Is S a Subspace of $\mathbb{R}^n$ Defined by Linear Combinations of Vectors?

[danago]

Let \vec{u},\vec{v},\vec{w} be fixed vectors in Rⁿ. Define S to be the set of all vectors in Rⁿ which are linear combinations of the form k_1 \vec{u}+k_2 \vec{v}+3 \vec{w}, where k_1,k_2 \in R. Is S a subspace of Rⁿ?

Im a little stuck with this one. I've tried defining two vectors, \vec{x},\vec{y} \in S and then forming a linear combination of the two, to get:

<br /> a\overrightarrow x + b\overrightarrow y = (ak_1 + bc_1 )\overrightarrow u + (ak_2 + bc_2 )\overrightarrow v + (3a + 3b)\overrightarrow w <br />

Where:

<br /> \begin{array}{l}<br /> \overrightarrow x = k_1 \overrightarrow u + k_2 \overrightarrow v + 3\overrightarrow w \\ <br /> \overrightarrow y = c_1 \overrightarrow u + c_2 \overrightarrow v + 3\overrightarrow w \\ <br /> \end{array}<br />

Thats where I am lost; I am not even sure if I've taken the right approach. From this i can see that the linear combination of vectors x and y results in an expression containing linear combinations of vectors u and v, but its the w vector that's causing me problems.

Any hints are greatly appreciated

Thanks,
Dan.

 

[n_bourbaki]

Why is it causing problems? Is the result of adding those two vectors together in the given set?

 

[danago]

Why is it causing problems? Is the result of adding those two vectors together in the given set?

Well i would be inclined to say no since the linear combination of x and y is not explicitly of the same form as the vectors in S, but i didnt think it was as simple as that.

 

[n_bourbaki]

When can you make it of the same form? When can't you make it of the same form?

 

[danago]

When can you make it of the same form? When can't you make it of the same form?

Well if a+b=1 then it will be of the same form. Is that what you are getting at?

 

[n_bourbaki]

No, since we have to do this for all a and b.

It may help you to think about whether the 0 vector is in the set or not.

 

[danago]

When can you make it of the same form? When can't you make it of the same form?

No, since we have to do this for all a and b.

It may help you to think about whether the 0 vector is in the set or not.

Well if w was the zero vector itself, then yes the zero vector is in the set (k₁,k₂=0). It is also in the set when w is linearly dependent on u and v?

Hmm alright if w is linearly dependent on u and v, then the expression CAN be written in the form required. I could write the (3a+3b)w=Kw (K=3a+3b) part as 3w + (K-3)w, and then write the w in terms of u and v, which would put it in the correct form.

If it is linearly independent, then would i be correct to say that it is not in the set?

So is that what you meant? There are two cases for this question?

 

[n_bourbaki]

Yes there are two cases. What is your geometric visualisation for this question? If you think about what that set describes, then it is clear that it is a subspace if and only if w is in the plane spanned by u and v.

What is the set? It is the plane spanned by u and v translated by 3w. This can only be a plane through the origin if 3w (equivalently w) is in the span of u and v