As @Julien pointed out, every square matrix admits a $PLU$ decomposition, where $P$ is a permutation matrix. We have: $A = P \cdot L \cdot U$, such that:
$A=\begin{bmatrix}1 & 2 & 3 & 4 \\5 & 6 & 7 & 8\\1 & -1 & 2 & 3 \\2 & 1 & 1 &2 \end{bmatrix}= \begin{bmatrix} 1 & 0 & 0 & 0\\0 & 0 & 1 & 0\\0 & 0 & 0 & 1\\0 & 1 & 0 & 0 \end{bmatrix} \cdot \begin{bmatrix} 1 & 0 & 0 & 0\\1 & 1 & 0 & 0\\2 & 1 & 1 & 0\\5 & \dfrac{4}{3} & \dfrac{5}{3} & 1 \end{bmatrix} \cdot \begin{bmatrix} 1 & 2 & 3 & 4\\0 & -3 & -1 & -1\\0 & 0 & -4 & -5\\0 & 0 & 0 & -\dfrac{7}{3} \end{bmatrix}$
You could try manually cranking this one to find its $LU$ factorization. We want:
$L \cdot U = \begin{bmatrix} 1 & 0 & 0 & 0\\l_{21} & 1 & 0 & 0\\l_{31} & l_{32} & 1 & 0\\l_{41} & l_{42} & l_{43} & 1 \end{bmatrix} \cdot \begin{bmatrix} u_{11} & u_{12} & u_{13} & u_{14}\\0 & u_{22} &u_{23} & u_{24}\\0 & 0 & u_{33} & u_{34}\\0 & 0 & 0 & u_{44} \end{bmatrix} = \begin{bmatrix}1 & 2 & 3 & 4 \\5 & 6 & 7 & 8\\1 & -1 & 2 & 3 \\2 & 1 & 1 &2 \end{bmatrix}$
We start off by solving the first row, so we get:
$$u_{11} = 1, u_{12} = 2, u_{13} = 3, u_{14} = 4$$
The portion of the multiplication that determines the remaining entries in the first column of $A$ yields:
$$l_{21}u_{11} = 5 \rightarrow l_{21} = 5$$
$$l_{31}u_{11} = 1 \rightarrow l_{31} = 1$$
$$l_{11}u_{11} = 2 \rightarrow l_{41} = 2$$
At this point rewrite all the variables you solved for and then continue the process and see if you can solve the remaining variables. Of course it is easy to check the result if you can solve all of the equations.
So, we currently have:
$L \cdot U = \begin{bmatrix} 1 & 0 & 0 & 0\\5 & 1 & 0 & 0\\1 & l_{32} & 1 & 0\\2 & l_{42} & l_{43} & 1 \end{bmatrix} \cdot \begin{bmatrix} 1 & 2 & 3 & 4\\0 & u_{22} &u_{23} & u_{24}\\0 & 0 & u_{33} & u_{34}\\0 & 0 & 0 & u_{44} \end{bmatrix} = \begin{bmatrix}1 & 2 & 3 & 4 \\5 & 6 & 7 & 8\\1 & -1 & 2 & 3 \\2 & 1 & 1 &2 \end{bmatrix}$
Try solving for $u_{22},u_{23}, u_{24}$, and then $l_{32}, l_{42}$ and continue this process.
Best Answer
Assume that $T_1$ is nonsingular. If $T_1$ is lower triangular and $T_2:=QT_1$ is also lower triangular, we have $Q=T_2T_1^{-1}$. Since both $T_1$ and $T_2$ are lower triangular, it follows that $Q$ is lower triangular. But the inverse of a lower triangular matrix is lower triangular and the transpose is upper triangular. Since $Q^T=Q^{-1}$ we have that $Q$ is diagonal.