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Orthogonal Basis for a subspace

  1. May 29, 2012 #1
    1. The problem statement, all variables and given/known data

    Let [itex]W = \begin{cases} \begin{pmatrix}x\\y\\z\\w\end{pmatrix} \in R^4 | w + 2x + 2y + 4z = 0 \end{cases}[/itex]

    A)Find basis for [itex]W[/itex].
    B)Find basis for [itex]W^{\perp}[/itex]
    C)Use parts (A) and (B) to find an orthogonal basis for [itex]R^4[/itex] with
    respect to the Euclidean inner product.

    2. Relevant equations

    ..

    3. The attempt at a solution

    So, part A is done and part B, I used the Gram-Shmidt process to find the basis of [itex]W^{\perp}[/itex]. So far okay? But part C is a little vague to me. How do I approach this?
     
  2. jcsd
  3. May 29, 2012 #2

    HallsofIvy

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    If you used Gram-Schmidt for A and B, then you have two orthonormal bases for W and W and the union of the two bases is a basis for R4.
     
  4. May 30, 2012 #3
    Oh thank you.

    For part A), the basis I found was [itex]B_W = \begin{pmatrix}1\\0\\0\\-2\end{pmatrix}, \begin{pmatrix}0\\1\\0\\-2\end{pmatrix}, \begin{pmatrix}0\\0\\1\\-4\end{pmatrix}[/itex]

    For part B), [itex]v = \begin{pmatrix}x\\y\\z\\w\end{pmatrix}[/itex] is in the orthogonal compliment of [itex]W[/itex] if and only if [itex]v \cdot u_is = 0[/itex] where the [itex]u_is[/itex] are the vectors of the basis of [itex]W[/itex].
    We find that the basis for [itex]W^{\perp} = \begin{pmatrix}2\\2\\4\\1\end{pmatrix}[/itex]. Is this the way to solve the first two parts?

    Now for part C), let [itex]S[/itex] be the set of vectors [itex]S = \{u_1, u_2, u_3, v\}[/itex]. We need to now use the Gram-Shmidt process on the basis of parts A and B then unionize the vectors as one set. Does this make [itex]S[/itex] the basis for [itex]R^4[/itex]?

    I might be confused here, but my teacher, my friend and the book all say different things.
     
    Last edited: May 30, 2012
  5. May 30, 2012 #4
    The choice of basis has a lot of room for choice, so it's possible that everyone has different answers but is correct. You might try to understand how they chose their bases.
     
  6. May 30, 2012 #5

    HallsofIvy

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    algebrat is completely correct- there are an infinite number of possible bases for any vector space.
    What you have done, for others who might want to follow this, is say that, since you are requiring that w+ 2x+ 2y+ 4z= 0, so that w= -2x- 2y- 4z, we can write
    [tex]\begin{pmatrix}x \\ y \\ z\\ w\end{pmatrix}= \begin{pmatrix}1 \\ y \\ z\\ -2x- 2y- 4z\end{pmatrix}[/tex]
    [tex]= \begin{pmatrix}x \\ 0 \\ 0 \\ -2x\end{pmatrix}+ \begin{pmatrix}0 \\ y \\ 0 \\ -2y\end{pmatrix}+ \begin{pmatrix}0 \\ 0 \\ z \\-4z\end{pmatrix}= x\begin{pmatrix}1 \\ 0 \\ 0 \\ -2\end{pmatrix}+ y\begin{pmatrix}0 \\ 1 \\ 0 \\ -2\end{pmatrix}+ z\begin{pmatrix}0 \\ 0 \\ 1 \\ -4\end{pmatrix}[/tex]
    showing clearly what the basis vectors are.

    Any vector (a, b, c, d) in the "orthogonal complement" of that space must have dot product with any vector in the space, and in particular with any basis vector, equal to 0. So we must have [itex](a, b, c, d)\cdot (1, 0, 0, -2)= a- 2d= 0[/itex], [itex](a, b, c, d)\cdot (0, 1, 0, -2)= b- 2d= 0[/itex], and [itex](a, b, c, d)\cdot (0, 0, 1, -4)= c- 4d= 0[/itex]. In other words, a= 2d, b= 2d, and c= 4d so that any vector in this space can be written as
    [tex]\begin{pmatrix}a \\ b \\ c \\ d\end{pmatrix}= \begin{pmatrix}2d \\ 2d \\ 4d \\ d\end{pmatrix}= d\begin{pmatrix}2 \\ 2 \\ 4\\ 1\end{pmatrix}[/tex].

    Use the Gram-Schmidt process to convert your given basis for W to an orthonormal basis. If your only purpose is to find an orthogonal basis for Rn, you don't really need to do anything to the single vector in the basis for W is already perpendicular to all three vectors in the basis for W. If you want an orthonormal basis you can divide (2, 2, 4, 1) by its length, 5.
     
  7. May 30, 2012 #6
    Finally, a CLEAR explanation. Thank you, HallsofIvy and algebrat.
     
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