I am trying to solve the following exercise:
Show that a $n \times n$ upper triangular matrix with zeros in the diagonal is always nilpotent.
Definition:
An Endomorphism $\phi$ is called nilpotent if there exists a $n$ such that,
$\phi^n \equiv 0$
My Calculations:
So let $\phi: V \rightarrow V$ be an endomorphism and let $B$ be a base for $V$.
To save some space, let $B=${$b_1,b_2,b_3$}
One can write the matrix as:
$_B[\phi]_B=(_B[\phi(b_1)]$ $_B[\phi(b_2)]$ $_B[\phi(b_3)])$
Now this matrix has to be a upper triangular matrix with zeros in the diagonal, this means, it has to be of the form:
$\left( \begin{array}{ }
0 & x_{12} & x_{13} \\
0 & 0 & x_{23} \\
0 & 0 & 0 \\
\end{array}\right) $
Thus,
$_B[\phi(b_1)]=0$
$_B[\phi(b_2)]=x_{12}b_1$
$_B[\phi(b_3)]=x_{13}b_1+x_{23}b_2$
(Back to the $n \times n$ case):
This means, $\phi$ is defined as:
$\phi(b_1) =0$
and
$\phi(b_{i+1})=\sum_{n=0}^{i} c_n b_n$ for (i=1,2,…,n-1)
This means, $_B[\phi^{n}(b_n)]=0$
In the $3 \times 3$ Case we had:
$_B[\phi(b_1)]=0$
$_B[\phi(b_2)]=x_{12}b_1$
$_B[\phi(b_3)]=x_{13}b_1+x_{23}b_2$
Now considering $\phi^2$:
$_B[\phi^2(b_1)]=0$
$_B[\phi^2(b_2)]=_B[x_{12} \phi(b_1)]=0$
$_B[\phi^2(b_3)]=_B[x_{13}\phi(b_1)+x_{23}\phi(b_2)]=0+x_{23} x_{12} b_1$
And for $\phi^3$ we get
$_B[\phi^3(b_1)]=0$
$_B[\phi^3(b_2)]=0$
$ _B[\phi^3(b_3)]=x_{23} x_{12} $ $ {_B}[\phi(b_1)]=0$
Thus, an upper triangular matrix with zeros in the diagonal is nilpotent.
Question:
Lineare Algebra isn't my strength, so I would like to know if my thoughts/calculations are correct.
Best Answer
Let $V=V_1=F\times F\times F$, $V_2=\{0\}\times F\times F$, and $V_3=\{0\}\times\{0\}\times F$, and $V_4$ to be the zero subspace. Notice that $V_4\subseteq V_3\subseteq V_2\subseteq V_1$.
One way to understand the action of an upper triangular matrix $T$'s action on $V$ (by multiplying on the right of row vectors) is that $T(V_i)\subseteq V_i$ for $i\in 1\ldots 4$. This maximal chain of subspaces is called a flag and triangular matrices have this connection to them.
Now when you assume the diagonal is zero, you get, even more strongly, $T(V_i)\subseteq V_{i+1}$.
From this it is obvious that $T^3(V)=\{0\}$, so that $T^3$ is the zero map on $V$.
Unfortunately this is one of those cases when defaulting to bases, matrices and coordinates just kind of gets in the way of seeing what's going on.