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 :)

• I think you mean that the sum of the off-diagonal entries in each column is $-1$, so that the entries in each columns sum up to zero. – daw Apr 29 at 8:24

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.
• its not a counter-example... since i asked off-diagonal entires to be not 0 and not -1. However i solved my own question. The answer is to used diagonal dominance of the Minor matrix $W_n$ (kill the last column and last row) to see that $rang$ of $\tilde W$ is at least n-1. – mathemagier May 10 at 16:30