Results 1 to 4 of 4

Math Help - Prove that an isomorphism produces a basis of W

  1. #1
    Member
    Joined
    Jan 2010
    Posts
    232

    Prove that an isomorphism produces a basis of W

    I think I may have the answer for this one, but I could use a second opinion.

    Suppose V and W are vector spaces over a field F.

    Now suppose \psi:V\rightarrow W is an isomorphism and \{ v_1,...,v_n\} is a basis of V. Prove that \{\psi(v_1),...,\psi(v_n)\} is a basis of W.

    Here is what I have so far:

    Let \{w_1,...,w_n\} be a basis for W where
    w_1=\psi(v_1),...,w_n=\psi(v_n), \forall n>0.
    Then we consider the map
    \psi(x)=x_1w_1+x_2w_2+...+x_nw_n where x=x_1v_1+x_2v_2+...+x_nv_n\in V.
    Since the bases are inverse to each other, we can conclude that \{\psi(v_1),...,\psi(v_n)\} is a basis of W.

    I don't know if I have any major holes in my proof, so if you see any, please let me know.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Member HappyJoe's Avatar
    Joined
    Sep 2010
    From
    Denmark
    Posts
    234
    The last part in particular sounds strange.

    What does it mean that the bases are inverse to each other, and why does it imply that \{\psi(v_1),\ldots,\psi(v_n)\} is a basis for W?

    I would do like this: We need to prove that \{\psi(v_1),\ldots,\psi(v_n)\} is a basis for W. So we need to check two things, namely that the vectors in the proposed basis are linearly independent and that they span all of W.

    Let x_1\psi(v_1)+\ldots+x_n\psi(v_n)=0 for some scalars x_i. Apply linearity of \psi to get

    \psi(x_1v_1+\ldots+x_nv_n)=0.

    Remember that \psi is an isomorphism, so what do you know about vectors that \psi map to 0?

    To see that the basis spans all of W, take some vector w\in W, and choose a vector v\in V, such that \psi(v)=w (why does this vector v exist?).

    You know that \{v_1,\ldots,v_n\} is a basis for V, so you can write v as a linear combination of these basis vectors. Try applying \psi to this relation and see what happens.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Jan 2010
    Posts
    232
    Quote Originally Posted by HappyJoe View Post
    The last part in particular sounds strange.

    What does it mean that the bases are inverse to each other, and why does it imply that \{\psi(v_1),\ldots,\psi(v_n)\} is a basis for W?

    I would do like this: We need to prove that \{\psi(v_1),\ldots,\psi(v_n)\} is a basis for W. So we need to check two things, namely that the vectors in the proposed basis are linearly independent and that they span all of W.

    Let x_1\psi(v_1)+\ldots+x_n\psi(v_n)=0 for some scalars x_i. Apply linearity of \psi to get

    \psi(x_1v_1+\ldots+x_nv_n)=0.

    Remember that \psi is an isomorphism, so what do you know about vectors that \psi map to 0?

    To see that the basis spans all of W, take some vector w\in W, and choose a vector v\in V, such that \psi(v)=w (why does this vector v exist?).

    You know that \{v_1,\ldots,v_n\} is a basis for V, so you can write v as a linear combination of these basis vectors. Try applying \psi to this relation and see what happens.
    Thanks a lot for the help. That "bases are inverse to each other" part, I actually got off Wikipedia. It applied to ordered bases, though, so that could've made a difference.

    Either way, I've gotten it now so, again, thanks a bunch.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor

    Joined
    Apr 2005
    Posts
    16,400
    Thanks
    1849
    Quote Originally Posted by Runty View Post
    I think I may have the answer for this one, but I could use a second opinion.

    Suppose V and W are vector spaces over a field F.

    Now suppose \psi:V\rightarrow W is an isomorphism and \{ v_1,...,v_n\} is a basis of V. Prove that \{\psi(v_1),...,\psi(v_n)\} is a basis of W.

    Here is what I have so far:

    Let \{w_1,...,w_n\} be a basis for W where
    w_1=\psi(v_1),...,w_n=\psi(v_n), \forall n>0.
    You appear to be assuming what you want to prove! That is that w_1=\psi(v_1),...,w_n=\psi(v_n) is a basis.

    Instead you need to show that this set (1) is independent and (2) spans the space.

    To show independence: Supppose that [tex]a\psi(v_1)+ b\psi(v_2)+ \cdot\cdot\cdot+ z\psi(v_n)= 0[/itex]
    Since \psi is a linear transformation, that is the same as \psi(av_1+ bv_2+ \cdot\cdot\cdot+ zv_n)= 0

    Since \psi is an isomorphism, its kernel is the 0 vector only. Therefore av_1+ bv_2+ \cdot\cdot\cdot+ av_n= 0. What does that tell you?

    To show that \{\psi(v_1), \psi(v_n), \cdot\cdot\cdot, \pis(v_n)\} span W, let w be any vector in W. Since \psi is an isomorphism, there exist v in V such that \psi(v)= w. Since [tex]\{v_1, v_2, \cdot\cdot\cdot, v_n} is a basis for V, there exist scalars a, b, ..., z such that av_1+ bv_2+ \cdot\cdot\cdot+ zv_n= v. Apply \psi to both sides of that equation.

    Then we consider the map
    \psi(x)=x_1w_1+x_2w_2+...+x_nw_n where x=x_1v_1+x_2v_2+...+x_nv_n\in V.
    Since the bases are inverse to each other, we can conclude that \{\psi(v_1),...,\psi(v_n)\} is a basis of W.

    I don't know if I have any major holes in my proof, so if you see any, please let me know.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Prove that Phi^-1:W->V is an isomorphism
    Posted in the Advanced Algebra Forum
    Replies: 2
    Last Post: September 22nd 2010, 08:30 AM
  2. Prove isomorphism
    Posted in the Advanced Algebra Forum
    Replies: 20
    Last Post: May 18th 2010, 01:55 AM
  3. Prove isomorphism
    Posted in the Advanced Algebra Forum
    Replies: 5
    Last Post: April 13th 2010, 08:46 PM
  4. Prove isomorphism
    Posted in the Advanced Algebra Forum
    Replies: 2
    Last Post: April 6th 2010, 06:21 PM
  5. Replies: 4
    Last Post: February 14th 2010, 04:05 AM

Search Tags


/mathhelpforum @mathhelpforum