1
$\begingroup$

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?

$\endgroup$
0
$\begingroup$

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 $M_{ij} = 0$.

$\endgroup$
  • $\begingroup$ I am really not tailored for proofs...perhaps, I don't put enough effort but, since I am the beginning, I see this as an insurmountable obstacle. $\endgroup$ – Always learning Nov 3 '15 at 19:28
  • $\begingroup$ summation notation helps. I might add something here $\endgroup$ – Omnomnomnom Nov 3 '15 at 19:30
  • $\begingroup$ See my latest edit $\endgroup$ – Omnomnomnom Nov 3 '15 at 19:39
  • $\begingroup$ Thank you, really! But, unfortunately, I can't focus on this moment...will try later... Always more sure I will abandon this degree, although just started.. $\endgroup$ – Always learning Nov 3 '15 at 19:49
  • $\begingroup$ is the proof completed? $\endgroup$ – Always learning Nov 4 '15 at 7:39

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.