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" ?


Adding some more information:

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

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.

  • $\begingroup$ 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. $\endgroup$ Commented Jan 10, 2022 at 17:53
  • $\begingroup$ I mean, it is a vector space with component-wise addition and scalar multiplication.. $\endgroup$
    – Kenta S
    Commented Jan 10, 2022 at 17:53
  • 2
    $\begingroup$ Tensor......... $\endgroup$ Commented Jan 10, 2022 at 17:55
  • $\begingroup$ 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. $\endgroup$ Commented Jan 10, 2022 at 18:30

2 Answers 2


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.


  • $\begingroup$ ${\bf p}$, 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. $\endgroup$ Commented Jan 10, 2022 at 17:54
  • $\begingroup$ Sorry but your definition of a vector is too narrow. $\endgroup$
    – maxmilgram
    Commented Jan 10, 2022 at 17:55
  • $\begingroup$ 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. $\endgroup$ Commented Jan 10, 2022 at 17:57
  • $\begingroup$ 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. $\endgroup$
    – maxmilgram
    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.


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

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 $j \neq i$.


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