# Vectors forming a Subspace

1. Dec 9, 2009

1. The problem statement, all variables and given/known data
Determine whether or not the set of vectors:

$$\left\{\bar{x}=t \left( \begin{array}{cc}1\\2\\1\end{array} \right) +s\left( \begin{array}{cc}1\\1\\1\end{array} \right) +\left( \begin{array}{cc}1\\0\\2\end{array} \right),-\infty < t,s < \infty \right\}$$

is a subspace.

I really have no idea how to go about this. Do I just check if they are linear combinations of each other?

2. Dec 9, 2009

### Dick

What properties does a set of vectors have to have to be a subspace? Look it up and start checking the properties.

3. Dec 9, 2009

1) If A belongs to W, and B belongs to W, then A+B belongs to W
2) If A belongs to W and r is in the space Rn, then rA belongs to W

I don't understand how this applies to the problem. How can apply property 1 for example if I don't know if A or B belongs to W to begin with?

4. Dec 9, 2009

### Dick

For the second property, you mean "r is in R", I think. That means if W is a vector space then 0*A=(0,0,0) is in the space. That's usually listed as one of the conditions. I would first try checking if (0,0,0) is in W. Is there any choice of s and t that will produce the vector (0,0,0)?

5. Dec 9, 2009

No, because when you row-reduce the matrix all the columns are linearly independent, which means they can only equal zero if their coefficients are zero. Since vector three doesn't have a coefficient, they can never equal zero. Therefore, they are not a subspace.

Is my reasoning correct?

6. Dec 9, 2009

### Dick

Sure. If the zero vector isn't in your space, then it's not a subspace.

7. Dec 9, 2009