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Math Help - 2 subspaces of R^4, bases, intersection

  1. #1
    MHF Contributor arbolis's Avatar
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    2 subspaces of R^4, bases, intersection

    Hi MHF,
    I appreciate very much all the help I got so far. If I didn't answered my last 2 posts it's because of lack of time. I feel overloaded with linear Algebra and the final exam is coming up next 26th of February.
    Here is the exercise I'd like to check out with you.
    Let S_1=\{ (1,0,1,2), (-1,2,3,8), (-1,1,1,3), (0,1,2,5) \} and S_2=\{ (1,1,0,1), (1,-1,0,3), (3,1,2,4), (1,1,2,0) \} and let W_1 be spanned by S_1 and W_2 be spanned by S_2.
    a)Give a base B_i of W_i contained in S_i, for i=1, 2.
    My attempt : I wrote each vector of the base into a matrix as being column vectors and reduced it. I got that 3 of the 4 vectors of S_1 were colinear and then I solve a 2 equations system, getting finally that a base of W_1 included into S_1 is \{ (1,0,1,-1), (0,1,-1,1) \}.
    Doing the same for S_2, I got that a base of W_2 included into S_2 is \{ (1,0,0,0), (0,-1,1,-1)  \}.
    b)Describe W_1, W_2, W_1 \cap W_2 and W_1+W_2 implicitly.
    My attempt : don't know how to start.
    c)Give a base of W_1 \cap W_2 and of W_1+W_2.
    My attempt : I wrote the bases I got in part a) as vector column in a matrix. I checked out that \det A \neq 0, hence the matrix is invertible and the 4 vectors are linear independent. So I think that W_1 \cap W_2 = ∅. So there's no base that generates W_1 \cap W_2, or I'm wrong?
    And as the 4 vectors are linear independent, a basis of W_1+W_2 would be the union of the bases of W_1 and W_2. Or : \{ (1,0,1,-1), (0,1,-1,1),(1,0,0,0), (0,-1,1,-1)  \}. Am I right?
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  2. #2
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    a. A_1=[\vec{S}_{11}\ \ \vec{S}_{12}\ \ \vec{S}_{13}\ \ \vec{S}_{14}]=\begin{bmatrix} 1&-1&-1&0\\0&2&1&1\\1&3&1&2\\2&8&3&5 \end{bmatrix}, A_2=[\vec{S}_{21}\ \ \vec{S}_{22}\ \ \vec{S}_{23}\ \ \vec{S}_{24}]=\begin{bmatrix} 1&1&3&1\\1&-1&1&1\\0&0&2&2\\1&3&4&0 \end{bmatrix}

    rref(A_1)=\begin{bmatrix} 1&0&-\frac{1}{2}&\frac{1}{2}\\0&1&\frac{1}{2}&\frac{1}{  2}\\0&0&0&0\\0&0&0&0 \end{bmatrix},\ rref(A_2)=\begin{bmatrix} 1&0&0&-1\\0&1&0&-1\\0&0&1&1\\0&0&0&0 \end{bmatrix}

    quantity\ of\ vectors\ in\ basis:\ dim(W_1)=rank(A_1)=2, dim(W_2)=rank(A_2)=3 ,
    you can select \mathfrak{B}_1=(\vec{S}_{11},\vec{S}_{12})\ as\ basis\ of \ W_1,\ \mathfrak{B}_2=(\vec{S}_{21},\vec{S}_{22},\vec{S}_  {23})\ as\ basis\ of \ W_2

    b. we suppose \vec{v}_1,\cdots ,\vec{v}_p, \vec{w}_1,\cdots ,\vec{w}_q , \vec{b}_1,\cdots ,\vec{b}_m \in R^n \ linear\ independent and W_1=span(\vec{v}_1,\cdots ,\vec{v}_p, \vec{b}_1,\cdots ,\vec{b}_m),\ W_2=span(\vec{w}_1,\cdots ,\vec{w}_q, \vec{b}_1,\cdots ,\vec{b}_m)

    Then, dim(W_1)=p+m,\ dim(W_2)=q+m,\ dim(W_1+W_2)=p+q+m,\ dim(W_1\cap W_2)=m
    Obviously, dim(W_1) + dim(W_2) -dim(W_1\cap W_2)=dim(W_1+W_2)=p+q+m

    c. A_4=\frac{[ \vec{S}_{11}\ \ \vec{S}_{12}}{\mathfrak{B}_1}\ \ \frac{\vec{S}_{21}\ \ \vec{S}_{22}\ \ \vec{S}_{23}]}{\mathfrak{B}_2}, rref(A_4)=\begin{bmatrix} 1&0&0&0&\frac{9}{4}\\0&1&0&0&-\frac{1}{12}\\0&0&1&0&\frac{11}{12}\\0&0&0&1&-\frac{1}{4} \end{bmatrix} , \ dim(W_1+W_2)=4=dim(R^4)
    There are only 4 vectors which are linear independent, so\ W_1+W_2=span(\vec{S}_{11},\vec{S}_{12},\vec{S}_{21  },\vec{S}_{22}) =span(\vec{e}_1,\vec{e}_2,\vec{e}_3,\vec{e}_4)=R^4

    \because\frac{9}{4}\vec{S}_{11}-\frac{1}{12}\vec{S}_{12}+\frac{11}{12}\vec{S}_{21}-\frac{1}{4}\vec{S}_{22}=\vec{S}_{23}\ \Rightarrow \frac{\frac{9}{4}\vec{S}_{11}-\frac{1}{12}\vec{S}_{12}}{in\ W_1}=\frac{-\frac{11}{12}\vec{S}_{21}+\frac{1}{4}\vec{S}_{22}+  \vec{S}_{23}}{in\ W_2}
    \therefore W_1 \cap W_2 = span(\frac{9}{4}\vec{S}_{11}-\frac{1}{12}\vec{S}_{12}) =span(-\frac{11}{12}\vec{S}_{21}+\frac{1}{4}\vec{S}_{22}+  \vec{S}_{23})=span(\begin{bmatrix} \frac{7}{3}\\ -\frac{1}{6}\\ 2 \\ \frac{23}{6}  \end{bmatrix})

    I will update frequently if I find any mistake. So you may find contents changed when refresh Web page.
    arbolis, to be careful! Linear Algebra will be a great challenge in final exam.
    Last edited by math2009; February 15th 2009 at 11:03 PM.
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  3. #3
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    Quote Originally Posted by arbolis View Post
    Hi MHF,
    I appreciate very much all the help I got so far. If I didn't answered my last 2 posts it's because of lack of time. I feel overloaded with linear Algebra and the final exam is coming up next 26th of February.
    Here is the exercise I'd like to check out with you.
    Let S_1=\{ (1,0,1,2), (-1,2,3,8), (-1,1,1,3), (0,1,2,5) \} and S_2=\{ (1,1,0,1), (1,-1,0,3), (3,1,2,4), (1,1,2,0) \} and let W_1 be spanned by S_1 and W_2 be spanned by S_2.
    a)Give a base B_i of W_i contained in S_i, for i=1, 2.
    My attempt : I wrote each vector of the base into a matrix as being column vectors and reduced it. I got that 3 of the 4 vectors of S_1 were colinear and then I solve a 2 equations system, getting finally that a base of W_1 included into S_1 is \{ (1,0,1,-1), (0,1,-1,1) \}.
    What? These vectors may well span W_1 but neither of them is in S_1. Did you misunderstand? If W_1 is 2 dimensional, then "a basis for W_1 contained in S_1" must be two of the vectors in S_1.

    Doing the same for S_2, I got that a base of W_2 included into S_2 is \{ (1,0,0,0), (0,-1,1,-1)  \}.
    Same point as before. Neither of these vectors is in S_2.

    b)Describe W_1, W_2, W_1 \cap W_2 and W_1+W_2 implicitly.
    My attempt : don't know how to start.
    c)Give a base of W_1 \cap W_2 and of W_1+W_2.
    My attempt : I wrote the bases I got in part a) as vector column in a matrix. I checked out that \det A \neq 0, hence the matrix is invertible and the 4 vectors are linear independent. So I think that W_1 \cap W_2 = ∅. So there's no base that generates W_1 \cap W_2, or I'm wrong?
    And as the 4 vectors are linear independent, a basis of W_1+W_2 would be the union of the bases of W_1 and W_2. Or : \{ (1,0,1,-1), (0,1,-1,1),(1,0,0,0), (0,-1,1,-1)  \}. Am I right?
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  4. #4
    MHF Contributor arbolis's Avatar
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    Thank you very much for the help to both.
    I'll check out the answers more in details in some hours.
    arbolis, to be careful! Linear Algebra will be a great challenge in final exam.
    It will be a massacre.

    What? These vectors may well span but neither of them is in . Did you misunderstand?
    I'm not even sure I misunderstood. I just made an error I'd say. I understand my error.
    But I don't know what I've done then.
    Yeah, linear Algebra is a tough challenge for me. I'm thinking of taking the final exam in July instead of this month, but if I do so I'll likely graduate later. If I fail too, and I'll get less chance for being accepted into a graduate program.
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  5. #5
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    I update solution. Please refresh web page.
    By the way, I admire your spirit of sacrifice
    Last edited by math2009; February 15th 2009 at 07:56 PM.
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