# Linear Algebra

1. Jul 17, 2008

### 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. Ive tried defining two vectors, $$\vec{x},\vec{y} \in S$$ and then forming a linear combination of the two, to get:

$$a\overrightarrow x + b\overrightarrow y = (ak_1 + bc_1 )\overrightarrow u + (ak_2 + bc_2 )\overrightarrow v + (3a + 3b)\overrightarrow w$$

Where:

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

Thats where im lost; im not even sure if ive 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 thats causing me problems.

Any hints are greatly appreciated

Thanks,
Dan.

2. Jul 17, 2008

### n_bourbaki

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

3. Jul 17, 2008

### danago

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.

4. Jul 17, 2008

### n_bourbaki

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

5. Jul 17, 2008

### danago

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

6. Jul 17, 2008

### 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.

7. Jul 17, 2008

### danago

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?

8. Jul 17, 2008

### 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