## Strictly upper (lower) triangular matrices are zero divisors

Let $S$ be any ring and let $n \geq 2$ be an integer. Prove that if $A$ is any strictly upper triangular matrix in $M_n(S)$ then $A^n = 0$. (A strictly upper triangular matrix is a matrix whose entries on and below the main diagonal are zero.)

Before approaching this problem, we will introduce some “structural” operations on matrices and prove some basic properties.

Definition: Let $X$ be a set.

1. Suppose $A = [a_{i,j}] \in \mathsf{Mat}_{k,n}(X)$ and $B = [b_{i,j}] \in \mathsf{Mat}_{k,m}(X)$. We define a matrix $[A | B] \in \mathsf{Mat}_{k,n+m}(X)$ as follows: $([A|B])_{i,j} = a_{i,j}$ if $1 \leq j \leq n$ and $b_{i,j-n}$ otherwise.
2. Suppose $A = [a_{i,j}] \in \mathsf{Mat}_{n,k}(X)$ and $B = [b_{i,j}] \in \mathsf{Mat}_{m,k}(X)$. We define a matrix $\left[ \frac{A}{B} \right] \in \mathsf{Mat}_{n+m,k}(X)$ as follows: $\left(\left[ \frac{A}{B} \right]\right)_{i,j} = a_{i,j}$ if $1 \leq i \leq n$ and $b_{i-n,j}$ otherwise.

Lemma: Let $X$ be a set, $A = [a_{i,j}] \in \mathsf{Mat}_{n,k}(X)$, $B = [b_{i,j}] \in \mathsf{Mat}_{n,\ell}(X)$, $C = [c_{i,j}] \in \mathsf{Mat}_{m,k}(X)$, and $D = [d_{i,j}] \in \mathsf{Mat}_{m,\ell}(X)$. Then

 $\left[ \dfrac{[A | B]}{[C | D]} \right] = \left[ \left[ \dfrac{A}{C} \right] \bigg| \left[ \dfrac{B}{D} \right] \right].$

Proof: Let $\mathcal{S}$ denote the matrix on the left hand side of the equals sign and $\mathcal{T}$ the matrix on the right. We consider four possibilities for $(i,j)$.

1. Suppose $0 < i \leq n$ and $0 < j \leq k$. Then $(\mathcal{S})_{i,j} = a_{i,j} = (\mathcal{T})_{i,j}$.
2. Suppose $0 < i \leq n$ and $k < j \leq j+\ell$. Then $(\mathcal{S})_{i,j} = b_{i,j} = (\mathcal{T})_{i,j}$.
3. Suppose $n < i \leq n+m$ and $0 < j \leq k$. Then $(\mathcal{S})_{i,j} = c_{i,j} = (\mathcal{T})_{i,j}$.
4. Suppose $n < i \leq n+m$ and $k < j \leq k+\ell$. Then $(\mathcal{S})_{i,j} = d_{i,j} = (\mathcal{T})_{i,j}$.

Thus $\mathcal{S} = \mathcal{T}$. $\square$

Since these two operators “abide”, we will drop the inner brackets and write (for example) $\left[ \dfrac{A}{C} \bigg| \dfrac{B}{D} \right]$ for brevity.

Lemma: Let $R$ be a ring.

1. If $A = [a_{i,j}] \in \mathsf{Mat}_{n,k}(R)$, $B = [b_{i,j}] \in \mathsf{Mat}_{k,m}(R)$, and $C = [c_{i,j}] \in \mathsf{Mat}_{k,\ell}(R)$, then $A \cdot [B|C] = [A \cdot B | A \cdot C]$.
2. If $A = [a_{i,j}] \in \mathsf{Mat}_{n,k}(R)$, $B = [b_{i,j}] \in \mathsf{Mat}_{m,k}(R)$, and $C = [c_{i,j}] \in \mathsf{Mat}_{k,\ell}(R)$, then $\left[ \frac{A}{B} \right] \cdot C = \left[ \frac{A \cdot C}{B \cdot C} \right]$.

Proof: The $(i,j)$ entry of $A \cdot [B|C]$ is $\sum_{p=1}^k a_{i,p}([B|C])_{p,j}$. If $0 < j \leq m$, then this sum is $\sum_{p=1}^k a_{i,p}b_{p,j} = (A \cdot B)_{i,j}$. If $m < j \leq m+\ell$, then this sum is $\sum_{p=1}^k a_{i,p}c_{p,j-m} = (A \cdot C)_{i,j-m}$. Thus $(A \cdot [B|C])_{i,j} = ([A \cdot B|A \cdot C])_{i,j}$ for all $(i,j)$. The proof of the second statement is analogous. $\square$

Lemma: Let $R$ be a ring. If $A = [a_{i,j}] \in \mathsf{Mat}_{n,k}(R)$, $B = [b_{i,j}] \in \mathsf{Mat}_{n,\ell}(R)$, $C = [c_{i,j}] \in \mathsf{Mat}_{k,m}(R)$, and $D = [d_{i,j}] \in \mathsf{Mat}_{\ell,m}(R)$, then $[A|B] \cdot \left[ \dfrac{C}{D} \right] = AC + BD$. Proof: For each $(i,j)$, note the following.

 $\left( [A|B] \cdot \left[ \dfrac{C}{D} \right] \right)_{i,j}$ = $\displaystyle\sum_{p=1}^{k+\ell} \left( [A|B] \right)_{i,p} \left( \left[ \dfrac{C}{D} \right] \right)_{p,j}$ = $\left( \displaystyle\sum_{p=1}^{k} \left( [A|B] \right)_{i,p} \left( \left[ \dfrac{C}{D} \right] \right)_{p,j} \right) + \left( \displaystyle\sum_{p=k+1}^{k+\ell} \left( [A|B] \right)_{i,p} \left( \left[ \dfrac{C}{D} \right] \right)_{p,j} \right)$ = $\left( \displaystyle\sum_{p=1}^k a_{i,p}c_{p,j} \right) + \left( \displaystyle\sum_{p=k+1}^{k+\ell} b_{i,p-k}d_{p-k,j} \right)$ = $(A \cdot C)_{i,j} + (B \cdot D)_{i,j}$ = $(A \cdot C + B \cdot D)_{i,j}$

Thus the two matrices are equal. $\square$

Lemma: Let $R$ be a ring. Let $A_1 \in \mathsf{Mat}_{n \times k}(R)$, $B_1 \in \mathsf{Mat}_{n \times \ell}(R)$, $C_1 \in \mathsf{Mat}_{t \times k}(R)$, $D_1 \in \mathsf{Mat}_{t \times \ell}(R)$, $A_2 \in \mathsf{Mat}_{k \times m}(R)$, $B_2 \in \mathsf{Mat}_{k \times p}(R)$, $C_2 \in \mathsf{Mat}_{\ell \times m}(R)$, and $D_2 \in \mathsf{Mat}_{\ell \times p}(R)$. Then

 $\left[ \dfrac{A_1 | B_1}{C_1 | D_1} \right] \cdot \left[ \dfrac{A_2 | B_2}{C_2 | D_2} \right] = \left[ \dfrac{A_1A_2 + B_1C_2 | A_1B_2 + B_1D_2}{C_1A_2 + D_1C_2 | C_1B_2 + D_1D_2} \right].$

Proof: Using the previous lemmas, we have the following.

 $\left[ \dfrac{A_1 | B_1}{C_1 | D_1} \right] \cdot \left[ \dfrac{A_2 | B_2}{C_2 | D_2} \right]$ = $\left[ \dfrac{A_1 | B_1}{C_1 | D_1} \right] \cdot \left[ \left[ \dfrac{A_2}{C_2} \right] \bigg| \left[ \dfrac{B_2}{D_2} \right] \right]$ = $\left[ \left[ \dfrac{A_1 | B_1}{C_1 | D_1} \right] \cdot \left[ \dfrac{A_2}{C_2} \right] \bigg| \left[ \dfrac{A_1 | B_1}{C_1 | D_1} \right] \cdot \left[ \dfrac{B_2}{D_2} \right] \right]$ = $\left[ \left[ \dfrac{[A_1 | B_1]}{[C_1 | D_1]} \right] \cdot \left[ \dfrac{A_2}{C_2} \right] \bigg| \left[ \dfrac{[A_1 | B_1]}{[C_1 | D_1]} \right] \cdot \left[ \dfrac{B_2}{D_2} \right] \right]$ = $\left[ \left[ \dfrac{[A_1|B_1] \cdot \left[ \dfrac{A_2}{C_2} \right]}{[C_1|D_1] \cdot \left[ \dfrac{A_2}{C_2} \right]} \right] \Bigg| \left[ \dfrac{[A_1|B_1] \cdot \left[ \dfrac{B_2}{D_2} \right]}{[C_1|D_1] \cdot \left[ \dfrac{B_2}{D_2} \right]} \right] \right]$ = $\left[ \dfrac{A_1A_2 + B_1C_2 | A_1B_2 + B_1D_2}{C_1A_2 + D_1C_2 | C_1B_2 + D_1D_2} \right]$. $\square$

We now introduce another definition.

Definition: Let $R$ be a ring, $n \geq 2$, and $1 \leq k \leq n$. A matrix $M \in \mathsf{Mat}_n(R)$ is called $k$-strictly upper triangular if $M = \left[ \dfrac{0 | M^\prime}{0_k | 0} \right]$ where $0_k$ is the $k \times k$ zero matrix, $M^\prime$ has dimensions $(n-k) \times (n-k)$, and $M^\prime$ is upper triangular.

For example, 1-strictly upper triangular matrices and strictly upper triangular matrices are the same, and an $n \times n$ matrix is zero if and only if it is $n$-strictly upper triangular.

Lemma: Let $R$ be a ring, $n \geq 2$, and $N$ a square matrix over $R$ of dimension $n$. If $N$ is strictly upper triangular and $N = \left[ \dfrac{N_1 | N_2}{N_3 | N_4} \right]$, where $N_4$ is square, then $N_4$ is strictly upper triangular. Proof: The elements on or below the main diagonal of $N_4$ are on or below the main diagonal of $N$, hence are zero. $\square$

Lemma: Let $R$ be a ring, $n \geq 2$, and $1 \leq k \leq n$. If $A = \left[ \begin{array}{c|c} 0 & A^\prime \\ \hline 0_k & 0 \end{array} \right]$ is $k$-strictly upper triangular and $A^\prime$ is strictly upper triangular, then $A$ is $k+1$-strictly upper triangular. Proof: We have $A^\prime = \left[ \begin{array}{c|c} 0 & A^{\prime\prime} \\ \hline 0_1 & 0 \end{array} \right]$, where $A^{\prime\prime}$ is upper triangular and $0_1$ has dimension $1 \times 1$. Thus we have the following.

$A = \left[ \begin{array}{c|c|c} 0 & 0 & A^{\prime\prime} \\ \hline 0 & 0_1 & 0 \\ \hline 0_k & 0 & 0 \end{array} \right] = \left[ \begin{array}{c|c} 0 & A^{\prime\prime} \\ \hline 0_{k+1} & 0 \end{array} \right],$

So that $A$ is $k+1$-strictly upper triangular. $\square$

Lemma: Let $R$ be a ring, let $n \geq 2$, and let $M,N \in \mathsf{Mat}_n(R)$. If $M$ is upper triangular and $N$ is strictly upper triangular, then $MN$ is strictly upper triangular. Proof: Recall that $(MN)_{i,j} = \sum_{k=1}^n m_{i,k}n_{k,j}$. Suppose $i \geq j$. Then if $k \geq j$, $n_{k,j} = 0$. If $k < i$, $m_{i,k} = 0$. Thus $(MN)_{i,j} = 0$, so that $MN$ is strictly upper triangular. $\square$

Lemma: Let $R$ be a ring, let $n \geq 2$, and let $1 \leq k < n$. If $M,N \in \mathsf{Mat}_n(R)$ such that $M$ is $k$-strictly upper triangular and $N$ is strictly upper triangular, then $MN$ is $k+1$-strictly upper triangular. Proof: Write $M = \left[ \begin{array}{c|c} 0 & M^\prime \\ \hline 0_k & 0 \end{array} \right]$ and $N = \left[ \begin{array}{c|c} N_1 & N_2 \\ \hline N_3 & N_4 \end{array} \right]$, where $M$ and $N_4$ have dimension $n-k \times n-k$. Evidently, $MN = \left[ \begin{array}{c|c} 0 & M^\prime N_4 \\ \hline 0_k & 0 \end{array} \right]$. By the previous lemma, since $M^\prime$ is upper triangular and $N_4$ is strictly upper triangular, $M^\prime N_4$ is strictly upper triangular. Thus $MN$ is $k+1$-strictly upper triangular. $\square$

Now to the main result.

If $A$ is an $n \times n$ matrix over a ring $R$, and $A$ is strictly upper triangular, then by an easy induction argument $A^k$ is $k$-strictly upper triangular. Thus $A^n = 0$.