Negative of Triangular Matrix

Theorem
Let $\mathbf A = \left[{a}\right]_{n}$ 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:


 * $\forall i, j \in \left[{1 .. n}\right]: \left[{-a}\right]_{ij} = - a_{ij}$

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

Hence:
 * $\forall i > j: \left[{-a}\right]_{ij} = -a_{ij} = 0$

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

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

Hence:
 * $\forall i < j: \left[{-a}\right]_{ij} = -a_{ij} = 0$

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