# Basis of a product vector space

Let $E$ a vector space of dimension $p$ with $(e_1, \ldots, e_p)$ as a basis. Define the cartesian product vector space $F = v_1^\top E \times v_2^\top E \times \ldots \times v_n^\top E$ where the superscript $\top$ denotes the transpose and $v_1, \ldots, v_n$ are vectors in $\mathbb{R}^p.$ Here the notation $v_i^\top E = \{ v_i^\top x: x \in E\}.$

My question is to find a basis of $F.$ I know that $\dim(F) = np.$ Does anyone have an idea of how to write a basis of $F$ using the basis $(e_1, \ldots, e_p)$ of $E$?

Thank you

• let me guess, you write vectors as rows? – user251257 Jul 31 '15 at 1:13
• no, I am using column vectors. – mathdatastatsml Jul 31 '15 at 1:28
• I am confused. Is $E$ actually $\mathbb R^p$? If that is true, then $v_i^\top E = \mathbb R$ or $\{ 0 \}$ depending on $v_i\ne 0$ or $v_i = 0$. – user251257 Jul 31 '15 at 1:38
• $E$ isn't $\mathbb{R}^p,$ in my case $E$ is a $p$-cartesian product of Hilbert space. – mathdatastatsml Jul 31 '15 at 9:04
• how is $v_i^\top x$ defined? – user251257 Jul 31 '15 at 9:06

If $\{e_i\}$ is a basis for $E$, we can define a basis $\{v_j^T e_i\}$ for $v_j^T E$. So it follows a suitable basis for $F=\prod_{j=1}^n v_j^T E$ could be $$\{(\underbrace{0,\dots,0}_{j-1\text{ zeros}},v_j^T e_i,\underbrace{0,\dots,0}_{n-j\text{ zeros}}):1\le i\le p,1\le j\le n\}$$
Let $V$ be n-dimensional .Consider the product $W:= V \times V \times ...\times V$ (k times product) and an element $(a_1, a_2,.., a_n)$, with {$v_1, v_2,..,v_n$} a basis for $V$, so that $W$ is $kn-$dimensional . Assume for simplicity that $V$ is $\mathbb R^n$ (yes, every finite-dimensional vector space is isomorphic to $\mathbb R^n$, but this is $\mathbb R^n$)Then the set {$(v_1,0,0,..,0), (v_2,0,...,0),..,(v_n,0,0,....,0)| (0,v_1,0,...,0), (0, v_2, 0,..0),...,(0, v_n, 0,...,0)|....., (0,..,0,v_1), (0,0,..,v_2), ... ( 0,0,..,v_n)$ , is a basis for $V^n$, where we "embed" $v_k$ as an n-ple in the $kn-$ ple $(v_{11}, v_{12}.., v_{1n}| v_{21} , v_{22},.., v_{2n}|.....| v_{k1}, v_{k2},.., v_{kn}$ , where $v_{jn}$ is in the $j-$th copy of $V$ in $V \times V \times...\times V$. Sorry, could not come up with a better notation.
Notice that the collection $(0,0,..,0, v_{k1},..,v_{kn}, 0,0,0,..0)$ allows you to generate all the vectors in the $kth$ copy of $V$. _ EDIT: Rewriting, please wait.