I was interested on the same question, so allow me to exploit my logic, hopping of course to get comments for possible flaws. Suppose you have two lower triangular matrices $\mathbf{L}_1$ and $\mathbf{L}_2$ illustrated bellow
$$\mathbf{L}_1 =
\begin{bmatrix}
l_{11}^{(1)} & l_{12}^{(1)} & \dots & \dots & l_{1n}^{(1)} &\\
& l_{22}^{(1)} & l_{23}^{(1)} & \dots & \vdots &\\
& & l_{33}^{(1)} & & \vdots &\\
& & & \ddots & \vdots &\\
& & & & l_{nn}^{(1)} &\\
\end{bmatrix}~~~~~\mathbf{L}_2 =
\begin{bmatrix}
l_{11}^{(2)} & l_{12}^{(2)} & \dots & \dots & l_{1n}^{(2)} &\\
& l_{22}^{(2)} & l_{23}^{(2)} & \dots & \vdots &\\
& & l_{33}^{(2)} & & \vdots &\\
& & & \ddots & \vdots &\\
& & & & l_{nn}^{(2)} &\\
\end{bmatrix}$$
We want to prove that the following product is a lower triangular matrix,
$$\mathbf{L}_1 \mathbf{L}_2 = \mathbf{L}_1 \big[ \mathbf{l}_1, \mathbf{l}_2^{(2)}, \dots, \mathbf{l}_n^{(2)} \big] = \big[ \mathbf{L}_1 \mathbf{l}_1^{(2)}, \mathbf{L}_1 \mathbf{l}_2^{(2)}, \dots, \mathbf{L}_1 \mathbf{l}_n^{(2)} \big]$$
As we can see, the $k$-th column of product matrix $\mathbf{L}_1 \mathbf{L}_2$ is given by $\mathbf{L}_1 \mathbf{l}_k^{(2)}$ which is the linear combination of the $\mathbf{L_1}$ matrix columns with coefficients defined by the $k$-th column vecor $\mathbf{l}_k^{(2)}$. Each of the product matrix columns $\big(\mathbf{L}_1 \mathbf{l}_k^{(2)}\big)$ have possible non-zeros entries only above the $k$-th element.
This is because, the new columns are linear combinations of the first $k$ columns $\mathbf{l}_k^{(1)}$ which by their turn have possible non-zero values above their $k$-th entry. This property comes form the fact that columns $\mathbf{l}_k^{(2)}$ have zero entries after their $k$-th element.
$\mathcal{Thanks~for~reading}$.
Proof of the equivalency
$(2) \implies (1)$
Let $\mathcal E' = \mathcal B U^*$, then $T (\mathcal B) = \mathcal B U^* A U $, or $T (\mathcal E' U) = \mathcal E' U U^* A U = \mathcal E' A U $, hence $T (\mathcal E') = \mathcal E' A$, i.e. $\mathcal M(T, \mathcal E') = A$ is upper triangular. Now apply Gram-Schmidt orthogonalization process to $\mathcal E'$ to obtain an orthonormal basis $\mathcal E$, then $\mathcal M(T, \mathcal E)$ is still upper triangular, hence the statement.
$(1) \implies (2)$
From now on, the bold-italic letters represent matrices, the regular letters are mappings, and the calligraphic letters are basis. Also $\mathbb C^n \cong \mathrm M_{n,1}(\mathbb C)$ is the collection of $n \times 1$ complex matrices.
Let $\boldsymbol S = \mathcal M(T, \mathcal B)$, on $W = \mathbb C^n$, define linear operator $S \colon \boldsymbol x \mapsto \boldsymbol {Sx}$ where $\boldsymbol x \in \mathbb C^n$, then apply $(1)$ to the operator $S$, there exists some ONB $\mathcal F$ of $\mathbb C^n$ that $\mathcal M(S, \mathcal F) = \boldsymbol A$ is upper triangular. Let $\mathcal E$ be the standard basis of $\mathbb C^n$, and let $\mathcal F = \mathcal E \boldsymbol U$, then $\boldsymbol U$ is unitary. Write $\boldsymbol U^*$ instead of $\boldsymbol U$, i.e. $\mathcal F = \mathcal E \boldsymbol U^*$ and $S(\mathcal F) = \mathcal F \boldsymbol A$ becomes $S (\mathcal E \boldsymbol U^*) = \mathcal E \boldsymbol U^* \boldsymbol A$, hence $S (\mathcal E) = \mathcal E (\boldsymbol U^*\boldsymbol {AU})$. But $S (\mathcal E) = \boldsymbol S$, hence $\boldsymbol S = \boldsymbol {U}^* \boldsymbol {AU}$. Hence the statement.
Best Answer
By Schur's theorem, there is a unitary $U$ such that $U^*A^*U$ is upper triangular. Hence $(U^*A^*U)^*=U^*AU^*$ is lower triangular.