Hollow matrix and n-th power proof I have this exercise but I don't have acquired yet all the information I need in order to deal with it, at least.
Show that if an upper (lower) triangular matrix has only zeros on the main diagonal, then there is some n ∈ N such that the n-th power of the matrix is zero. Can such a matrix (and the corresponding linerar operator) be invertible (why, or why not)?.
As far as I know, in order for a triangular matrix to be invertible, can't be any zero in its diagonal. More in general, I know that when a matrix has an inverse, the product of the matrix and its inverse is equal to the identity matrix - I know that the identity matrix is the one with 1s as its diagonal. Thus, I know the answer to the last question in bold character. Namely, I think that a matrix with 0s as its diagonal is not invertible. Why? I can only give the reasons above as still I am not confident with proof-writing.
However, the task also asks to show that if an upper (lower) triangular matrix has only zeros on the main diagonal, then there is some n ∈ N such that the n-th power of the matrix is zero. How can I prove this?

Hint: Show (by looking at the matrix product as a summation) that if $S$ and $T$ are lower-triangular with $0$s on the diagonal, then $M = ST$ will have zeros on the diagonal and below the diagonal. That is, if $i \geq j-1$, then $M_{ij} = 0$.
How does this pattern continue?  Note that we keep getting more $0$s.

Note that $S$ and $T$ satisfy $S_{ij}=0$ and $T_{ij} = 0$ if $i \geq j$.  So, we have
$$
M_{ij} = \sum_{k=1}^n S_{ik}T_{kj} = \sum_{k = i+1}^{j-1} S_{ik}T_{kj}
$$
When $i \geq j - 1$, the above is an empty summation so that $M_{ij} = 0$.
