User:Tonyxc600/sandbox

$$

S = \{(1,1,1), (3,1,4), (3,1,5)\}

$$

Putting the vectors in set S into rows of a matrix A:

$$ A = \begin{bmatrix} 1 & 1 & 1 \\ 3 & 1 & 4 \\ 3 & 1 & 5 \end{bmatrix}

\sim \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}

$$

rank(A) = 3, and there are 3 elements in S. Hence, S is linearly independant.

Now let x be some vector in $R^{3}$, where

$$x = (a, b , c)$$

Now we set up the linear equation:

$$

\alpha (1,1,1) + \beta (3,1,4) + \gamma (3,1,5) = (a,b,c) $$

In an augmented matrix,

$$ \left[ \begin{array}{ccc|c} 1 & 3 & 3 & a \\ 1 & 1 & 1 & b \\ 1 & 4 & 5 & c \end{array} \right]

\sim \left[ \begin{array}{ccc|c} 1 & 0 & 0 & \frac{-a}{2} + 2b - \frac{3c}{2} \\ 0 & 1 & 0 & \frac{3a}{2}-b+ \frac{c}{2} \\ 0 & 0 & 1 & c-b \end{array} \right]

$$

Hence it is clear that a solution exists for any $(a,b,c) ∈ R^{3}$.

Therefore, the set S is a spanning set of $R^{3}$.

Now we have S is both linearly independant and a spanning set of $R^{3}$. Therefore, by definition, S is a basis of $R^{3}$

tonyxc600    comms        logs     11:59, 3 September 2016 (UTC)