Transpose of Upper Triangular Matrix is Lower Triangular

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Theorem

The transpose of an upper triangular matrix is a lower triangular matrix.


Proof

Let $\mathbf U$ be an upper triangular matrix:

$\mathbf U = \begin{bmatrix} a_{11} & a_{12} & a_{13} & \cdots & a_{1,n-1} & a_{1n} \\ 0 & a_{22} & a_{23} & \cdots & a_{2,n-1} & a_{2n} \\ 0 & 0 & a_{33} & \cdots & a_{3,n-1} & a_{3n} \\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\ 0 & 0 & 0 & \cdots & a_{n-1,n-1} & a_{n-1,n} \\ 0 & 0 & 0 & \cdots & 0 & a_{nn} \\ \end{bmatrix}$

By definition:

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

Let $\mathbf U^\intercal$ be the transpose of $\mathbf U$.

That is:

$\mathbf U^\intercal = \left[{b}\right]_n: \forall i \in \left[{1 \,.\,.\, n}\right], j \in \left[{1 \,.\,.\, n}\right]: b_{i j} = a_{j i}$

Thus:

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

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

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

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

$\blacksquare$