Finding the characteristic polynomial of this matrix

Question

I have a question about this post.

How can we know that the geometric multiplicity of the eigenvalue $\lambda=0$ is $n-1$? I get that $0$ is an eigenvalue of $A$ because $\textrm{det}A=0$, but I can't get why its geometric multiplicity is $n-1$.

Answer 1

Because $(1, -\frac{1}{2}, 0, ..., 0)$, $(1, 0, -\frac{1}{3}, 0, ..., 0)$, ..., $(1, 0,..., 0, -\frac{1}{n})$ are linearly independent engenvectors.

Another way of seeing it is by simply noting that all the rows are equal. (The thing written in the paragraph above illustrates why it is enough to note this).