## 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

1. richyw Group Title

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

2. abb0t Group Title

I think that's dot product.

3. richyw Group Title

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

4. Thomas9 Group Title

It's not a true statement, is it?

5. Thomas9 Group Title

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

6. richyw Group Title

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

7. richyw Group Title

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 Group Title

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 Group Title

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

10. ZeHanz Group Title

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 Group Title

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 Group Title

$|x+y|^2=(x+y)\cdot (x+y)$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 Group Title

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