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.

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Comments

  • Gobi Ree  On January 17, 2012 at 9:55 pm

    It will be easier if we use the Cayley-Hamilton theorem.

    • nbloomf  On August 20, 2012 at 2:13 pm

      True, but that hasn’t been proved yet. The ease of proving this result with C-H is a strong argument that this exercise is in the wrong place. đŸ˜›

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