# Thread: A Proof Involving Upper Triangular Matrices

1. ## A Proof Involving Upper Triangular Matrices

I'm trying to think of a way to start this "proof":

Let $\displaystyle S$ be an $\displaystyle n\times n$ strictly upper triangular matrix. Show that $\displaystyle \left(I-S\right)^{-1}=I+S^2+S^3+\dots+S^{n-1}$
Source: Matrices and Linear Algebra, 2nd Ed. by Schneider and Barker

I don't want a complete proof. All I would appreciate would be an explanation on how I can start this. I will be able to take it from there.

Thank you in advance!!

2. Let $\displaystyle S$ be an $\displaystyle n\times n$ strictly upper triangular matrix. Show that $\displaystyle \left(I-S\right)^{-1}=I+S^2+S^3+\dots+S^{n-1}$
We just need to show $\displaystyle (I - S)(I + S + S^2 + ... + S^{n-1}) = I$.
Open LHS to get, $\displaystyle I + S + S^2 + ... + S^{n-1} - S - S^2 - ... - S^n = I - S^n$.
We want for $\displaystyle I - S^n = I$ which is true if and only if $\displaystyle S^n = \bold{0}$ where $\displaystyle \bold{0}$ is $\displaystyle n\times n$ zero matrix.

Thus, it remains to prove $\displaystyle S^n = \bold{0}$.
(Since $\displaystyle S$ is strictly upper triangular it is nilpotent).

3. Originally Posted by ThePerfectHacker
Thus, it remains to prove $\displaystyle S^n = \bold{0}$.
(Since $\displaystyle S$ is strictly upper triangular it is nilpotent).
I've been trying to think of a way to prove this for hours now.

This is brought up in the "Matrix Multiplication" section of my textbook. They ask me to prove this as well...but I can't use eigenvalues and determinants.

I think that mathematical induction can be used...but I'm not sure what I'm supposed to let my $\displaystyle P\left(n\right)$ be.

I was thinking that $\displaystyle P\left(n\right):A^n=0$, but $\displaystyle P\left(1\right): A=0$, but that's not necessarily true.

I would appreciate any suggestions!!!

4. Originally Posted by Chris L T521
I was thinking that $\displaystyle P\left(n\right):A^n=0$, but $\displaystyle P\left(1\right): A=0$, but that's not necessarily true.
One way round that is to start the induction at n=2, where the result is easily verified.

Another way is to think more carefully about the n=1 case. A 1×1 matrix has just a single element. This lies on the main diagonal (which also contains just the one element!). So if the matrix is strictly upper triangular then that element must be 0. Therefore the matrix is the zero matrix.

5. Observe that A = $\displaystyle \begin{pmatrix} 0 & v \\ \textbf{0} & A' \end{pmatrix}$ where the bold faced 0 stands for the column vector 0 and v stands for a "n-1" length row vector and A' is a (n-1) x (n-1) strictly upper triangular matrix.

finally for induction to work, you just need to see that:

$\displaystyle A^n = \begin{pmatrix} 0 & v(A')^{n-1} \\ \textbf{0} & (A')^n \end{pmatrix}$

Actually the above can be proved by another induction

6. an easy way is to see that since $\displaystyle \det (xI - S)=x^n,$ all eigenvalues of $\displaystyle S$ are 0 and thus by Cayley-Hamilton: $\displaystyle S^n=0.$

7. Originally Posted by NonCommAlg
an easy way is to see that since $\displaystyle \det (xI - S)=x^n,$ all eigenvalues of $\displaystyle S$ are 0 and thus by Cayley-Hamilton: $\displaystyle S^n=0.$
I do not know how to solve this problem, but just to make sure others are aware, I talked to Chris about this problem (outside MHF) and the difficulty he was having with it was that it was in a chapter preceeding both determinants and eigenvalues.

8. Originally Posted by Isomorphism
finally for induction to work, you just need to see that:

$\displaystyle A^n = \begin{pmatrix} 0 & (vA')^{n-1} \\ \textbf{0} & (A')^n \end{pmatrix}$
That is almost correct, except that the entry $\displaystyle (vA')^{n-1}$ should be $\displaystyle v(A')^{n-1}$.