# Prove that $B=\{(1,1,4), (1,1,3), (1,-1,2)\}$ is a basis of vector space $\Bbb R^{3}$.

Prove that $B=\{(1,1,4), (1,1,3), (1,-1,2)\}$ is a basis of vector space $\Bbb R^{3}$.

To prove this, I have to prove whether this set is linearly independent and whether the vectors in $B$ span $\Bbb R^{3}$.

$1)$ $\alpha(1,1,4)+\beta(1,1,3)+\gamma(1,-1,2)=0$

$(\alpha, \alpha, 4\alpha)+(\beta,\beta,3\beta)+(\gamma,-\gamma,2\gamma)=0$

The system

$\alpha+\beta+\gamma=0$

$\alpha+\beta-\gamma=0$

$4\alpha+3\beta+2\gamma=0$ has the solution $\alpha=\beta=\gamma=0$

So these vectors $(1,1,4), (1,1,3), (1,-1,2)$ are linearly independent.

$2)$ Now I'm not sure how to check these vectors span $\Bbb R^{3}$. I was told to do this:

$\alpha(1,1,4)+\beta(1,1,3)+\gamma(1,-1,2)=(x,y,z)$ $\Rightarrow$

$\alpha+\beta+\gamma=x$

$\alpha+\beta-\gamma=y$

$4\alpha+3\beta+2\gamma=z$

I'm showing that any vector in $\Bbb R^{3}$ can be represented as linear combination of those three given vectors, is that right?

What should I do with this system of equations?

$2)$ Now I'm not sure how to check these vectors span $\Bbb R^{3}$. I was told to do this:

$\alpha(1,1,4)+\beta(1,1,3)+\gamma(1,-1,2)=(x,y,z)$ $\Rightarrow$

$\alpha+\beta+\gamma=x$

$\alpha+\beta-\gamma=y$

$4\alpha+3\beta+2\gamma=z$

I'm showing that any vector in $\Bbb R^{3}$ can be represented as linear combination of those three given vectors, is that right?

What should I do with this system of equations?

Now you have to solve this system for $\alpha,\beta,\gamma$ (see below). If this system has a solution, for arbitrary $x,y,z$; then you have shown that you can find suitable values of $\alpha,\beta,\gamma$ to 'make' any $(x,y,z)$ as a linear combination of the given vectors. They then span $\mathbb{R^3}$.

Note that this is the hard way, staying close to the definition of 'basis', checking if the vectors:

• are linearly independent;
• span the space.

When or if you have seen some relevant properties/theorems, you'll know that 3 linearly independent vectors will always span a 3-dimensional vector space - a lot easier.

For future reference and perhaps you can check:

$$\left\{ \begin{array}{l} \alpha+\beta+\gamma=x \\ \alpha+\beta-\gamma=y \\ 4\alpha+3\beta+2\gamma=z \end{array} \right. \iff \left\{ \begin{array}{l} \alpha=-\tfrac{5}{2}x-\tfrac{1}{2}y+z \\ \beta=3x+y-z \\ \gamma=\tfrac{1}{2}x-\tfrac{1}{2}y \end{array} \right.$$

Three linearly independents vectors in $\mathbb R^3$ obviously span $\mathbb R^3$. You just need to verify the linear independence.

• What do you mean obviously? – lmc Nov 7 '16 at 10:11
• Well the dimension of $\mathbb{R}^3$ is three, so if you have a set of three linearly independent vectors in $\mathbb{R}^3$, then they will always span this set. If this isn't sufficient for your proof, then you can, like StackTD was explaining, show that any arbitrary vector in $\mathbb{R}^3$ can be constructed by these three linearly independent vectors. – AndroidFish Nov 7 '16 at 18:04