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richyw Group Title

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

  • one year ago
  • one year ago

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

    • one year ago
  2. abb0t Group Title
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    I think that's dot product.

    • one year ago
  3. richyw Group Title
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    it is. the dots are dot products and the bold are vectors

    • one year ago
  4. Thomas9 Group Title
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    It's not a true statement, is it?

    • one year ago
  5. Thomas9 Group Title
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    Pick x=(1,0), y=(0,1): 0=\=1+0+1

    • one year ago
  6. richyw Group Title
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    oops! you caught a mistake. i'll change it. sorry!

    • one year ago
  7. richyw Group Title
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    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!

    • one year ago
  8. Thomas9 Group Title
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    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.

    • one year ago
  9. Thomas9 Group Title
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    \[|x+y|^2=(x_1+y_1)^2+...+(x_n+y_n)^2=...\] that is.

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

    • one year ago
  11. joemath314159 Group Title
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    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.

    • one year ago
  12. joemath314159 Group Title
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    \[|x+y|^2=(x+y)\cdot (x+y)\]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: \[|x|^2+2(x\cdot y)+|y|^2\]

    • one year ago
  13. richyw Group Title
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    thanks everyone. looks like i've got a lot of stuff to learn!

    • one year ago
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