I am given a matrix $A$ to decompose onto its $LU$ form. From this I find that $PA = LU$ where $P$ is a permutation matrix. Now the problem is that they define a matrix $B$ as $B = U^TAL^T$ and ask to find the $LU$ of $B$ without additional computations. Anyone can give me a tip where to start? I get that by substituting $A$ into $B$ the last part will be an upper triangular matrix, but can't get a way to make the lower triangular part.
[Math] Properties of LU decomposition
linear algebramatricesmatrix decomposition
Best Answer
Define the matrices
$$P = \begin{pmatrix} 0&1\\1&0\end{pmatrix}, \ L = \begin{pmatrix} 1&0\\l_{21}&1\end{pmatrix}, \ U = \begin{pmatrix} u_{11}&u_{12}\\0&u_{22}\end{pmatrix}$$
Substituting now $A=P^{-1}LU$ in $B=U^T A L^T$ gives $B=(U^T P^{-1}L) (U L^T)$.
Using the property that the set of upper triangular matrices is closed under multiplication,
The matrix $UL^T$ is itself an upper triangular matrix.
Thus we should check the matrix $M=U^T P^{-1}L$ which gives
$$M = \begin{pmatrix} u_{11}&0\\u_{12}&u_{22}\end{pmatrix} \begin{pmatrix} 0&1\\1&0\end{pmatrix} \begin{pmatrix} 1&0\\l_{21}&1\end{pmatrix} =\begin{pmatrix} 0&u_{11}\\u_{22}&u_{12}\end{pmatrix} \begin{pmatrix} 1&0\\l_{21}&1\end{pmatrix} =\begin{pmatrix} l_{21}u_{11}&u_{11}\\u_{22}+l_{21}u_{12}&u_{12}\end{pmatrix} $$
Which for generic parameters $l_{ij},u_{ij}$ is NOT a triangular matrix.
So this constitutes a counterexample to your question if $P\ne I$.