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I am solving problems in Axler's Linear Algebra done right, and this one has me stumped. It says

Prove $\langle x, y\rangle = \lVert x\rVert\, \lVert y\rVert\cos \theta$ using the law of cosines and drawing the triangle formed by $x$, $y$, $x-y$.

Both $x$ and $y$ are coming out of origin, and we are working in $\mathbb{R}^2$. How can this hint be used to answer the proof?

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    Note that if one of $x$, $y$ is a multiple of the other, we do not get a genuine triangle, so a full proof of the result needs to deal with this situation.2011-10-05

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The vectors $x$, $y$, and $x-y$ form a triangle with vertices at the origin, the endpoint of $x$, and the endpoint of $y$. The law of cosines says that for any triangle with sides of length $a$, $b$, and $c$, $c^2 = a^2+b^2 - 2ab\cos(\theta),$ where $\theta$ is the angle opposite the side $c$.

Apply the Law of cosines to the angle formed between $x$ and $y$ at the origin. The lengths of the sides that form that angle are precisely $\lVert x\rVert$ and $\lVert y\rVert$. So from the law of cosines you would have $\lVert x-y\rVert^2 = \lVert x\rVert^2 + \lVert y\rVert^2 - 2\lVert x\rVert\,\lVert y\rVert\cos\theta.$

Now use the fact that $\lVert x-y\rVert^2 = \langle x-y,x-y\rangle,\quad \lVert x\rVert^2 = \langle x,x\rangle,\quad\text{and}\quad \lVert y\rVert^2 = \langle y,y\rangle.$ Expand $\langle x-y,x-y\rangle$, and simplify.

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    @RamanaVenkata: There is no distributive property at play ("distributive property" is when one operation distributes over another, as in $a\times(b+c) = (a\times b)+(a\times c)$. What we have here is *bilinearity* of the inner product.2011-10-05
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The law of cosines says that for a triangle with sides of length $a,b,$ and $c$, $c^2 = a^2+b^2-2ab \cos \theta,$ where $\theta$ is the angle opposite the side of length $c$. Rewriting this in vector notation (so $a = |x|$, $b = |y|$, and $c = |x-y|$) gives $ |x-y|^2 = |x|^2 + |y|^2 - 2|x||y| \cos \theta. $ Expanding the left hand side gives: $ |x-y|^2 = \langle x-y, x-y \rangle = |x|^2-2 \langle x, y \rangle + |y|^2. $

Subtracting $|x|^2+|y|^2$ from both sides and dividing by $-2$ then yields the result.

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    Yes, I am in the habit of using the absolute value symbols to emphasize being in Euclidian space! Also, I had typically used `\|` to typeset norms. Is there a difference?2011-10-05