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...
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...
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?
$\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.
$\displaystyle A= \begin{bmatrix} 2 & -7 & 3 \end{bmatrix}, \vec{a}=\begin{bmatrix} a_1\\a_2\\a_3 \end{bmatrix}, W=\{A\vec{a}=\vec{0}\}=ker(A)$
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.