# Series of short questions intended to clarify $p$-norm of vectors with complex elements

Note:

My central question is Q.2.B. I'd really like an answer to that question.

Context:

From [1], we find that

Let $p \geq 1$ be a real number. The $p$-norm of vectors $\mathbf{x} = (x_1, \ldots, x_n)$ is $$\left\| \mathbf{x} \right\| _p := \bigg( \sum_{i=1}^n \left| x_i \right| ^p \bigg) ^{1/p}.$$

Questions:

[Q.1.A]

Is the $1$-norm of vector $\mathbf{x} = ([1 + i], [3 + 2\,i], [6-20\,i])$ given as follows: \begin{align} \left\| \mathbf{x} \right\| _1 = & \bigg( \sum_{i=1}^3 \left| x_i \right| ^1 \bigg) ^{1/1} \\ = & \bigg( \left| 1 + i \right| + \left| 3 + 2\,i \right|+ \left| 6-20\,i \right|\bigg) \\ = & \bigg( \left| 1 \right|+ \left|i \right| + \left| 3 \right|+ \left| 2\,i \right|+ \left| 6 \right| + \left|- 20\,i \right|\bigg) \\ = & \bigg( 1 + 1 + 3 + 2 + 6 + 20 \bigg) \\ = & \, 33? \end{align}

[Q.1.B]

Or, is the $1$-norm of vector $\mathbf{x} = ([1 + i], [3 + 2\,i], [6-20\,i])$ given as follows: \begin{align} \left\| \mathbf{x} \right\| _1 = & \bigg( \sum_{i=1}^3 \left| x_i \right| ^1 \bigg) ^{1/1} \\ = & \bigg( \left| 1 + i \right| + \left| 3 + 2\,i \right|+ \left| 6-20\,i \right|\bigg) \\ = & \bigg( \sqrt{1^2 + 1^2} + \sqrt{3^2 + 2^2} + \sqrt{6^2 + 20^2} \bigg) ? \end{align}

[Q.2.A]

Is there a named norm such that the named norm of vector $\mathbf{x} = ([1 + i], [3 + 2\,i], [6-20\,i])$ is equal to 33 (see Q.1.A)?

[Q.2.b]

If there is a such a named norm, what is the name of the norm?

[Q.3]

Is the $2$-norm of vector $\mathbf{x} = ([1 + i], [3 + 2\,i], [6-20\,i])$ given as follows: \begin{align} \left\| \mathbf{x} \right\| _2 = & \bigg( \sum_{i=1}^3 \left| x_i \right| ^2 \bigg) ^{1/2} \\ = & \sqrt{ \left| 1 + i \right|^2 + \left| 3 + 2\,i \right|^2+ \left| 6-20\,i \right|^2} \\ = & \sqrt{ \left( \sqrt{ 1^2 + 1^2 }\right)^2 + \left( \sqrt{ 3^2 + 2^2 }\right)^2 + \left( \sqrt{ 6^2 + (-20)^2 }\right)^2} \\ = & \sqrt{ 1^2 + 1^2 + 3^2 + 2^2 + 6^2 + (-20)^2 } \\ = & \sqrt{ 1 + 1 + 9 + 4 + 36 + 400 } \\ = & \sqrt{ 451 }? \end{align}

Bibliography

• $|A+B|\neq|A|+|B|$ – zipirovich Mar 19 '18 at 21:49

The wording on your question seems odd (not sure what a 'named norm' means) but I feel your real question translates to see if $$\rho(z_1,\dots,z_n)=\sum_{i=1}^n |\textrm{Re}(z_i)|+|\textrm{Im}(z_i)|$$ is a norm on $\mathbb{C}^n$. To see that this isn't true just consider $v=(1+i,0,\dots,0)$ and $\lambda=1-i$. In order to have $\rho$ be a norm it shoud satisfy $\rho(\lambda v)=|\lambda|\rho(v)$, but we actually have $$\rho(\lambda v)=\rho(2,0,\dots,0)=2$$ and $$|\lambda|\rho(v)=2\sqrt 2$$ This is basically the argument you used in your answer but with a concrete example to show it doesn't always hold.

• I agree with you that the nomenclature 'named norm' is not as clear as it could be. I mean to highlight that some norms have names. For example, consider 'Euclidean norm' or 'Manhattan norm', where one notes the proper noun indicating a name (cf. en.wikipedia.org/wiki/Norm_(mathematics)). – Michael Levy Apr 30 '18 at 17:19

No, the 1-norm of $x=([1+i],[3+2i],[6−20i])$ is not equal to 33.

Yes, the 1-norm of $x=([1+i],[3+2i],[6−20i])$ is as given.

Pursuant to [2], "given a vector space $V$ over a subfield $F$ of the complex numbers, a norm on $V$ is a function $p$: $V → R$ with the following properties: For all $a \in F$ and all $\mathbf{u}, \mathbf{v} \in V$,

$p(a\,\mathbf{v}) = |a| \, p(\mathbf{v})$ (being absolutely homogeneous or absolutely scalable).

$p(\mathbf{u} + \mathbf{v}) \leq p(\mathbf{u}) + p(\mathbf{v})$ (being subadditive or satisfying the triangle inequality).

$p(\mathbf{v}) \geq 0$ (being positive or more precisely non-negative).

If $p(\mathbf{v}) = 0$ then $\mathbf{v}=0$ is the zero vector (being definite or being point-separating)."

Here I define a function $p_o(\mathbf{v})$ as

$$p_o(\mathbf{v}) := \left\| \textrm{Re}(\mathbf{v}) \right\|_1 + \left\| \textrm{Im}(\mathbf{v}) \right\|_1;$$

where

$$\left\| \mathbf{x} \right\| _p := \bigg( \sum_{i=1}^n \left| x_i \right| ^p \bigg) ^{1/p}$$

and $\textrm{Re}(\mathbf{v})$ and $\textrm{Im}(\mathbf{v})$ are the real parts of $\mathbf{v}$ and imaginary parts of $\mathbf{v}$ , respectively.

Now, I have to check that the function $p_o$ has the properties of a norm. If so, the function is a norm. If not, the function is not a norm.

Is function $p_o$ absolutely homogeneous or absolutely scalable? \begin{align} p_o(a\,\mathbf{v}) = & \left\| \textrm{Re}(a\,\mathbf{v}) \right\|_1 + \left\| \textrm{Im}(a\,\mathbf{v}) \right\|_1 \\ & \left\| \textrm{Re}(a)\, \textrm{Re}(\mathbf{v}) - \textrm{Im}(a)\, \textrm{Im}(\mathbf{v}) \right\|_1 + \left\| \textrm{Im}(a)\, \textrm{Re}(\mathbf{v}) + \textrm{Re}(a)\, \textrm{Im}(\mathbf{v}) \right\|_1 \\ & \sum_{i=1}^{n} \left| \textrm{Re}(a)\, \textrm{Re}( {v_i}) - \textrm{Im}(a)\, \textrm{Im}({v_i}) \right| + \sum_{i=1}^{n}\left| \textrm{Im}(a)\, \textrm{Re}({v_i}) + \textrm{Re}(a)\, \textrm{Im}( {v_i}) \right| \\ & \sum_{i=1}^{n}\left( \left| \textrm{Re}(a)\, \textrm{Re}( {v_i}) - \textrm{Im}(a)\, \textrm{Im}({v_i}) \right| + \left| \textrm{Im}(a)\, \textrm{Re}({v_i}) + \textrm{Re}(a)\, \textrm{Im}( {v_i}) \right| \right) \end{align} while \begin{align} \left|a\right|\,p_o(\mathbf{v}) = & \left|a\right|\,\left( \left\| \textrm{Re}(\mathbf{v}) \right\|_1 + \left\| \textrm{Im}(\mathbf{v}) \right\|_1\right) \\ & \left|a\right|\,\left(\sum_{i=1}^3\left| \textrm{Re}( {v_i}) \right| + \sum_{i=1}^3\left| \textrm{Im}( {v_i}) \right| \right) \\ & \left|a\right|\,\sum_{i=1}^3\left(\left| \textrm{Re}( {v_i}) \right| + \left| \textrm{Im}( {v_i}) \right| \right) \\ & \sqrt{\left(\textrm{Re}(a)\right)^2 + \left(\textrm{Im}(a)\right)^2} \,\sum_{i=1}^3\left(\left| \textrm{Re}( {v_i}) \right| + \left| \textrm{Im}( {v_i}) \right| \right) \end{align}

I find that, in general,

$$p_o(a\,\mathbf{v})\neq a\,p_o(\mathbf{v})$$

Resulting from the fact that the function $p_o$ is not scalable, $p_o$ is not a norm at all. Therefore, by definition, the function cannot be a named norm. Specifically, within the context of question Q.2, though it is possible to define a function such that the function of $\textbf{x}$ equals to 33, such a function will not be a norm.

Yes, the 2-norm of $x=([1+i],[3+2i],[6−20i])$ is as given.
• I'd suggest avoid using the 'metric' in this context, as it usually refers to a function of the form $d:X\times X \rightarrow \mathbb{R}_{\geq 0}$ that satisfies some properties. – Julio Cáceres Apr 30 '18 at 17:02
• I'd just change metric for function, and then proceed to check whether $\rho$ is a norm or not. – Julio Cáceres Apr 30 '18 at 17:31