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Math Help - proof that dim V + dim V perp = n

  1. #1
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    proof that dim V + dim V perp = n

    I am trying to prove that the dimension of a subspace V of R^n + the dimension of the orthogonal complement of that subspace V is equal to n for any subspace V in R^n.

    So i assumed the basis for the subspace V consisted of vectors v1, v2,..., vk. now i extend the basis so that the new basis is a basis for R^n. This new basis is v1, v2, ..., vk, w1, w2, ..., w_m. where m + k = n. So in order to prove this theorem i have to show that w1, w2, ..., w_m forms a basis for the orthogonal complement of V. What i have trouble showing is that these vectors span the orthogonal complement of V. i have that any vector in R^n let's say u can be expressed as a linear combination of v1, v2, ..., vk, w1, w2, ..., w_m since that is a basis for R^n. so u = c1v1 + ... +ckvk + d1w1 + ... + d_m(w_m). i tried taking the dot product on both sides with an arbitrary vector w in the orthogonal complement of V and i got u * w = w * (d1w1 + ...+ d_m(w_m)) since w is orthogonal to every vector in V so w * v_i = 0. now i'm stuck and i'm not sure how to continue on.

    I can show that w1, ... w_m are linearly independent since v1, v2, ..., vk, w1, w2, ..., w_m form a basis which means that the only linear combination of them that forms the 0 vector is for all constants to be 0. Then since i know that v1, ... vk is also a basis, the only linear combination of those vectors that equal 0 is for all the constants to equal 0 as well. and therefore the only linear combination of w1, w2, ..., w_m that equal 0 is all the constants being 0 so those vectors are independent.

    now all i need to show is that those vectors span the orthogonal complement of V and the proof will be done. is my approach to showing that the vectors span correct? if anyone could give a hint or 2 to help me move on in this problem i'd greatly appreciate it.
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by oblixps View Post
    I am trying to prove that the dimension of a subspace V of R^n + the dimension of the orthogonal complement of that subspace V is equal to n for any subspace V in R^n.

    So i assumed the basis for the subspace V consisted of vectors v1, v2,..., vk. now i extend the basis so that the new basis is a basis for R^n. This new basis is v1, v2, ..., vk, w1, w2, ..., w_m. where m + k = n. So in order to prove this theorem i have to show that w1, w2, ..., w_m forms a basis for the orthogonal complement of V. What i have trouble showing is that these vectors span the orthogonal complement of V. i have that any vector in R^n let's say u can be expressed as a linear combination of v1, v2, ..., vk, w1, w2, ..., w_m since that is a basis for R^n. so u = c1v1 + ... +ckvk + d1w1 + ... + d_m(w_m). i tried taking the dot product on both sides with an arbitrary vector w in the orthogonal complement of V and i got u * w = w * (d1w1 + ...+ d_m(w_m)) since w is orthogonal to every vector in V so w * v_i = 0. now i'm stuck and i'm not sure how to continue on.

    I can show that w1, ... w_m are linearly independent since v1, v2, ..., vk, w1, w2, ..., w_m form a basis which means that the only linear combination of them that forms the 0 vector is for all constants to be 0. Then since i know that v1, ... vk is also a basis, the only linear combination of those vectors that equal 0 is for all the constants to equal 0 as well. and therefore the only linear combination of w1, w2, ..., w_m that equal 0 is all the constants being 0 so those vectors are independent.

    now all i need to show is that those vectors span the orthogonal complement of V and the proof will be done. is my approach to showing that the vectors span correct? if anyone could give a hint or 2 to help me move on in this problem i'd greatly appreciate it.
    Do you know the fact that if V\oplus V^{\perp}=\mathbb{R}^n? I'm guessing not.

    EDIT: Upon closer inspection I see that my question was stupid, since that's what it looks like you're trying to do.
    Last edited by Drexel28; November 6th 2010 at 12:20 PM.
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  3. #3
    MHF Contributor Also sprach Zarathustra's Avatar
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    How to prove the above?


    Proof:

    There exist an orthogonal basis \{u_1,...,u_r\} to V, also by some theorem we can extend it to orthonormal basis of \mathbb{R}^n: \{u_1,...,u_n\}.Therefor u_{r+1},...,u_n \in V^{\perp}. If v\in\mathbb{R}^n, v is from the form: v=a_1u_1+...+a_nu_n when a_1u_1+...+a_ru_r \in V and a_{r+1}u_{r+1}+...+a_nu_n \in V^{\perp}, the conclusion is \mathbb{R}^n=V+V^{\perp}.

    Now, if w\in V\bigcap  V^{\perp}, then <w,w>=0, therefor w=0, hence V\bigcap  V^{\perp}=\{0\}

    \mathbb{R}^n=V+V^{\perp} and V\bigcap  V^{\perp}=\{0\} gives us V\oplus V^{\perp}=\mathbb{R}^n.
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  4. #4
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    Actually the easiest method to prove this is to use the idea that for any m \times n matrix rank(A)+nullity (A)=n:

    If V=\{ 0 \}, then V^{\perp} =R^n, in this case dim(V)+dim(V^{\perp})=0+n=n. If V  \ne \{ 0 \}, then choose a basis for V and form the matrix A which has these basis as its row vectors. The matrix has n columns since its row vectors come from R^n. Moreover, the row space of A is V and the nullspace of A is V^{\perp}, so you will have

    dim(V)+dim(V^{\perp})= rank(A)+nullity(A)=n
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  5. #5
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    so basically in your proof you didn't have to show that u_{r+1},...,u_n \in V^{\perp} formed a basis for V perp? and how did you know that the vectors u_r+1, ... u_n are in V perp? maybe this is supposed to be obvious and i'm just thinking too much but we start with a basis for V, extend to a basis of R^n so the left over vectors must be in V perp? also you never showed that the vectors were a basis so we don't know if the vectors span V perp so how do you know that a_{r+1}u_{r+1}+...+a_n u_n is any vector is V perp?
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  6. #6
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    For Also sprach Zarathustra's proof I am confused on the part when he said that u_r+1,...u_n are in V perp. how did you know that the vectors u_r+1, ... u_n are in V perp?
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