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:

$\mathbf U = \begin{bmatrix} a_{1 1} & a_{1 2} & a_{1 3} & \cdots & a_{1, n - 1} & a_{1 n} \\ 0 & a_{2 2} & a_{2 3} & \cdots & a_{2, n - 1} & a_{2 n} \\ 0 & 0 & a_{3 3} & \cdots & a_{3, n - 1} & a_{3 n} \\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\ 0 & 0 & 0 & \cdots & a_{n - 1, n - 1} & a_{n - 1, n} \\ 0 & 0 & 0 & \cdots & 0 & a_{n n} \\ \end{bmatrix}$

By definition:

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

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

That is:

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

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.

$\blacksquare$