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I have to show that:

For an inner product space $V$, $\|x + y \| = \|x\| + \|y\|$, for all $x$, $y \in V$

if and only if one of the vectors $x$ or $y$ is a scalar multiple of the other.

I am thinking, if $x= cy$, for some scalar then the above equality holds.

I am unsure how to proceed the other way. Please help.

Edit : I guess have to look at the proof of the triangle inequality which holds as an equality if

$|\langle x, y\rangle | = \|x\|\,\|y\|.$

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That's not quite true -- the scalar multiple has to be a positive scalar multiple. –  joriki Mar 1 '12 at 22:28
    
@joriki: Does it follow from the equality | < x,y >| = ||x|| ||y|| ? Please suggest. –  Tav Mar 1 '12 at 23:07
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2 Answers

up vote 5 down vote accepted

The result is trivial if $x=\mathbf{0}$, as we then have that $x$ is a nonnegative real scalar multiple of $y$ and we have equality. Assume $x\neq \mathbf{0}$.

Look carefully at the proof of the Cauchy-Schwartz inequalty:

Express $y$ as $\alpha x + z$, where $\langle x,z\rangle = 0$. This can always be done by letting $\alpha = \frac{\langle x,y\rangle}{\langle x,x\rangle}$ and $z=y-\alpha x$.

Then $$\begin{align*} \langle y,y\rangle &= \langle \alpha x+z,\alpha x+z\rangle\\ &= \alpha\overline{\alpha}\langle x,x\rangle + \alpha\langle x,z\rangle + \overline{\alpha}\langle z,x\rangle + \langle z,z\rangle\\ &= |\alpha|^2\lVert x\rVert^2 + \lVert z\rVert^2\\ &\geq |\alpha|^2\lVert x\rVert^2, \end{align*}$$ with equality if and only if $z=\mathbf{0}$. Since $$\alpha = \frac{\langle x,y\rangle}{\langle x,x\rangle} = \frac{\langle x,y\rangle}{\Vert x\rVert^2},$$we conclude that $$\lVert y\rVert^2 \lVert x\rVert^2 \geq |\langle x,y\rangle|^2$$ with equality if and only if $z=\mathbf{0}$ (i.e., if and only if $y$ is a scalar multiple of $x$).

Now for the triangle inequalty, we have $$\begin{align*} \lVert x+y\rVert^2 &= \langle x+y,x+y\rangle\\ &= \langle x,x\rangle + \langle y,y\rangle + 2\mathscr{R}e\langle x,y\rangle\\ &\leq \lVert x\rVert^2 + \lVert y\rVert^2 + 2|\langle x,y\rangle| \tag{1}\\ &\leq \lVert x\rVert^2 + \lVert y\rVert^2 + 2\lVert x\rVert\lVert y\rVert \tag{2}\\ &= (\lVert x\rVert + \lVert y\rVert)^2. \end{align*}$$

Now, $(1)$ is an equality if and only if $|\langle x,y\rangle|=\mathscr{R}e\langle x,y\rangle$, which occurs if and only if $\langle x,y\rangle = \alpha\lVert x\rVert^2$ is a nonnegative real; which occurs if and only if $\alpha$ is a nonnegative real; and $(2)$ is an equality if and only if $z=\mathbf{0}$. Thus, writing $y=\alpha x +z$ with $\langle x,z\rangle = 0$, we conclude that we have equality if and only if $z=\mathbf{0}$ and $\alpha$ is a nonnegative real number.

In summary, equality holds if and only if $x=\mathbf{0}$ (hence is equal to a nonnegative real scalar multiple of $y$), or if $y=\alpha x$ with $\alpha$ a nonnegative real number. That is, if and only if one of $x$ and $y$ is a nonnegative real scalar multiple of the other.

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Many thanks for the reply. –  Tav Mar 1 '12 at 23:17
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We assume that $x$ and $y$ are not $0$. If we have a relationship of the form $x=cy$ where $c$ is a scalar then $\|x\|=|c|\cdot \|y\|$ so $|c|=\frac{\|x\|}{\|y\|}$. So we try to compute $\Bigl\| \|x\|y-\|y\|x\Bigr\|^2$: $$\begin{align*} \Bigl\lVert \|x\|y-\|y\|x\Bigr\rVert^2&=\|x\|^2\|y\|^2+\|y\|^2\|x\|^2-\|x\|\cdot \|y\|(\langle x,y\rangle+\langle y,x\rangle)\\ &=\|x\|\cdot \|y\|(2\|x\|\cdot \|y\|-(\|x+y\|^2-\|x\|^2-\|y\|^2))\\ &=\|x\|\cdot \|y\|(2\|x\|\cdot \|y\|-2\|x\|\cdot \|y\|)\\&=0\end{align*} $$ which gives the result.

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This sort of thing becomes a lot more legible if you use \| instead of || :-) –  joriki Mar 1 '12 at 23:00
1  
@joriki: And even better with \lVert and \rVert, which are spaces as left and right delimiters... –  Arturo Magidin Mar 1 '12 at 23:06
    
@DavideGiraudo: Thanks for your reply. I am confused about the other way. That is, how to prove that if the triangle equality holds as an equality then x =cy. –  Tav Mar 1 '12 at 23:10
    
Thanks to Joriki and Arturo Magidin for editing. I used the fact that if the triangular equality holds then we can go from the second line to the third. –  Davide Giraudo Mar 2 '12 at 8:48
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