# Name of a "vector" where the components are vectors?

This is my first question on the forum and I am not a native speaker of english, so i apologize if i miss something of the post rules.

I have a tuple $$\boldsymbol{p} = \left(\boldsymbol{p_1}, \ldots,\boldsymbol{p_i}, \ldots \boldsymbol{p_N}\right)$$ where the components $$\boldsymbol{p_i}$$ are vectors of different dimensions; that is, $$\boldsymbol{p_i} = \left(x_1, \ldots, x_{n(i)}\right)$$, where $$n(i)$$ depends of $$i$$, for all $$i = 1, \ldots, N$$.

How $$\boldsymbol{p}$$ can be called? It is right to say that $$\boldsymbol{p}$$ is a vector? If so, should i called it as a "vector of vectors" ?

EDIT:

$$x_1, \ldots, x_{n(i)} \in \mathbb{R}$$.

And adding some context, this is for an statistical paper; my intention is define the parametric space and explain it for the reader, so i don't need to define operations over $$\boldsymbol{p}$$. Nevertheless, thank you to everyone who replied.

• Technically it is not a vector, but we can consider it as a vector by rewriting it as $p=(x_1,\ldots ,x_{n(1)},x_2,\ldots ,x_{n(2)},\ldots )$. In this sense, also a matrix is just a vector. Commented Jan 10, 2022 at 17:53
• I mean, it is a vector space with component-wise addition and scalar multiplication.. Commented Jan 10, 2022 at 17:53
• Tensor......... Commented Jan 10, 2022 at 17:55
• The way we index vectors in different spaces is immaterial as long as they're over the same field and the operations are done component-wise. Also, for any finite dimensional combinations this vector space you mention will be isomorphic to $\mathbb{R}^k$ where $k$ is the sum of the dimensions of each space. Unless we gain some insight by breaking it into parts it makes more sense to work over the more familiar space. Commented Jan 10, 2022 at 18:30

Yes, any tuple of vectors over the same scalar field defines another viable vector space with the obvious choice of multiplication with a scalar and addition of vectors. The eight axioms can be easily verified. This holds true even for general "non-euclidian" choices of vector spaces. Regarding the naming I am not aware of a standard name for this, to avoid confusion I would suggest to not rely on a catchy short name but to explain it in more words. In your special case e.g.: we have a vector which is a tuple of different dimensional euclidean vectors.

en.m.wikipedia.org/wiki/Vector_space

• ${\bf p}$, as the OP wrote, is only a vector if the entries are from the underlying field. A vector is not an element of the field. What you mean is, that we can rewrite it like that. Commented Jan 10, 2022 at 17:54
• Sorry but your definition of a vector is too narrow. Commented Jan 10, 2022 at 17:55
• Yes, but I assume the definition the OP has given, namely $p_i=(x_1,\ldots ,x_{n(1)})$ with scalars $x_i$. You should read what he wrote about a vector. Commented Jan 10, 2022 at 17:57
• There is a standard definition of a vector. I use that. Furthermore I differentiate in the answer between the standard euclidean vector spaces and the general definition to make sure I get my pont across. Lastly if we use only vectors as given in the question your comment is void as it is automatically fulfilled. Commented Jan 10, 2022 at 18:01

I'm going to assume you mean all the $$x$$'s involved are real numbers and that all of $$n(i)$$ and $$N$$ are finite natural numbers greater than or equal to 1.

If $$\mathbb{p}_i \in \mathbb{R}^{n(i)}$$ for all $$i$$ from $$1$$ to $$N$$. Then what you are forming is a direct sum

$$V \equiv \bigoplus_{i=1}^N \mathbb{R}^{n(i)}$$

So $$\mathbb{p} \in V$$. It is a vector in a bigger vector space.

To perform the vector space operations on $$\mathbb{p}$$ and $$\mathbb{q}$$ which are both tuples of the same format, do them in each of the $$N$$ indices.

So

$$\alpha \mathbb{p} + \beta \mathbb{q} = (\alpha \mathbb{p}_1 + \beta \mathbb{q}_1 , \cdots \alpha \mathbb{p}_N + \beta \mathbb{q}_N)$$

and each of the factors are defined because the $$\mathbb{R}^{n(i)}$$ were all vector spaces.

For each $$i$$ you can include $$\mathbb{R}^{n(i)}$$ into $$V$$ by taking the inclusion map $$i$$ to be defined by

$$i (\mathbb{p}_i) \equiv (0,\cdots, \mathbb{p}_i , \cdots 0)$$

where each of those 0's indicate the zero vector in $$\mathbb{R}^{n(j)}$$ when $$j \neq i$$.