1. ## Dimension and basis

Find the dimension and a basis of the following vector spaces V over the given field K.
i). V is the set of all vectors $(\alpha,\beta,\gamma)$ in $\mathbb{R}^3$ with $\alpha+4\beta-3\gamma=0; \ K= \mathbb{R}$
$dim_{\mathbb{R}} \mathbb{R}^3=3$.

Is the basis just $V=A(2,0,0)+B(0,1,0)+C(0,0,2)$?
I can't really see how to include the condition in a way better than this.

ii). V is the set of all vectors $(\alpha,\beta, \gamma)$ in $\mathbb{R}^3$ with $\alpha+\beta=\gamma$ and $\alpha-2\beta=-\gamma; \ k= \mathbb{R}$.
Once again $dim_{\mathbb{R}} \mathbb{R}^3=3$.

However, i'm a little puzzled regarding this basis. Would it be:

$V=\alpha(1,0,0) + \beta(0,2,0) + \gamma (0,0,3)$?

2. Originally Posted by Showcase_22
$dim_{\mathbb{R}} \mathbb{R}^3=3$.

Is the basis just $V=A(2,0,0)+B(0,1,0)+C(0,0,2)$?
I can't really see how to include the condition in a way better than this.
But you haven't included the condition at all. A general vector of that form would be (2A, B, 2C) and those do not satisfy (2A)+4(B)-3(2C)= 0. In order that (A, B, C) satisfy A+ 4B- 3C= 0 then you must have A= 3C- 4B. In particular, if C= 1 and B= 0, A= 3. <3, 0, 1> satisfies (3)+ 4(0)-3(1)= 0. Also if C= 0 and B= 1, A= -4. <-4, 1, 0> satisfies (-4)+ 4(1)- 3(0)= 0.
The dimension is NOT 3, it is 2. And I know that because I just wrote a basis.

Once again $dim_{\mathbb{R}} \mathbb{R}^3=3$.

However, i'm a little puzzled regarding this basis. Would it be:

$V=\alpha(1,0,0) + \beta(0,2,0) + \gamma (0,0,3)$?
Once again, such a vector does NOT satisfy the condition you gave.