Suppose $X$ is a $k$-regular graph with adjacency matrix $A$. I wish to show that if $k$ has multiplicity $1$ as an eigenvalue of $A$ then $X$ is connected.
By way of contradiction I assume that X is not connected. Now A can be block diagonalized, $A=diag(A_1,A_2\cdots A_t)$ where each $A_i$ is the adjacency matrix of a connected component and is of order $n_i\times n_i$. Furthermore if $v$ is the eigenvector corresponding to $k$ then the first $n_1$ entries in $Av$ are obtained from linear combinations of the columns $A_1$, the next $n_2$ entries from linear combinations of the columns of $A_2$ and so on. At this point I am stuck. Ideally I would like to construct two linearly independent eigenvectors corresponding to $k$ to derive a contradiction. Its not clear to me how can I do that.
Can someone help please.
PS: My intuition is that the vector consisting of first $n_1$ entries of $v$ followed by zeroes, is an eigenvector, and $n_1$ zeroes followed by the corresponding $n_2$ entries of $v$, followed by zeroes is another eigenvector, independent of the first. Is this correct?