Does $\mathbf x\cdot \mathbf y = 0$ imply that $\lVert x+y\rVert_1 = \lVert x\rVert_1 + \lVert y\rVert_1$?

Question

If x and y are orthogonal vectors and we define $\lVert x\rVert_1 =\sum^{n}_{j=1} |x_j|$,

is it possible to express $\lVert x+y\rVert_1$ in terms of $\lVert x\rVert_1$ and $\lVert y\rVert_1$ ?

So I get that $\lVert x+y\rVert_1 = \sum^{n}_{j=1}|x_j+ y_j|$ and by the Triangle Inequality, $|x_j + y_j| \leq |x_j| + |y_j|$ so we must have that $\sum^{n}_{j=1}|x_j+ y_j|\leq\sum^{n}_{j=1}|x_j|$ + $\sum^{n}_{j=1}|y_j|$.

My question then is does x $ \cdot $ y $ = 0$ (x and y are orthogonal) imply that $\lVert x+y\rVert_1 = \lVert x\rVert_1 + \lVert y\rVert_1$ and so the Triangle Inequality is equal, and if so why?

Answer 1

Since $\|(a,b)\|_1=|a|+|b|$, we get that $$ \begin{align} \|(a,b)+(c,d)\|_1 &=\|(a+c,b+d)\|_1\\ &=|a+c|+|b+d| \end{align} $$ However, $$ \begin{align} \|(a,b)\|_1+\|(c,d)\|_1 &=|a|+|b|+|c|+|d| \end{align} $$ Thus, $\|(-1,1)+(1,1)\|_1=2$, yet $\|(-1,1)\|_1+\|(1,1)\|_1=4$

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What is true is that if $x\cdot y=0$ then $$ \begin{align} \|x+y\|_2^2 &=(x+y)\cdot(x+y)\\ &=x\cdot x+y\cdot y+2x\cdot y\\ &=\|x\|_2^2+\|y\|_2^2 \end{align} $$

Answer 2

Consider $\mathbb{R}^2$ and $x = (1,1), y = (1,-1)$. Then $x \cdot y = 0$, but $\|x\|_1 = \|y\|_1 = \|x+y\|_1 = 2$.