I would like to compute the kernel of the following matrix.
Let $W \in \mathbb{R}^n$ have only $1$'s on its main diagonal and $w_{i,j} \in (-1,0)$ (open interval) off the main diagonal, so that the columns sum up to $-1$. I would like to show that
$$\dim \ker W = 1$$
If this is not true, I ask for a counterexample.
Idea:
It is clear that $\dim\ker W \geq 1$, since adding for all $i=1,...,n-1$ the $i$-th row to the last row of $W$ kills the last row.
Consider the Minor $W_n$, which is $W$ without the last column and the last row. It is left to show that $W_n$ has $\ker = 0$. Consider the decomposition (set $\tilde W_n = W_n - I_{n-1})$ $W_n = I_{n-1} + \tilde W_n$. Now $W_n x=0$ implies $\tilde W_n x = -x$, e.g., $\tilde W_n$ has eigenvalue $-1$.
So the goal is to show that $\tilde W_n$ has eigenvalue $-1$. I wanted to estimate the spectral raduius and show that is (in absolute value) strictly smaller $1$. This would finish the proof, but I dont know how to do that
I need help here or a different approach :)
The minor $W_n$ has diagonal entries $1$, and the sum of the off-diagonal entries in each columns is strictly larger than $-1$ (because deleting the last row of the matrix increases this sum). Then Gershgorin's theorem tells you that the eigenvalues of the remaining matrix are all non-zero.
If the off-diagonal entries would allowed to be taken from $[-1,0]$ then $$ \pmatrix{ 1 & -1 &0&0\\ -1 & 1 &0&0\\ 0&0&1&-1\\0&0&-1&1} $$ would be a counter-example.