You're right.

A matrix, or linear transformation, being singular means that it collapses the space onto a lower dimensional subspace. From n to n-1 and so on.

This kind of linear transformation would then have at most n eigenvalues. If one of these were 0, then the corresponding eigenvector would be non zero vector that by the linear transformation gets mapped to the null space.

This suggests that some of the rows of this matrix are linearly dependent.

If there were a linear transformation that had all the eigenvalues 0. That would mean all the non zero eigenvectors get mapped to null space. Or all the rows are linearly dependent. Which can't happen because at least one of the rows would have to have an eigenvalue. Unless it was all 0 matrix.

Basically I'm saying rank has to be at least 1 . So all eigenvalues being 0 is not possible.

And only one or more have to be 0, not all of them.

I think you can find many circular explanations why you're right, from many different perspectives.