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Linear Algebra

  1. Jul 17, 2008 #1


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

    Im a little stuck with this one. Ive tried defining two vectors, [tex]\vec{x},\vec{y} \in S[/tex] 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


    \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 \\

    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 :smile:

  2. jcsd
  3. Jul 17, 2008 #2
    Why is it causing problems? Is the result of adding those two vectors together in the given set?
  4. Jul 17, 2008 #3


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    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.
  5. Jul 17, 2008 #4
    When can you make it of the same form? When can't you make it of the same form?
  6. Jul 17, 2008 #5


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    Well if a+b=1 then it will be of the same form. Is that what you are getting at?
  7. Jul 17, 2008 #6
    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.
  8. Jul 17, 2008 #7


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    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?
  9. Jul 17, 2008 #8
    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
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