Quantcast

A community for students.

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

richyw

  • 2 years ago

Given \( \mathbf{x,y}\in\mathbb{R}^n\) show that \[|\mathbf{x}+\mathbf{y}|^2=|\mathbf{x}|^2+2\mathbf{x}\cdot\mathbf{y}+|\mathbf{y}|^2\]

  • This Question is Closed
  1. richyw
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    I have no idea how to "show" this. it just seems obvious to me?

  2. abb0t
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    I think that's dot product.

  3. richyw
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    it is. the dots are dot products and the bold are vectors

  4. Thomas9
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    It's not a true statement, is it?

  5. Thomas9
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    Pick x=(1,0), y=(0,1): 0=\=1+0+1

  6. richyw
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    oops! you caught a mistake. i'll change it. sorry!

  7. richyw
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    it's now edited in the original question. Now you can see why I said it "seems" obvious to me. still don't know how to show it!

  8. Thomas9
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    I think it's easiest to do use the components of the vectors: \[(x+y)^2=(x_1+y_1)^2+...+(x_n+y_n)^2=...\] Now write out the squares, group the terms differently and go back to vectors again.

  9. Thomas9
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    \[|x+y|^2=(x_1+y_1)^2+...+(x_n+y_n)^2=...\] that is.

  10. ZeHanz
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    It looks more difficult than it is... Just do what Thomas9 says: (I prefer sigma-notation instead of ...)\[|\overline x \cdot \overline y|^2=\sum_{i=1}^{n}(x_i+y_i)^2=\sum_{i=1}^{n}(x_i^2+2x_iy_i+y_i^2)=\]\[\sum_{i=1}^{n}x_i^2+2\sum_{i=1}^{n}x_iy_y+\sum_{i=1}^{n}y_i^2=|\overline x|^2+2 \overline x \overline y + |\overline y|^2\]

  11. joemath314159
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    The easiest way to show this is to use the fact that for any vector x:\[|x|^2=x\cdot x\]and the fact that the dot product has bilinear properties (it has two slots, and it is linear in each slot). By linear, i mean, \[\forall x,y,z\in \mathbb {R}^n,\forall \lambda\in \mathbb{R}\]\[(\lambda x+y)\cdot z=\lambda (x\cdot z)+(y\cdot z)\] Using these two properties, you can take the expression:\[|x+y|^2\]and change it to the desired equation.

  12. joemath314159
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    \[|x+y|^2=(x+y)\cdot (x+y)\]Using linearity in the "first slot":\[(x+y)\cdot (x+y)=\left(x\cdot(x+y)\right)+\left(y\cdot (x+y)\right)\]Now we use linearity on the second slot on both terms:\[\left(x\cdot(x+y)\right)+\left(y\cdot (x+y)\right)=(x\cdot x)+(x\cdot y)+(y\cdot x) +(y\cdot y)\] Since the dot product is symmetric, x*y = y*x, so we end up with 2(x*y) in the middle, and from earlier, we note that:\[x\cdot x = |x|^2\]So we get: \[|x|^2+2(x\cdot y)+|y|^2\]

  13. richyw
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    thanks everyone. looks like i've got a lot of stuff to learn!

  14. Not the answer you are looking for?
    Search for more explanations.

    • Attachments:

Ask your own question

Sign Up
Find more explanations on OpenStudy
Privacy Policy

Your question is ready. Sign up for free to start getting answers.

spraguer (Moderator)
5 → View Detailed Profile

is replying to Can someone tell me what button the professor is hitting...

23

  • Teamwork 19 Teammate
  • Problem Solving 19 Hero
  • You have blocked this person.
  • ✔ You're a fan Checking fan status...

Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.