Negative of Triangular Matrix

Theorem
Let $\mathbf A = \sqbrk a_n$ be a square matrix of order $n$.

Let $-\mathbf A$ be the negative of $\mathbf A$.

If $\mathbf A$ is an upper triangular matrix, then so is $-\mathbf A$.

If $\mathbf A$ is a lower triangular matrix, then so is $-\mathbf A$.

Proof
From the definition of negative matrix, we have:


 * $\forall i, j \in \closedint 1 n: \sqbrk {-a}_{i j} = -a_{i j}$

If $\mathbf A$ is an upper triangular matrix, we have:
 * $\forall i > j: a_{i j} = 0$

Hence:
 * $\forall i > j: \sqbrk {-a}_{i j} = -a_{i j} = 0$

and so $-\mathbf A$ is itself upper triangular.

Similarly, if $\mathbf A$ is a lower triangular matrix, we have:
 * $\forall i < j: a_{i j} = 0$

Hence:
 * $\forall i < j: \sqbrk {-a}_{i j} = -a_{i j} = 0$

and so $-\mathbf A$ is itself lower triangular.