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Math Help - Orthogonal Subspaces

  1. #1
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    Orthogonal Subspaces

    Let U and W be orthogonal subspaces.

    a) Prove that U\cap W=\{\underline{0} \}

    b) If \{u_1,u_2,...,u_m\} is a linearly independent subset of U and \{w_1,w_2,...,w_n\} is a linearly independent subset of W, prove that \{u_1,u_2,...,u_m,w_1,w_2,...,w_n\} is a linearly independent set in V.

    I find this bit a little hard to understand, and I have to multitask, so I'll have difficulty if I have to do this on my own. Any help would be greatly appreciated.
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  2. #2
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    Quote Originally Posted by Runty View Post
    Let U and W be orthogonal subspaces.

    a) Prove that U\cap W=\{\underline{0} \}


    If x\in U\cap W then as x\in U\,\,\,and\,\,\,x\in W we get <x,x>=0\Longleftrightarrow x=0 , with <,> the inner product in the v.s.

    b) If \{u_1,u_2,...,u_m\} is a linearly independent subset of U and \{w_1,w_2,...,w_n\} is a linearly independent subset of W, prove that \{u_1,u_2,...,u_m,w_1,w_2,...,w_n\} is a linearly independent set in V.


    Suppose \sum^m_{i=1}a_1u_1+\sum^n_{j=1}b_jw_j=0\,,\,\,a_i\  ,,\,b_j \in\mathbb{F}= the definition field, then \sum^m_{i=1}a_1u_1=-\sum^n_{j=1}b_jw_j\Longrightarrow \sum^m_{i=1}a_1u_1\,,\,\sum^n_{j=1}b_jw_j \in U\cap W \Longrightarrow \sum^m_{i=1}a_1u_1=0=\sum^n_{j=1}b_jw_j\Longrighta  rrow . Deduce from here and (a) above that  a_i=0=b_j\,\,\forall\,i,j .

    Tonio

    I find this bit a little hard to understand, and I have to multitask, so I'll have difficulty if I have to do this on my own. Any help would be greatly appreciated.
    .
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  3. #3
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    Those are good answers, Tonio, but our material does not make use of summation notation. Still, I guess I can use it and it'll provide a correct answer, or I could just go long-hand.

    Two questions, however.

    First, for part a), when you say inner product, you mean x\bullet x, right? I just want to make sure.

    Second, for part b), when you say definition field, would you mean something like real numbers (i.e. for scalars)?
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  4. #4
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    Quote Originally Posted by Runty View Post
    Those are good answers, Tonio, but our material does not make use of summation notation. Still, I guess I can use it and it'll provide a correct answer, or I could just go long-hand.

    Two questions, however.

    First, for part a), when you say inner product, you mean x\bullet x, right? I just want to make sure.


    Yes, I meant that: inner, scalar or dot product.


    Second, for part b), when you say definition field, would you mean something like real numbers (i.e. for scalars)?

    I meant the field over which the vector space containing U, W is defined.

    Tonio
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  5. #5
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    Quote Originally Posted by Runty View Post
    Let U and W be orthogonal subspaces.

    a) Prove that U\cap W=\{\underline{0} \}
    For every u\in U and every w \in W, the inner product of u and w is 0. Suppose v is in both subspaces. What must be true of the inner product of v with itself? Now think about the definition of "inner product".

    b) If \{u_1,u_2,...,u_m\} is a linearly independent subset of U and \{w_1,w_2,...,w_n\} is a linearly independent subset of W, prove that \{u_1,u_2,...,u_m,w_1,w_2,...,w_n\} is a linearly independent set in V.[/quote]
    Remember the definition of "linearly independent". That set is linearly independent if and only if a_1u_1+ a_2u_2+ \cdot\cdot\cdot\+ a_mu_m+ b_1w_1+ b_1w_2+ \cdot\cdot\cdot+ a_nw_n= 0 implies that a_1= a_2= \cdot\cdot\cdot\= a_m= b_1= b_2= \cdot\cdot\cdot= w_n= 0. What do you get if you take the dot product of u_1 with math]a_1u_1+ a_2u_2+ \cdot\cdot\cdot\+ a_mu_m+ b_1w_1+ b_1w_2+ \cdot\cdot\cdot+ a_nw_n= 0[/tex]? What about the dot product with u_2, w_1, etc.

    I find this bit a little hard to understand, and I have to multitask, so I'll have difficulty if I have to do this on my own. Any help would be greatly appreciated.
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  6. #6
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    Quote Originally Posted by HallsofIvy View Post
    For every u\in U and every w \in W, the inner product of u and w is 0. Suppose v is in both subspaces. What must be true of the inner product of v with itself? Now think about the definition of "inner product".

    b) If \{u_1,u_2,...,u_m\} is a linearly independent subset of U and \{w_1,w_2,...,w_n\} is a linearly independent subset of W, prove that \{u_1,u_2,...,u_m,w_1,w_2,...,w_n\} is a linearly independent set in V.

    Remember the definition of "linearly independent". That set is linearly independent if and only if a_1u_1+ a_2u_2+ \cdot \cdot \cdot + a_mu_m+ b_1w_1+ b_1w_2+ \cdot \cdot \cdot + a_nw_n= 0 implies that a_1= a_2= \cdot \cdot \cdot = a_m= b_1= b_2= \cdot \cdot \cdot = b_n= 0. What do you get if you take the dot product of u_1 with a_1u_1+ a_2u_2+ \cdot \cdot \cdot + a_mu_m+ b_1w_1+ b_1w_2+ \cdot \cdot \cdot + a_nw_n= 0? What about the dot product with u_2, w_1, etc.
    Corrected the mistakes in Latex. This certainly makes things complicated, however. I'll see if I can figure it out.
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