Is the Given Set of Vectors a Subspace?

[Raddy13]

Homework Statement

Determine whether or not the set of vectors:

\left\{\bar{x}=t<br /> \left(<br /> \begin{array}{cc}1\\2\\1\end{array}<br /> \right)<br /> <br /> +s\left(<br /> \begin{array}{cc}1\\1\\1\end{array}<br /> \right)<br /> <br /> +\left(<br /> \begin{array}{cc}1\\0\\2\end{array}<br /> \right),-\infty &lt; t,s &lt; \infty<br /> \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?

 

[Dick]

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

 

[Raddy13]

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?

 

[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)?

 

[Raddy13]

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?

 

[Dick]

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?

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

 

[Raddy13]

But that is how I would go about solving it on a test? Just row reduce the vectors and determine whether or not a zero vector is possible?

 

[Dick]

That's not the only thing. But it's a good starting point. Then you also have to prove the set has closure under addition, if a and b are in W then a+b is in W, and under scalar multiplication, a is in W then r*a is in W. Just what you said. But the zero vector is a good starting point. If zero isn't in the set, you can forget the other two.