Can anyone help me with this elementary problem? How do I prove the following is a subspace?

W = {(a1, a2, a3) | 2*a1 - 7*a2 +a3 = 0}

I know I have to prove its closed under addition and multiplication... But the condition on W is confusing me...

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- Jun 1st 2009, 05:17 PMpaupsersProving this is a subspace
Can anyone help me with this elementary problem? How do I prove the following is a subspace?

W = {(a1, a2, a3) | 2*a1 - 7*a2 +a3 = 0}

I know I have to prove its closed under addition and multiplication... But the condition on W is confusing me... - Jun 1st 2009, 05:26 PMpaupsers
I think I just solved it, actually!

I let a3 = 7*a2 - 2*a1, and made that the new condition on W.

I then let a = (a1, a2, 7*a2 - 2*a1) and similarly for b.

Then I did a + cb = (a1 + c*b1, a2 + c*b2, 7(a2 + c*b2) - 2(a1 + c*b1))

Which satisfies my new condition on W I believe...

Can anyone verify this? - Jun 1st 2009, 05:36 PMRandom Variable
$\displaystyle W = ({a_{1},a_{2},-2a_{1}+7a_{2}} $)

Let $\displaystyle X = (x_{1},x_{2},-2x_{1}+7x_{2}) $ and $\displaystyle Y = (y_{1},y_{2},-2y_{1}+7y_{2}) $ be in $\displaystyle W $

then $\displaystyle X+Y= (x_{1}+y_{1},x_{2}+y_{2},-2x_{1}+7x_{2}-2y_{1}+7y_{2}) $ $\displaystyle = \Big (x_{1}+y_{1},x_{2}+y_{2},-2(x_{1}+y_{1})+7(x_{2}+y_{2}) \Big) $ which is in $\displaystyle W $

I'm sure you can then prove that it's also closed under scalar multiplication. - Jun 1st 2009, 05:42 PMRandom Variable
If you want to prove both at the same time, you have to show that $\displaystyle \alpha X + \beta Y $ is in $\displaystyle W $

- Jun 1st 2009, 06:37 PMmath2009
$\displaystyle A= \begin{bmatrix} 2 & -7 & 3 \end{bmatrix}, \vec{a}=\begin{bmatrix} a_1\\a_2\\a_3 \end{bmatrix}, W=\{A\vec{a}=\vec{0}\}=ker(A)$

Obviously, $\displaystyle T(\vec{x})=A\vec{x}$ is linear transformation

$\displaystyle 1.\ \vec{x}_1,\vec{x}_2 \in W, T(\vec{x}_1+\vec{x}_2)=A(\vec{x}_1+\vec{x}_2)=A(\v ec{x}_1)+A(\vec{x}_2)=\vec{0}+\vec{0}=\vec{0}, so\ \vec{x}_1+\vec{x}_2 \in W$

$\displaystyle 2.\ k\in R, T(k\vec{x})=A(k\vec{x})=kA\vec{x}=k\vec{0}=\vec{0} , so\ k\vec{x}\in W$

$\displaystyle 3.\ A\vec{0}=\vec{0}, so\ \vec{0}\in W$

Finally, W is a subspace. - Jun 1st 2009, 06:47 PMIsomorphism