Are strictly upper triangular matrices nilpotent?

Question

An $n\times n$ matrix $A$ is called nilpotent if $A^m = 0$ for some $m\ge1$.

Show that every triangular matrix with zeros on the main diagonal is nilpotent.

Answer 1

Its characteristic polynomial is $T^n$, so by Cayley-Hamilton, $A^n=0$.

Answer 2

WLOG assume that $A$ is upper-triangular (otherwise, a similar argument works with the basis reversed).

Regard $A$ as a linear transformation on $\mathbb{F}^n$ with basis $e_1, \ldots, e_n$. Let $U_i$ be the span of $e_1, e_2, \ldots, e_i$ for $i = 0, 1, \ldots, n$. Observe that $0 = U_0 \subseteq \ldots \subseteq U_n = \mathbb{F}^n$ and also note that $AU_i \subseteq U_{i - 1}$ for $i = 1, \ldots, n$ since $A$ is strictly upper-triangular. Therefore, $A^n = 0$.

Answer 3

As therealak12 mentioned, you can prove this by induction. Let $T$ be an arbitrary matrix that is strictly upper triangular.

Induction basis ($k = 1$): $T_{ij} = 0$ for $i \geq j$ follows from the fact that $T$ is a strictly upper triangular matrix.

Induction hypothesis: $T^k_{ij}=0$ for $i + k - 1 \geq j$ for $k \in \mathbb{N}^+$

Induction step ($k \rightarrow k + 1$): Let $i, j \in \{1, \cdots, n\}$ be arbitrary such that $i + k \geq j$.

$$ T^{k+1}_{ij} = (TT^k)_{ij} = \sum_{l=1}^n T_{il} T^k_{lj} = \sum_{l=1}^i T_{il} T^k_{lj} + \sum_{l=i+1}^n T_{il} T^k_{lj} $$

For our left sum, $T_{il} = 0$ since $i \geq l$ and $T$ is a strictly upper triangular matrix.

For our right sum, we have $l \geq i + 1 \geq j - k + 1$ (since we assumed $i + k \geq j$), from which it follows that $l + k - 1 \geq j$. If we apply the induction hypothesis to this condition, we get $T^k_{ij} = 0$.

Thus:

$$T^{k+1}_{ij} = \sum_{l=1}^i 0 \cdot T^k_{lj} + \sum_{l=i+1}^n T_{il} \cdot 0 = 0 $$

from which the statement then follows.

Answer 4

I think you can prove this using induction.

Prove that for each $1 \leq i \leq n$, all elements of the first $i$ lines of $A^{i}$ are zero and for the $(i+1)th$ line, only the first element is not zero and the other elements in this line are zero.

Answer 5

Yes, all strictly upper triangular matrices are nilpotent and for good reason. It's the same reason why differentiating a finite polynomial function eventually yields zero.

If I end the answer here, I'll attract downvotes, so I'll explain more :)

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Consider the vectorspace $V$ of polynomials of degree less than $n$ with real coefficients. $V = \{a_1x^{n-1}+a_2x^{n-2}+\ldots+a_{n-1}x+a_n : a_i\in \mathbb R\}\tag*{}$

Let $D$ be the differentiation operator on $V$ which maps a polynomial $f(x)$ to $\frac{\mathrm d f(x)}{\mathrm dx}$.

Show that $D$ is linear (I leave the proof to the reader). Consider the matrix of $D$ w.r.t. the standard ordered basis $\beta=\{1,x,\ldots,x^{n-1}\}$. Show that $[D]_{\beta}$ is an upper triangular matrix... This isn't hard to do, the $D1=0$ so the first column is all zeroes. $Dx=1$ so the second column is $(1,0,\ldots,0)$. Now go on... $Dx^2=0+2x$.

$$[D]_\beta =\begin{pmatrix}0&1&0&\ldots&0\\ 0&0&1&\ldots&0 \\ \vdots&\vdots&\vdots&\ddots&\vdots\\0&0&0&\ldots&0\end{pmatrix}$$

You have 1's on the super-diagonal and other entries are all zeroes.

For any polynomial $f$ in the space $V$, $D^{n}f=0$ (that's what successive differentiation does to $x^{n-1}$ and in fact, to any polynomial in $V$). In terms of matrices, $([D]_{\beta})^{n}=\bf 0$.

And you can adjust your definition of 'differentiating'... You could very well define your linear operator to have $Dx^2=10x$, as long as it reduces the degree of polynomials, its matrix (w.r.t. $\beta$) will be strictly upper triangular. This covers all strictly upper triangular matrices.

You can interpret any given $n\times n$ upper triangular as a 'differentiating' linear operator $D$ on the space of polynomials of degree $\leq n-1$. You know that, no matter what, $D^nf=0$ for all polynomials $f$ in this space.

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This was not really a rigorous proof, I just wanted to give this intuition which I got from my professor in an introductory linear algebra class.