Compute eigenvalues of a square matrix given LU decomposition I have heard from here, Eigenvalues for LU decomposition , that the eigenvalues of A are NOT the same ones for U, given $A=LU$. However, if A is a symmetric matrix, is it possible to still use the LU decomposition to compute the eigenvalue, especially when it it known that $U=DL^T$? 
There were some cases where I was able to get an LU decomposition of a symmetric matrix, test the diagonals of U for them being eigenvalues, only to find that they satisfy $|A-u_{kk}I| = 0$, but then the only solution for $(A-u_{kk}I)\overrightarrow x = \overrightarrow 0$ has $\overrightarrow x = \overrightarrow 0$ as the only solution! Is it then true that the fact that $c:|A-cI|=0$ doesn't necessarily make c and eigenvalue of A?
 A: The entries in $U$ have little or no relation to the eigenvalues of $A$ so it is not surprising that 
$$(A-u_{kk}I)\overrightarrow x = \overrightarrow 0 $$ 
gives $\overrightarrow x = \overrightarrow 0 $.
However $c:|A-cI|=0$ does give the eigenvalues.
Only thing you can claim is that $LU$ and $UL$ have the same non-zero eigenvalues. If $L$ and $U$ are both non-singular then $UL$ will usually (but not always) have a "better" eigenstructure. Beyond that, not much can be said even if $A$ is symmetric and $L^T = U$
A: By the Sylvester theorem of inertia, the eigenvalues of A will have the same sign structure as the diagonal entries of $U$ resp. $D$. 
More can not be said, as per user44197. 
One can use the LU decomposition in a similar manner as the QR decomposition to make an GR algorithm
$L_kU_k=A_k$, $A_{k+1}=U_kL_k$
which will converge for the same reason the QR algorithm converges. However, this generalized GR algorithm will often be numerically very unstable.
A: By definition, the LU decomposition of a matrix $A \in \mathbb{R}^{M \times M}$ is a lower diagonal matrix $L \in \mathbb{R}^{M \times M}$ and an upper diagonal matrix $U \in \mathbb{R}^{M \times M}$ such that,
$$ A = LU $$
Without loss of generality we can assume that either $U$ or $L$ has a full diagonal of ones, by the uniqueness of the $LDU$ decomposition algorithm, and we know that,
$$ \det A = \det LU = (\det L) (\det U) $$
Suppose that $L$ has ones on the diagonal, then $\det A = \det U = \prod_{i=1}^M u_{ii}$. Which is what you might expect.
The problem is that LU decomposition algorithms are not stable for every square matrix. Instead people use algorithms which first permute the rows and columns of the input matrix, in order to ensure that it produces a stable result. For example, in partial pivoting, a permutation matrix $P$ is computed, with the property that,
$$ PA = LU $$
In this case, in order to compute the determinant of $A$, we must first compute the determinant of $P$. Fortunately, $ P = \prod_{i = 1}^K P_i$, where $P_i$ are single row permutations, and so $ \det(P) \in \{-1,1\} $. We have that, $ \det(P) = \prod_{i=1}^K \det(P_i) $, and so, $\det(P) = -1$ if $K$ is odd, and $|P| = 1$ if $K$ is even. 
The problem now is finding $K$, which can easily be computed if you have access to the implementation. So far as I know, it is more difficult in third party libraries.
