2,3),(-1,2,9))\}$. I tried to find basis for first one $W = (0,-3,3)$ and $(3,-3,0)$ by using guideline from here and this process is totally based on plugging in values for different variables and then controlling the effect on other variables.

So is there any better way to find basis of a vector-space same as $W$?

For the second part do I need to do anything?

I ask this because suppose I have three vectors (which span a vector-subspace) then I have to control whether they are linearly independent or not and the number of retained vectors is dimension of space spanned by those vectors but above I have only two vectors so I think i don't need to control!

$\endgroup$

1 Answer 1

Reset to default
2
$\begingroup$

Guide:

  • Check whether $(1,2,3)$ and $(-1,2,9)$ satisfies $z-3y+3x=0$.

  • Check whether $(1,2,3)$ and $(-1,2,9)$ are linearly independent.

  • Find the dimension of the solution space to $z-3y+3x=0$.

$\endgroup$
2
  • $\begingroup$ Can you add a little bit explaination by step 3 , please? $\endgroup$
    – Luai Ghunim
    Dec 23, 2017 at 17:48
  • $\begingroup$ sorry for the late reply as I am travelling. we can treat $y$ and $x$ as free variable and the solution can be written as $(x,y,z)=x(1,0,-3) + y(0,1,3).$ The dimension are $2$ as we have two free variables. $\endgroup$
    – Siong Thye Goh
    Dec 29, 2017 at 4:58

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .