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.