How the eigenvectors look like of graph $G=G_1\cup G_2$?

Question

I have a Bilateral graph $K_{2,2}$ where $|L|=|R|=2$. The adjacency matrix of this graph is: $$ A=\begin{bmatrix}0 & 0 & 1 & 1\\ 0 & 0 & 1 & 1\\ 1 & 1 & 0 & 0\\ 1 & 1 & 0 & 0 \end{bmatrix} $$ The eigenvalues are $(2,0,0,-2)$ and the eigenvectors are: $$ \left\{ \begin{bmatrix}1\\ 1\\ 1\\ 1 \end{bmatrix},\begin{bmatrix}0\\ 0\\ 1\\ -1 \end{bmatrix},\begin{bmatrix}1\\ -1\\ 0\\ 0 \end{bmatrix},\begin{bmatrix}-1\\ -1\\ 1\\ 1 \end{bmatrix}\right\} $$ Now the adjacency matrix of $G=K_{2,2}\cup K_{2,2}$ is: $$ A'=\begin{bmatrix}0 & 0 & 1 & 1 & 0 & 0 & 0 & 0\\ 0 & 0 & 1 & 1 & 0 & 0 & 0 & 0\\ 1 & 1 & 0 & 0 & 0 & 0 & 0 & 0\\ 1 & 1 & 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 0 & 1 & 1\\ 0 & 0 & 0 & 0 & 0 & 0 & 1 & 1\\ 0 & 0 & 0 & 0 & 1 & 1 & 0 & 0\\ 0 & 0 & 0 & 0 & 1 & 1 & 0 & 0 \end{bmatrix} $$ And the eigenvalues are $\left(2,2,0,0,0,0,-2,-2\right)$ and eigenvectors are: $$ \left\{ \begin{bmatrix}1\\ 1\\ 1\\ 1\\ 0\\ 0\\ 0\\ 0 \end{bmatrix},\begin{bmatrix}0\\ 0\\ 0\\ 0\\ 1\\ 1\\ 1\\ 1 \end{bmatrix},\begin{bmatrix}0\\ 0\\ 1\\ -1\\ 0\\ 0\\ 0\\ 0 \end{bmatrix},\begin{bmatrix}1\\ -1\\ 0\\ 0\\ 0\\ 0\\ 0\\ 0 \end{bmatrix},\begin{bmatrix}0\\ 0\\ 0\\ 0\\ 0\\ 0\\ 1\\ -1 \end{bmatrix},\begin{bmatrix}0\\ 0\\ 0\\ 0\\ 1\\ -1\\ 0\\ 0 \end{bmatrix},\begin{bmatrix}1\\ 1\\ -1\\ -1\\ 0\\ 0\\ 0\\ 0 \end{bmatrix},\begin{bmatrix}0\\ 0\\ 0\\ 0\\ 1\\ 1\\ -1\\ -1 \end{bmatrix}\right\} $$ Given graph $G=G_1\cup G_1$, if the eigenvectors of $G_1$ are $\lambda_1,\ldots\lambda_n$ of eigenvectors $\vec{v}_1,\ldots\vec{v}_n$, then $\lambda_1,\ldots\lambda_n$ are eigenvectors of $G$ but how $\vec{v}'_1,\ldots\vec{v}'_n$ look like?

Answer 1

If you look at your example:

You have the eigenvectors of (the adjacency matrix of) $G_1$ as $$\left\{ \begin{bmatrix}1\\ 1\\ 1\\ 1 \end{bmatrix},\begin{bmatrix}0\\ 0\\ 1\\ -1 \end{bmatrix},\begin{bmatrix}1\\ -1\\ 0\\ 0 \end{bmatrix},\begin{bmatrix}1\\ 1\\ -1\\ -1 \end{bmatrix}\right\} $$

And the eigenvectors of $G_2$ are exactly the same (Because $G_1=G_2=K_{2,2}$).

Now the eigenvectors of $G_1\cup G_2$ are the following:

$$ \left\{ \begin{bmatrix}1\\ 1\\ 1\\ 1\\ 0\\ 0\\ 0\\ 0 \end{bmatrix},\begin{bmatrix}0\\ 0\\ 1\\ -1\\ 0\\ 0\\ 0\\ 0 \end{bmatrix},\begin{bmatrix}1\\ -1\\ 0\\ 0\\ 0\\ 0\\ 0\\ 0 \end{bmatrix},\begin{bmatrix}1\\ 1\\ -1\\ -1\\ 0\\ 0\\ 0\\ 0 \end{bmatrix},\begin{bmatrix}0\\ 0\\ 0\\ 0\\ 1\\ 1\\ 1\\ 1 \end{bmatrix},\begin{bmatrix}0\\ 0\\ 0\\ 0\\ 0\\ 0\\ 1\\ -1 \end{bmatrix},\begin{bmatrix}0\\ 0\\ 0\\ 0\\ 1\\ -1\\ 0\\ 0 \end{bmatrix},\begin{bmatrix}0\\ 0\\ 0\\ 0\\ 1\\ 1\\ -1\\ -1 \end{bmatrix}\right\} $$

(as you have already found)

Note that I changed the order of these vectors slightly, to make the point more clear. We see that the first 4 vectors are the eigenvectors of $G_1$, with zeroes on the spots belonging to $G_2$, and the last 4 vectors are the eigenvectors of $G_2$, with zeroes on the spots belonging to $G_1$.

This is true in general: Given two linear maps $A_1\colon V_1\to V_1$ and $A_2\colon V_2\to V_2$, the eigenvectors of $A_1\times A_2\colon V_1\times V_2\to V_1\times V_2$ are exactly $v_{11}\times0,\dots v_{1n}\times 0,0\times v_{21},\dots 0\times v_{2m}$ where $v_{11},\dots v_{1n}$ are eigenvectors of $A_1$ and $v_{21},\dots v_{2m}$ are eigenvectors of $A_2$.