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Math Help - [SOLVED] 2 proofs

  1. #1
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    Arrow [SOLVED] 2 proofs

    I was wondering if anyone could help me get started on these 2 proofs. I'm having trouble getting through them. They're to help me prep for my upcoming final.

    1) Let u and v be vectors in an inner product space. Prove that ||u|| = ||v|| if and only if (u + v) and (u - v) are orthogonal.

    2) Let A be an (n n) matrix, and let λ be an eigenvalue of A. Prove that if k is any scalar, then λ+k is an eigenvalue of A+kI.

    Any help is greatly appreciated.
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by leverin4 View Post
    I was wondering if anyone could help me get started on these 2 proofs. I'm having trouble getting through them. They're to help me prep for my upcoming final.

    1) Let u and v be vectors in an inner product space. Prove that ||u|| = ||v|| if and only if (u + v) and (u - v) are orthogonal.
    \begin{aligned}\left|\left|u\right|\right|=\left|\  left|v\right|\right|&\iff \left|\left|u\right|\right|^2=\left|\left|v\right|  \right|^2\\&\iff\left<u,u\right>=\left<v,v\right>\  \ &\iff\left<u,u\right>-\left<v,v\right>=0\\ &\iff\left<u,u\right>+\left<u,v\right>-\left<u,v\right>-\left<v,v\right>=0\\ &\iff\left<u+v,u\right>+\left<u+v,-v\right>=0\\&\iff\left<u+v,u-v\right>=0\qquad\square\end{aligned}

    Does this make sense? Some steps are skipped -- see if you can fill them in.

    2) Let A be an (n n) matrix, and let λ be an eigenvalue of A. Prove that if k is any scalar, then λ+k is an eigenvalue of A+kI.

    Any help is greatly appreciated.
    If \lambda is an eigenvalue of A, then Ax=\lambda x. But

    \begin{aligned}Ax=\lambda x&\implies Ax+kx=\lambda+kx\\ &\implies\ldots\end{aligned}

    Can you finish this off?
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  3. #3
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    Quote Originally Posted by Chris L T521 View Post
    \begin{aligned}\left|\left|u\right|\right|=...\\ &\iff\left<u,u\right>=\left<v,v\right>\\ &\iff\left<u,u\right>-\left<v,v\right>=0\\ &\iff\left<u,u\right>+\left<u,v\right>-\left<u,v\right>-\left<v,v\right>=0\\ &\iff\left<u,u\right>+\left<v,u\right>+\left<u,-v\right>+\left<v,-v\right>=0\\ &\iff\left<u+v,u\right>+\left<u+v,-v\right>=0\\&\iff\left<u+v,u-v\right>=0\qquad\square\end{aligned}

    Does this make sense? Some steps are skipped -- see if you can fill them in.
    It makes a lot of sense, thanks for your help. I added in what I think you meant to be the missing step; I only saw one, I think. I'm not very good at latex, sorry. Since the problem statement has "if and only if", I also have to prove it backwards, right? Which is essentially the same steps backwards?
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  4. #4
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by leverin4 View Post
    It makes a lot of sense, thanks for your help. I added in what I think you meant to be the missing step; I only saw one, I think. I'm not very good at latex, sorry. Since the problem statement has "if and only if", I also have to prove it backwards, right? Which is essentially the same steps backwards?
    You filled in the missing step correctly. The nice thing about this question is that we can take care of both proof directions with each step!
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    Quote Originally Posted by Chris L T521 View Post
    If \lambda is an eigenvalue of A, then Ax=\lambda x. But

    \begin{aligned}Ax=\lambda x&\implies Ax+kx=\lambda x+kx\\ &\implies\ \left(A+k\right)x=\left(\lambda+k\right)x\end{alig  ned}

    Can you finish this off?
    Still stuck on this one though, I added the only logical thing I can see. Don't see how it helps me all that much though.
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  6. #6
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by leverin4 View Post
    Still stuck on this one though, I added the only logical thing I can see. Don't see how it helps me all that much though.
    Close! The LHS contains matricies! So when you factor out x from Ax+kx, you don't get (A+k)x because you can't take a constant and add it to a matrix. Instead you get...(try to finish the statement)
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    Quote Originally Posted by Chris L T521 View Post
    Close! The LHS contains matricies! So when you factor out x from Ax+kx, you don't get (A+k)x because you can't take a constant and add it to a matrix. Instead you get...(try to finish the statement)
    (A+kI)x?
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  8. #8
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by leverin4 View Post
    (A+kI)x?
    Correct. Now what conclusion can be drawn from (A+kI)x=(\lambda+k)x?
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    Quote Originally Posted by Chris L T521 View Post
    Correct. Now what conclusion can we draw from (A+kI)x=(\lambda+k)x?
    We can conclude that (\lambda+k) is a scalar that has the same effect on x as (A+kI) and therefore (\lambda+k) is, by definition, an eigenvalue for (A+kI).
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  10. #10
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by leverin4 View Post
    We can conclude that (\lambda+k) is a scalar that has the same effect on x as (A+kI) and therefore (\lambda+k) is, by definition, an eigenvalue for (A+kI).
    Correct!

    Does this make better sense now?
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  11. #11
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    Quote Originally Posted by Chris L T521 View Post
    Correct!

    Does this make better sense now?
    A lot, I was actually stuck at the point where I factor out the x for awhile before I decided to post it here.. You're help has been great. I thanked you in each post.
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