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I'm trying to prove that the triangle inequality is equal when the two vectors $a, b$ are linearly dependent, but I'm failing to do that. I have as follows,

\begin{align*} |x + y | = |x| + |y| &\iff |x|^2 + 2(x\cdot y) + |y|^2 = |x|^2 + 2|x||y| + |y|^2\\ &\iff x\cdot y = |x||y|\\ \end{align*}

Not too sure where to go from here... Any help is appreciated!

sharkeater123
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  • First, you must use correct the hypothesis you have. What does it mean mathematically for 2 vectors to be linearly dependent ? – 領域展開 Oct 01 '21 at 17:08
  • $x = ay$ for some $a\in F$ WLOG – sharkeater123 Oct 01 '21 at 17:12
  • Good now let's take $|x+y|=...$ and use what you said – 領域展開 Oct 01 '21 at 17:13
  • oooooo, then by the Cauchy Schwarz, we have $ay \cdot y = |ay \cdot y|$. The algebra works out, but I'm still on the edge about it. Namely, I was wondering if we can prove it given $ay \cdot y = |ay \cdot y|$ and derive $|x + y| = |x| + |y|$. My main concern with this is I'm not sure if $ay \cdot y = |ay \cdot y|$ is true given the assumption $x, y$ are linearly dependent. – sharkeater123 Oct 01 '21 at 17:17

2 Answers2

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That statement is not true: If $y=-x$, then $||x+y||=0\neq ||x||+||y||=2||x||$.

Not assuming that the norm is induced by a scalar product, but for any norm:

If $y=\lambda x$, then $$||x+y|| = ||x+\lambda x|| = |\lambda +1|\,||x||$$ This is only $||x||+||y||$ if $|1+\lambda|=1+|\lambda|$, which is only the case if $\lambda\geq0$ (else either $1+\lambda = 1-\lambda$, thus $\lambda =0$ or $-1-\lambda = 1-\lambda$, thus $1=0$).

Lazy
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  • sorry if this is a rookie question, but why is $|\lambda x| = |\lambda||x|$ and not just $\lambda |x|$? – sharkeater123 Oct 01 '21 at 17:31
  • If $\lambda <0$ then $\lambda ||x||$ would be negative. A norm is defined to be non-negative, so instead of homogenity ($||\lambda x||=\lambda ||x||$) we demand absolute-homogenity ($||\lambda x||= |\lambda|, ||x||$). – Lazy Oct 01 '21 at 17:40
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I am going to write my proof because I think you are still a little confused

let $x=ay$

then $|x+y|=|ay+y|=|(a+1)y|=(a+1)|y|=a|y|+|y|=|ay|+|y|= |x|+|y|$

領域展開
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    Ahh, that makes sense. I just kept trying to square it. This makes so much sense. Thank you! – sharkeater123 Oct 01 '21 at 17:24
  • That proof is not correct, as the statement is not correct. Note that $||\lambda x|| = |\lambda|,||x||$, not $\lambda ||x||$. – Lazy Oct 01 '21 at 17:26
  • I missed that you are correct, we have to assume that $a\geq 0$ – 領域展開 Oct 01 '21 at 17:29
  • Yes, but read my answer: The whole statement is not correct and applies only if $x=ay$ or $y=ax$ with $a\geq0$. Else the statement is wrong, as can be seen by setting $y=-x$. – Lazy Oct 01 '21 at 17:41