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Thread: Subspaces

  1. #1
    Super Member Showcase_22's Avatar
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    Subspaces

    Show that if $\displaystyle dim \ W_1 \geq \ dim \ W_2$ and $\displaystyle dim(W_1+W_2)=1+dim(W_1 \cap W_2)$, then $\displaystyle W_1+W_2=W_1$ and $\displaystyle W_1 \cap W_2=W_2$.
    Proving $\displaystyle W_1+W_2=W_1$
    $\displaystyle dim \ W_1 \geq \ dim \ W_2 \Rightarrow dim \ W_1-dim \ W_2 \geq 0 \Rightarrow \ dim \ W_1-dim \ W_2 \geq k \geq 0$

    The value of k represents the dimension of a basis for the vectors in $\displaystyle W_1$ that are not in $\displaystyle W_2$.

    Hence $\displaystyle W_2 \subset W_1$ (Was the proof above enough to justify this step?).

    Therefore $\displaystyle \forall \ u \in W_1$ and $\displaystyle \forall v \in W_2$, $\displaystyle u+v \in W_1+W_2 \subset W_1$.

    Conversely, $\displaystyle \underline{0} \in W_2$ so $\displaystyle W_1 \subset W_1+W_2$.

    This gives that $\displaystyle W_1+W_2=W_1$.

    Proving $\displaystyle W_1 \cap W_2=W_1$
    $\displaystyle dim(W_1 +W_2)=dim(W_1)+dim(W_2)-dim(W_1 \cap W_2)$

    but we know that $\displaystyle dim(W_1+W_2)=dim(W_1)$

    $\displaystyle dim(W_1)=dim(W_1)+dim(W_2)-dim(W_1 \cap W_2) \Rightarrow dim(W_1 \cap W_2)=dim(W_2) $$\displaystyle \Rightarrow \ W_1 \cap W_2=W_2$

    (this is implied because they have the same dimension over the same vector space so they must be the same subspace).
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  2. #2
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    Quote Originally Posted by Showcase_22 View Post

    $\displaystyle dim \ W_1 \geq \ dim \ W_2 \Rightarrow dim \ W_1-dim \ W_2 \geq 0 \Rightarrow \ dim \ W_1-dim \ W_2 \geq k \geq 0$

    The value of k represents the dimension of a basis for the vectors in $\displaystyle W_1$ that are not in $\displaystyle W_2$.

    Hence $\displaystyle W_2 \subset W_1$ (Was the proof above enough to justify this step?). No!

    Therefore $\displaystyle \forall \ u \in W_1$ and $\displaystyle \forall v \in W_2$, $\displaystyle u+v \in W_1+W_2 \subset W_1$.

    Conversely, $\displaystyle \underline{0} \in W_2$ so $\displaystyle W_1 \subset W_1+W_2$.

    This gives that $\displaystyle W_1+W_2=W_1$.



    $\displaystyle dim(W_1 +W_2)=dim(W_1)+dim(W_2)-dim(W_1 \cap W_2)$

    but we know that $\displaystyle dim(W_1+W_2)=dim(W_1)$

    $\displaystyle dim(W_1)=dim(W_1)+dim(W_2)-dim(W_1 \cap W_2) \Rightarrow dim(W_1 \cap W_2)=dim(W_2) $$\displaystyle \Rightarrow \ W_1 \cap W_2=W_2$

    (this is implied because they have the same dimension over the same vector space so they must be the same subspace).
    if $\displaystyle W_1 \subseteq W_2$ or $\displaystyle W_2 \subseteq W_1,$ then we're done. otherwise, $\displaystyle W_1 \cap W_2$ would be a proper subspace of both $\displaystyle W_1$ and $\displaystyle W_2.$ thus: $\displaystyle \dim W_1 \geq 1 + \dim W_1 \cap W_2$ and $\displaystyle \dim W_2 \geq 1 + \dim W_1 \cap W_2.$

    therefore: $\displaystyle \dim (W_1+W_2)=\dim W_1 + \dim W_2 - \dim W_1 \cap W_2 \geq 2 + \dim W_1 \cap W_2,$ contradicting our assumption that $\displaystyle \dim(W_1 + W_2)=1 + \dim W_1 \cap W_2.$ Q.E.D.
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  3. #3
    Member Nacho's Avatar
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    $\displaystyle W_1$ and $\displaystyle W_2$ are in the same vectorial space?

    a question: If $\displaystyle V_1$ and $\displaystyle V_2$ are in the same vectorial space, I can tell: $\displaystyle
    \dim \left( {V_1 } \right) \leqslant \dim \left( {V_2 } \right) \Rightarrow V_1 \leqslant V_2
    $ ?

    where the second "mayor o equal" mean subspace
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  4. #4
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    Quote Originally Posted by Nacho View Post
    $\displaystyle W_1$ and $\displaystyle W_2$ are in the same vectorial space?
    of course! otherwise $\displaystyle W_1 + W_2$ would be undefined!


    a question: If $\displaystyle V_1$ and $\displaystyle V_2$ are in the same vectorial space, I can tell: $\displaystyle
    \dim \left( {V_1 } \right) \leqslant \dim \left( {V_2 } \right) \Rightarrow V_1 \leqslant V_2
    $ ?

    where the second "mayor o equal" mean subspace
    not necessarily!
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  5. #5
    Moo
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    Quote Originally Posted by Nacho View Post
    $\displaystyle W_1$ and $\displaystyle W_2$ are in the same vectorial space?

    a question: If $\displaystyle V_1$ and $\displaystyle V_2$ are in the same vectorial space, I can tell: $\displaystyle
    \dim \left( {V_1 } \right) \leqslant \dim \left( {V_2 } \right) \Rightarrow V_1 \leqslant V_2
    $ ?

    where the second "mayor o equal" mean subspace
    The command is \subseteq
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