If a non-singular square matrix can be reduced to row-echelon form without row interchanges, then it can be factored uniquely into $LDU,$ where $L$ is lower triangular, $U$ is upper triangular and $D$ is diagonal. Moreover, diagonal entries of $L$ and $U$ are $1.$
May I know if my proof is correct? Thank you.
It is clear that $A = LU.$ Let $L=(a_{ij})$ and $U=(b_{ij}).$
Since $L$ is non-singular, $a_{ii} \neq 0.$
Then $L$ can be factored into: $\pmatrix{1& 0 & \cdots & 0\\\dfrac{a_{21}}{a_{11}}&1 & \cdots & 0 \\\dfrac{a_{31}}{a_{11}} & \dfrac{a_{32}}{a_{22}} & 1 \cdots & 0 \\\vdots & \vdots & \vdots & \vdots\\ \dfrac{a_{n1}}{a_{11}} & \dfrac{a_{n2}}{a_{22}}& 0 \cdots &1 } $ $\pmatrix{a_{11}\\ & a_{22}\\ && a_{33} \\ & & & \ddots a_{nn}}$.
Hence, we can express $L=XD_L,$ where $X$ is lower triangular with $1$ on the diagonals and $D_L$ is diagonal.
$U$ can be factored into: $\pmatrix{b_{11}\\ & b_{22}\\ && b_{33} \\ & & & \ddots b_{nn}}$$\pmatrix{1& \dfrac{b_{12}}{b_{11}} & & \cdots & \dfrac{b_{1n}}{b_{11}}\\ 0 & 1 & \dfrac{b_{23}}{b_{22}} & \cdots & \dfrac{b_{2n}}{b_{22}} \\ 0 & 0 & 1 & \cdots & \dfrac{b_{3n}}{b_{33}}\\ \vdots & \vdots & \vdots & \vdots & \vdots \\ 0 & & \cdots & & 1 } $.
Hence, we can express $U=D_UY,$ where $D_U$ is diagonal and $Y$ is upper triangular with $1$ on the diagonals.
Hence, $LU=XD_LD_UY,$ where $D_LD_U$ is clearly diagonal.
Suppose $LDU=L'D'U'.$ $$DU=L^{-1}L'D'U' \implies DU(U')^{-1}(D')^{-1}=L^{-1}L'.$$ Since $DU(U')^{-1}(D')^{-1}$ is product of upper triangular matrices, $L^{-1}L'$ is upper triangular. Since $L^{-1}L'$ is product of lower triangular matrices, $L^{-1}L'$ is also lower triangular. It is clear that diagonal entries of $L^{-1}L'$ is $1.$ Hence, $L^{-1}L'=I,$ which means $L=L'.$ Similarly, $U=U'.$
I think your conclusion is false. Here is a counterexample. Let $A=(a_{ij})$ be a $2\times 2$ matrix with $a_{11}=a_{22}=0$ and $a_{12}=a_{21}=1$. Clearly $A$ is invertible. However, assume that $D=\operatorname{diag}(d_1,...,d_n)$, and we'd have $A_{11}=d_1$ (you can do the calculation). In our example, $A_{11}=d_1=0$, so $D$ is singular, which is wrong.
I think permutation matrices are needed to have such a decomposition.