Consider a matrix $E \in \mathbb{R}^{m \times n}$ with $m\geq n$ and a nullspace $\text{ker}(E) = \{ \alpha 1_n , \, \alpha \in \mathbb{R} \}$, where $1_n$ is a column vector of ones of appropriate length.
I want to prove that the kernel of the matrix $E \otimes I_p$, where $I_p$ is the identity matrix of dimension $p \in \mathbb{R}$, is $1_n \otimes z_p$ with $z_p$ an arbitrary vector of length $p \in \mathbb{R}$. I came up with a way to prove it, but I am very unsure if it is mathematically completely correct (I have an engineering background).
Consider \begin{equation} \label{eq:1}(E \otimes I_p)v = 0 \end{equation} with partition $$ E \otimes I_p = \begin{bmatrix} e_{11} I_p & e_{12} I_p & \dots & e_{1n}I_p \\ e_{21} I_p & \dots & & \vdots \\ \vdots & & & \\ e_{m1}I_p & \dots & & e_{mn}I_p \end{bmatrix}, \quad v = \begin{bmatrix} v_{1:p} \\ v_{p+1:2p} \\ \vdots \\ v_{(n-1)p+1:np} \end{bmatrix}. $$
By evaluating the individual block-rows of $(E \otimes I_p)v = 0$ and considering that $e_i^\top 1_n = 0$, where $e_i^\top = (e_{i1}, \dots, e_{in})$ is some row of $E$, we see that the different partitions in $v$ (i.e. $v_{1:p}, ...$) have to be equal.
This meand that the kernel of $(E \otimes I_p)$ has to be $1_n \otimes z_p$, with $z_p \in \mathbb{R}^p$.
Can someone give me some indications whether it is correct or maybe a more simple way to prove that exists? Thanks in advance!
Let $e_1,\dots,e_n$ be the basis of $\mathbb R^n$, then $e_{ij} = e_i \otimes e_j$ is the basis of $\mathbb R^n \otimes \mathbb R^p$, on which $E \otimes I_p$ acts.
To find the kernel of $E \otimes I_p$, we have to solve
$$ (E \otimes I_p) \sum\limits_{i,j} v_{ij}(e_i \otimes e_j) = \sum\limits_{i,j} v_{ij} (E e_i \otimes e_j) = 0. $$ Grouping summands by $j$, we get
$$ \sum\limits_j \left(\sum\limits_i v_{ij} E e_i \right) \otimes e_j = 0. $$
Since Kronecker products with $e_j$ are linearly independent from each other for distinct $j$, the only way this identity is true is if each individual LHS of Kronecker products is a nullvector of $E$, that is $\alpha_j 1_n$. Thus,
$$ \sum\limits_j \alpha_j 1_n \otimes e_j = 1_n \otimes \left(\sum\limits_j \alpha_j e_j\right) = 1_n \otimes z_p. $$