Eigenvalues for LU decomposition In general I know that the eigenvalues of A are not the same as U for the decomposition but for one matrix I had earlier in the year it was.  Is there a special reason this happened or was it just a coincidence?  The matrix was
$A = \begin{bmatrix}-1& 3 &-3 \\0 &-6 &5 \\-5& -3 &1\end{bmatrix}$
with
$U = \begin{bmatrix}-1& 3 &-3 \\0 &-6 &5 \\0& 0 &1\end{bmatrix}$
if needed 
$L = \begin{bmatrix}1& 0 &0 \\0 &1 &0 \\5& 3 &1\end{bmatrix}$
The eigenvalues are the same as $U$ which are $-1$,$-6$ and $1$. When I tried to do it the normal way I ended up with a not so nice algebra problem to work on which took way to long.  Is there some special property I am missing here?  If not is there an easy way to simplify $\mathrm{det}(A-\lambda I)$ that I am missing?  Thank you! 

It's hard to say if this is mere coincidence or part of a larger pattern. This is like asking someone to infer the next number to a finite sequence of given numbers. Whatever number you say, there is always some way to explain it.
Anyway, here's the "pattern" I see. Suppose
$$
A = \begin{pmatrix}B&u\\ v^T&\gamma\end{pmatrix},
$$
where


*

*$B$ is a 2x2 upper triangular matrix;

*the two eigenvalues of $B$, say $\lambda$ and $\mu$, are distinct and $\neq\gamma$;

*$u$ is a right eigenvector of $B$ corresponding to the eigenvalue $\mu$;

*$v$ is a left eigenvector of $B$ corresponding to the eigenvalue $\lambda$.


Then $A$ has the following LU decomposition:
$$
A = \begin{pmatrix}B&u\\ v^T&\gamma\end{pmatrix}
=\underbrace{\begin{pmatrix}I_2&0\\ kv^T&1\end{pmatrix}}_{L}
\quad
\underbrace{\begin{pmatrix}B&u\\0&\gamma\end{pmatrix}}_{U}
$$
where $k=\frac1\lambda$ if $\lambda\neq0$ or $0$ otherwise.
The eigenvalues of $U$ are clearly $\lambda,\mu$ and $\gamma$. Since $u$ and $v$ are right and left eigenvectors of $B$ corresponding to different eigenvalues, we have $v^Tu=0$. Therefore
\begin{align}
(v^T, 0)A
&=(v^T, 0)\begin{pmatrix}B&u\\ v^T&\gamma\end{pmatrix}
=(v^TB,\, v^Tu)=\lambda(v^T,0),\\
A\begin{pmatrix}u\\0\end{pmatrix}
&=\begin{pmatrix}B&u\\ v^T&\gamma\end{pmatrix}\begin{pmatrix}u\\0\end{pmatrix}
=\begin{pmatrix}Bu\\v^Tu\end{pmatrix}
=\mu\begin{pmatrix}u\\0\end{pmatrix},\\
A\begin{pmatrix}\frac{1}{\gamma-\mu}u\\1\end{pmatrix}
&=\begin{pmatrix}B&u\\ v^T&\gamma\end{pmatrix}
\begin{pmatrix}\frac{1}{\gamma-\mu}u\\1\end{pmatrix}
=\gamma\begin{pmatrix}\frac{1}{\gamma-\mu}u\\1\end{pmatrix}.
\end{align}
So, the eigenvalues of $U$ are also the eigenvalues of $A$.
