# Transpose of Upper Triangular Matrix is Lower Triangular

## Proof

Let $\mathbf U = \sqbrk a_{m n}$ be an upper triangular matrix.

By definition:

$\forall a_{i j} \in \mathbf U: i > j \implies a_{i j} = 0$

Let $\mathbf U^\intercal = \sqbrk b_{n m}$ be the transpose of $\mathbf U$.

That is:

$\mathbf U^\intercal = \sqbrk b_{n m}: \forall i \in \closedint 1 n, j \in \closedint 1 n: b_{i j} = a_{j i}$

Thus:

$\forall b_{j i} \in \mathbf U^\intercal: i > j \implies b_{j i} = 0$

By exchanging $i$ and $j$ in the notation of the above:

$\forall b_{i j} \in \mathbf U^\intercal: i < j \implies b_{i j} = 0$

Thus by definition it is seen that $\mathbf U^\intercal$ is a lower triangular matrix.

$\blacksquare$