$A^n=0$ if $a_{ij}=0$ for all $i\geq j$.

Question

Prove: $A^n=0$ if $a_{ij}=0$ for all $i\geq j$ where $A\in M_n(K)$.

I tested it for a few matrices and I guess I have to do it by induction. How can I do that?

Answer 1

Observe that such a matrix is strictly upper triangular, i.e., all entries on and below the main diagonal are zero. Our claim is therefore that any strictly upper triangular matrix is nilpotent.

Consider a strictly upper triangular $n \times n$ matrix $A.$ Observe that the matrix $xI - A$ is upper triangular. Particularly, all of the diagonal entries are $x.$ Considering that the determinant of an upper triangular matrix is the product of the diagonal entries, we have that the characteristic polynomial of $A$ is given by $\chi_A(x) = \det(xI - A) = x^n.$ By the Cayley-Hamilton Theorem, we have that $0 = \chi_A(A) = A^n,$ as desired.

Answer 2

Since $A$ is upper triangular, all its eigenvalues lie on the main diagonal which are zero since $a_{ii}=0$. The characteristic equation of this matrix is $$|\lambda I-A|=\lambda^n=0$$which by using Cayley-Hamilton's theorem leads to $A^n=0$.

Answer 3

$\newcommand{\R}{\mathbb{R}}\newcommand{\N}{\mathbb{N}}$Let me extend my comment into an answer. As you may have noticed in your example calculations, for every power of the original matrix $A$ another off-diagonal full of zeros appears.

So here is the claim which we will prove by fighting against a battalion of indices (the argument given in the other two answers is much more elegant!).

---

Lemma: Suppose $A\in\R^{n\times n}$ is such that $A_{ij}=0$ for all $i\geq j$. Then for every $k\in \N$ the powers $A^k$ have entries $(A^k)_{ij}=0$ for all $i+k-1\geq j$.

Proof: The case $k=1$ is covered by the assumption. So suppose the claim is true for $k$ and lets show it for $k+1$. Let $B=A^k$ and calculate for $i+(k+1)-1\geq j$ $$(A^{k+1})_{ij}=(BA)_{ij}=\sum_{l=1}^nB_{il}A_{lj}.$$ Now we know that $A_{lj}=0$ if $l\geq j$ so the sum actually only runs until $l=j-1$. On the other hand we know by induction that $B_{il}=0$ if $i+k-1\geq l$ so the sum actually only starts at $l=i+k$. Hence $$(A^{k+1})_{ij}=\sum_{l=i+k}^{j-1}B_{il}A_{lj}.$$ But by assumption $i+k\geq j$ and so this is actually an empty sum and equals $0$.

I'm sure one could write this down more elegantly without having to use an empty sum.

---

Since $i+n-1\geq j$ is true for all $i,j\in\{1,\ldots,n\}$ we have the result you actually wanted to prove.

Corollary: If $A\in\R^{n\times n}$ is as above then $A^n=0$.