Prove that if $A$ is symmetric and has a LU-decomposition then $A=LDU' \Rightarrow U'=L^T$ Suppose the matriz $A$ has a LU-decomposition, in other words, suppose there exists matrices $L$ and $U$ such that $A=LU$ where $L$ is lower triangular and $U$ is upper triangular.
We can to prove that $A$ has a LDU'-decomposition, where $D$ is a diagonal (to do so, put $d_{ii}=u_{ii}$ and $u'_{kj}=u_{kj}/u_{kk}$).
Prove that if $A$ is symmetric then $A=LDU' \Rightarrow  U'=L^T$, where $L^T$ is the transpose of matrix $L$.
Yes, $A$ is invertible. Therfore, $LDU'=U'^TDL^T\Rightarrow U'=L^T$. Here is the solution (question 9).
 A: First we need to show that the LU-decomposition of $A$ is unique. 
Assume there are two LU-decompositions of $A$. 
$LU = \tilde{L}\tilde{U} \Leftrightarrow L\tilde{L}^{-1} = \tilde{U}U^{-1}$. $L$ and $\tilde{L}$ are lower left triangular matrices (with $l_{i,i} = 1$) and $U$ and $\tilde{U}$ are upper right triangular matrices. The inverse matrices of triangular matrices are again triangular matrices and the product of two (lower left/upper right) triangular matrices is again a (lower left/upper right) triangular matrix (You can proof this by calculating the inverse/product). But the only matrix which is a lower left and an upper right triangular matrix is a diagonal matrix. Now $l_{i,i} = 1 \Rightarrow L\tilde{L}^{-1} = I$. Hence $L = \tilde{L}$ and $U = \tilde{U}$. This means the decomposition is unique. 
Let $A$ be symmetric and $A = LDU'$ be a LU-decomposition of $A$ where $D$ is a diagonal matrix and $u'_{i,i} = 1$. 
$A = LDU' = (LDU')^T = U'^TDL^T$. $U^T$ is a lower left triangular  matrix and $L^T$ is a upper right triangular matrix. This means $(LDU')^T$ is another LU-decomposition of $A$. But the decomposition is unique which implies $L = U'^T$ and $L^T = U'$. Hence $A = LDL^T$. 
