Matrices: LDU Decomposition

Question

Can someone please help me with the following:

Let $L_1$ and $L_2$ be nonsingular lower triangular matrices and let $U_1$ and $U_2$ be nonsingular upper triangular matrices. If $L_1$$U_1$ = $L_2$$U_2$, show that there exists a nonsingular diagonal matrix D such that -

$L_2D$ = $L_1$ and $U_2$ = $D$$U_1$.

Thanks in advance!

Edit: I know that this has something to do with the condition of the uniqueness of LU decomposition of a matrix, but I can't seem to go any further than that.

Answer 1

I only have a partial answer. Since $L_1$, $L_2$, $U_1$ and $U_2$ are nonsingular, their inverses exist. We could change your premise to $$L_1U_1=L_2U_2\iff L_2^{-1}L_1=U_2U_1^{-1}$$ Lets name this matrix $A$, then $$L_2^{-1}L_1=A\iff L_1=L_2A$$ $$U_2U_1^{-1} = A\iff U_2=AU_1$$ It left to show that $A$ is a diagonal matrix. This is as far as I can get.

Edit With the help of OP, here is the final step of the proof.

If $L$ is a lower triangular matrix, his inverse, if it exist, is also lower triangular. Thus, $A=L_2^{-1}L_1$ is the product of two lower triangular matrix, $A$ is a lower triangular matrix.

Similar argument with upper triangular matrix. If $U$ is a upper triangular matrix, his inverse, if it exist, is also upper triangular. Thus, $A=U_2U_1^{-1}$ is the product of two upper triangular matrix, $A$ is a upper triangular matrix.

Since $A$ is both lower triangular and upper triangular, it is diagonal.

Answer 2

Suppose $L_1, L_2$ are lower triangular, invertible. Suppose further that $U$ is upper triangular and $L_1 = L_2 U$.

Show that $U$ is diagonal by computing the first row of $L_2U$ and comparing with same in $L_1$. Note that $[L_2]_{kk} \neq 0$. Repeat for the remaining rows.

Or, show that the invertible lower triangular matrices form a group under multiplication and so $L_1 L_2^{-1}$ is lower triangular and equal to an upper triangular matrix.