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Let $ z = ( z_1, z_2, \dots, z_m) \in \mathbb{C}^m$ and $z_j>0$ for all $ 1 \leq j \leq m.$ Then $\| \sum_{j} z_j \|_2 = \sum_j \| z_j \|_2$ if and only if $ z_j = \alpha_j z_1$ for some $ \alpha_j>0.$

This side is clear: If $ z_j = \alpha_j z_1$ for $ \alpha_j>0,$ then $ \| \sum_{j} z_j \|_2 = \| z_1 + \alpha_2 z_1 + \dots + \alpha_m z_1 \|_2 = \| ( 1 + \alpha_2 + \alpha_3 + \dots + \alpha_m )z_1\|_2 = ( 1 + \alpha_2 + \alpha_3 + \dots + \alpha_m ) \|z_1\|_2 = \|z_1\|_2 + \|z_2\|_2 + \dots + \| z_m\|_2 = \sum_{j} \|z_j\|_2.$

For the other side of the proof, I know that I need to use the case of equality in the Cauchy-Bunyakovskii-Schwarz(CBS) Inequality, which says that: For all $x,y \in \mathbb{C}^{m \times 1 }$ we have \begin{align} |x^ * y| = \|x\| \|y\| \text{, where $x^*$ is the conjugate transpose of x,} \end{align} if and only if $ y = \alpha x$ for $ \alpha = \dfrac{x^ * y}{x^ * x}.$

Any directions or help with the proof?

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    If the $z_j$ are real numbers as you says in your first line, then $\| \sum_{j} z_j \|_2$ neither $\|z_j\|_2$ makes sense. I think you mean something like $\|(z_1,z_2,\ldots,z_n)\|_2=\sum_j |z_j|\iff z_j=\alpha_j z_1$ for some $\alpha_j\gt 0$2012-04-01
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    So what is the difference, is not $\|z_j\|_2 = |z_j|$?2012-04-01
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    I think you want the condition to be $\alpha_j\geq0$, because otherwise the assertion is not true. Also, you probably want to say that each $z_j\in\mathbb{R}^n$; otherwise, the proof doesn't need CBS. Finally, the statement of CBS is wrong, as the condition you mention is the one that characterizes equality (the inequality holds no matter which $x,y$ you choose).2012-04-01
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    Given $z=(z_1,z_2,\ldots,z_n)\in\mathbb{R}^n$, $\|(z_1,z_2,\ldots,z_n)\|_2=\sqrt{\sum_{j=1}^n z_j^2}$. Usually $|x|$ is just the absolute value of $x\in\mathbb{R}$.2012-04-01
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    @leo: I thought if we define a norm of $z_j \in \mathbb{R}$ by $\|z_j\|_p$ then as $\| \cdot\|_p \geq 0,$ we have $\|z_j\|_p = \left( z_j^p \right)^{ \frac{1}{p} } = |z_j|.$2012-04-01
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    @MartinArgerami: The assertion holds for $ \alpha_j = 0$ but I want to show it holds for some $\alpha_j >0.$ Also, $z_j \in \mathbb{R}.$ I agree the CBS inequality holds no matter $x,y$ we choose, in particular $x,y \in \mathbb{R}^m$ when $m=1.$2012-04-01
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    @Sulaiman: then at least you need to improve your notation, as "$z\ne0$" does not usually mean that every component of $z$ is nonzero. My comment about the CBS inequality referred to the fact that you wrote the inequality (which holds for arbitrary vectors) together with the condition for equality (which holds only when you have equality on CBS).2012-04-01
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    @MartinArgerami: I slightly changed the question as I realized its the new question that I actually need the answer for. Thanks for all your help and good discussion. I read in Meyer's book of Matrix Analysis and Applied Linear Algebra that $x,y$ of my CBS inequality are in $\mathbb{C}^{m\times 1}$ but I don't understand, I thought they must be in $\mathbb{C}^{m}.$2012-04-01
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    @MartinArgerami: I see that instead of $z>0$ I should also write that $z=a+ib,$ where $a,b \in \mathbb{R}, a>0$ and $b>0$ but I thought its clear.2012-04-01

2 Answers 2

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Here is an answer for the newly formulated question. I will actually assume that $z_j\in\mathbb{C}^k$, as this does not change the proof.

Let us start with $m=2$. Then we have $\|z_1+z_2\|_2=\|z_1\|_2+\|z_2\|_2$; using that $\|x\|_2=x^*x$ and squaring we get $$ \|z_1\|_2+\|z_2\|_2+2\text{Re }z_1^*z_2=\|z_1\|_2+\|z_2\|_2+2\|z_1\|_2\|z_2\|_2. $$ So $$ \text{Re }z_1^*z_2=\|z_1\|_2\|z_2\|_2. $$ But then $$ \|z_1\|_2\|z_2\|_2=\text{Re }z_1^*z_2\leq|z_1^*z_2|\leq\|z_1\|_2\|z_2\|_2, $$ which implies that $$ z_1^*z_2=\|z_1\|_2\|z_2\|_2. $$ This is equality in the CBS inequality, so $z_2=\alpha_2\,z_1$, with $\alpha_2=z_1^*z_2/\|z_1\|_2^2\geq0$. Note that if $z_1,z_2$ are complex numbers, then $\alpha_2>0$.

Now we proceed by induction. If we have the equality for $h+1$ vectors, then $$ \sum_1^{h+1}\|z_j\|_2=\|\sum_1^{h+1}z_j\|_2\leq\|\sum_1^{h}z_j\|_2+\|z_{h+1}\|_2\leq\sum_1^{h+1}\|z_j\|_2. $$ Then we get equality for the first $h$ vectors and we can apply the inductive hypothesis to get $z_j=\alpha_jz_1$ for $j=1,\cdots,h$. Now $\sum_1^hz_j=(\sum_1^h\alpha_j)z_1$ and we can apply the $m=2$ case to get the final result.

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We assume $$\left|\sum_{k=1}^mz_k\right|=\sum_{k=1}^m|z_k|. $$ We consider two cases. The first one is when $\sum_{k=1}^mz_k\geq0$. Then we have $$\sum_{k=1}^mz_k=\sum_{k=1}^m|z_k|. $$ Then $$0=\sum_{k=1}^m|z_k|-z_k. $$ As every term in the sum is nonnegative, we get that $|z_k|=z_k$ for all $k$, so $z_k>0$ for all $k$. Now take $\alpha_k=z_k/z_1$.

The second case, when $\sum_{k=1}^mz_k<0$, is treated similarly and we get $|z_k|+z_k=0$ for all $k$. Now $z_k=-|z_k|<0$ for all $k$. So we can define $\alpha_k=z_k/z_1>0$ (quotient of two negative numbers).

This problem is definitely more interesting when $z_1,\ldots,z_m$ are vectors. Then the $\|\cdot\|_2$ notation is then justified, and the proof indeed requires the use of the case of equality in the CBS inequality.

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    This is just let everybody knows that this answer actually answered of my first format of the question, that's when $z \in \mathbb{R}^m.$ Thanks @Martin2012-04-01