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Math Help - Give a basis of a subspace

  1. #1
    MHF Contributor arbolis's Avatar
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    Give a basis of a subspace

    Hi MHF,
    We have the following vectorial subspace W pertaining to \mathbb{R}^4 such that W=\{ (x,y,z,w) \in \mathbb{R}^4  : 2x-z+w=0, x-y-z=0 \}.
    1)Find a basis of W.
    2)Find a basis of W orthogonal.
    My attempt :
    1)I just wrote the 2 equations defining W as a matrix and chosen the 2 first column vectors of the matrix as being a basis of W.
    Precisely I got that a basis of W is {(2,1),(0,-1)}. (0,-1) being the last column vector of the matrix I formed.
    2)I'm not 100% sure about what I've done here. I chosen the basis {(2,1),(1,0)}.

    How can I check if I'm right on this? I guess I could form a matrix with the vectors in the basis I gave and try to get a null row or column when reducing it. But I'm not sure, and not sure how to form the matrix too.
    Can you help me?
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  2. #2
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    Quote Originally Posted by arbolis View Post
    Hi MHF,
    We have the following vectorial subspace W pertaining to \mathbb{R}^4 such that W=\{ (x,y,z,w) \in \mathbb{R}^4 : 2x-z+w=0, x-y-z=0 \}.
    1)Find a basis of W.
    2)Find a basis of W orthogonal.
    My attempt :
    1)I just wrote the 2 equations defining W as a matrix and chosen the 2 first column vectors of the matrix as being a basis of W.
    Precisely I got that a basis of W is {(2,1),(0,-1)}. (0,-1) being the last column vector of the matrix I formed.
    2)I'm not 100% sure about what I've done here. I chosen the basis {(2,1),(1,0)}.

    How can I check if I'm right on this? I guess I could form a matrix with the vectors in the basis I gave and try to get a null row or column when reducing it. But I'm not sure, and not sure how to form the matrix too.
    Can you help me?
    to find a basis for W, just solve the equations: \begin{cases}2x-z+w=0 \\ x-y-z=0 \end{cases}, which gives you: z=x-y, \ w=-x-y. so a typical element of W is: \begin{bmatrix}x \\ y \\ x-y \\ -x-y \end{bmatrix}=x \begin{bmatrix}1 \\ 0 \\ 1 \\ -1 \end{bmatrix} + y \begin{bmatrix} 0 \\ 1 \\ -1 \\ -1 \end{bmatrix}=xu_1+yu_2.

    thus \{u_1,u_2\} is a basis for W. to find a basis for W^{\perp}, an easy way is to look for all vectors v=\begin{bmatrix} x \\ y \\ z \\ w \end{bmatrix} such that v \cdot u_1=v \cdot u_2=0, i.e. \begin{cases}x+z-w=0 \\ y-z-w=0 \end {cases}. that gives us: z=\frac{y-x}{2}, \ w=\frac{x+y}{2}.

    so a typical element of W^{\perp} is: \begin{bmatrix} x \\ y \\ \frac{y-x}{2} \\ \frac{x+y}{2} \end{bmatrix}=\frac{x}{2}\begin{bmatrix} 2 \\ 0 \\ -1 \\ 1 \end{bmatrix}+\frac{y}{2}\begin{bmatrix} 0 \\ 2 \\ 1 \\ 1 \end{bmatrix}=\frac{x}{2}u_3 + \frac{y}{2}u_4. thus \{u_3,u_4 \} is a basis for W^{\perp}.
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  3. #3
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    2x-z+w=\begin{bmatrix} 2 \\ 0 \\-1 \\ 1 \end{bmatrix}.\begin{bmatrix} x \\ y \\z \\ w \end{bmatrix}=0,normal vector \vec{v}_1=\begin{bmatrix} 2 \\ 0 \\-1 \\ 1 \end{bmatrix}<br />
,x-y-z=0, normal vector \vec{v}_2=\begin{bmatrix} 1 \\ -1 \\ -1 \\ 0 \end{bmatrix}<br />

    W=\{ (x,y,z,w) \in \mathbb{R}^4 : 2x-z+w=0, x-y-z=0 \}<br />
, so W is intersection line of two hyperplanes.

    We solve equation A\vec{x}=\vec{0}, A=\begin{bmatrix} 2&0&-1&1\\ 1&-1&-1&0  \end{bmatrix},<br />
W=ker (A) ,rref (A)=\begin{bmatrix} 1&0&-1/2&1/2\\ 0&1&1/2&1/2  \end{bmatrix},
    rank(A) = Dim(W)=2, so there are two linear independent vectors in basis.

    W=\begin{bmatrix} (z-w)/2 \\ -(z+w)/2 \\ z \\ w \end{bmatrix}<br />
=z\begin{bmatrix} 1/2 \\ -1/2 \\ 1 \\ 0 \end{bmatrix}+w\begin{bmatrix} -1/2 \\ -1/2 \\ 0 \\ 1 \end{bmatrix}, basis = (\begin{bmatrix} \frac{1}{2}\\ -\frac{1}{2} \\ 1 \\ 0 \end{bmatrix},\begin{bmatrix} -\frac{1}{2}\\ -\frac{1}{2} \\ 0 \\ 1 \end{bmatrix}) <br />

    by the way, dimension is four. if it's 3, it's easy to calc by W = span(v1 x v2)

    **********
    <br />
W^{\perp}<br />
is hyperplane orthogonal to intersection line W. Its basis is two linear independent normal vectors (\vec{v}_1,\vec{v}_2).<br />
W^{\perp} basis = (\begin{bmatrix} 2 \\0 \\ -1 \\ 1 \end{bmatrix},\begin{bmatrix} 1 \\ -1 \\ -1 \\ 0 \end{bmatrix})<br />

    There is no union solution. This answer is the same with **NonCommAlg's Avatar** above.
    Last edited by math2009; February 10th 2009 at 05:06 PM.
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  4. #4
    MHF Contributor arbolis's Avatar
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    Thank you very much to both. I'll study this in detail today.
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  5. #5
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    Without using matrices: You have two equations, 2x-z+w=0 and x-y-z=0 in 4 variables. The equations are independent so we can solve for two of the variables in terms of the other 2 meaning that this is a two dimensional subspace of the four dimensional R^4.

    We get z= x- y from the second equation and, putting that into the first, 2x-(x-y)+ w= x+ y+ w= 0 so w= -x-y. If we take x= 1, y= 0, z= 1, w= -1 so (1, 0, 1, -1) is a vector in that subspace. Taking x= 0, y= 1, z= -1, w= -1 so (0, 1, -1, -1) is another vector in that subspace. Taking x=1, y= 0 and x= 0, y= 1 ensures that they are independent vectors so {(1, 0, 1, -1), (0, 1, -1, -1)} is a basis for the subspace W.

    Vectors in the orthogonal subspace to W must be perpendicular to every vector in W. Since we already know that 2x- z+ w= 2(x)+ 0(y)+ (-1)z+ (1)w= 0 and x-y-z= 1(x)+ (-1)y+ (-1)z+ 0(w)= 0, it is clear that (2, 0, -1, 1) and (1, -1, -1, 0) are in that orthogonal orthogonal space and it is clear that they are independent so {(2, 0, -1, 1), (1, -1, -1, 0)} is a basis for the orthogonal subspace to W
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  6. #6
    MHF Contributor arbolis's Avatar
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    To NCA : I could follow all you put until here :
    that gives us:
    . Could you please tell me how did you get these values for z and w?

    To HallsofIvy : Nicely saw. Especially the second part. Thank you.
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  7. #7
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    To add two equations: \begin{cases}x+z-w=0 \\ y-z-w=0 \end {cases} \rightarrow (x+z-w) + (y-z-w)=0 \rightarrow x+y-2w=0 \rightarrow w=\frac{x+y}{2}
    Last edited by math2009; February 10th 2009 at 03:55 PM.
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