Let T be a nilpotent operator on an $n$-dimensional vector space $V$, and suppose that $p$ is the smallest positive integer for which $T^P=T_0$ (zero-transformation).
I have so far proved the following subquestions:
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$N(T^i) \subseteq N(T^{i+1})$ for every positive integer $i$
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There is a sequence of ordered bases $\beta_1, \beta_2, …, \beta_p$ such that $\beta_i$ is a basis for $N(T^i)$ and $\beta_{i+1}$ contains $\beta_i$ for $1 \leq i \leq p-1$
Now, I must show that if $\beta = \beta_p$ is the ordered basis for $N(T^p) = V$ constructed so that it contains all the previous $\beta_i$, then $[T]_\beta$ is upper-triagular with diagonal entries being zero
I'm having a hard time showing this, and even seeing why this is true.
The two statements I was asked to prove above implies that I can have a scenario where $T^1$ has an empty nullspace. This doesn't violate $N(T) \subseteq N(T^2)$ nor $\{\emptyset\} = \beta_1 \subseteq \beta_2$.
But in that case, the first element of $\beta$ (denoted $v_1$) is not in the kernel of $T$. So $T(v_1) \neq 0$, and this contradicts what I am supposed to prove.
Could someone help me understand where my logic is breaking down, and how one could show the above?
Best Answer
Your logic is breaks down when you try to show $T$ is nilpotent. If $T$ has trivial nullspace (we can never have an empty nullspace), then $T$ has full rank and is invertible. Invertible matrices can never be nilpotent; we can just multiply $T^n = 0$ both sides by $T^{-1}$ $n$ times to get $I = 0$, a contradiction.
To prove this result, note that $T$ maps $N(T^{i+1})$ into $N(T^i)$. So, if $v \in \beta_{i+1}$, then $Tv \in \operatorname{span}(\beta_i)$. So, when you form your basis $\beta$, the image of each basis vector lies in the span of the basis vectors before it in the basis (not including itself). That is, if $\beta = (v_1, \ldots, v_n)$, then $$T(v_i) = a_{1i} v_1 + a_{2i} v_2 + \ldots + a_{(i-1)i}v_{i-1} + \color{red}0v_i + 0v_{i+1} + \ldots + 0v_n.$$ When forming the matrix for $T$ with respect to basis $\beta$, this turns into a column vector, placed in the matrix so that the red $0$ lies on the diagonal. This makes the matrix upper triangular, with a $0$ diagonal.