Question: I was reading the following
[How do i prove that every open set in $\mathbb{R}^2$ is a union of at most countable open rectangles?](https://math.stackexchange.com/questions/271420/how-do-i-prove-that-every-open-set-in-mathbbr2-is-a-union-of-at-most-count)

The answer by "user4594" basically considers a k-cell with rational endpoints, and I am guessing that when he says that "E would be the union of these cells" the axiom of choice is needed because for each x there may be more than one k-cell with rational endpoints.

Am I correct?

Answer: 
The diagonal entries of a triangular matrix are its eigenvalues. But it doesn't follow that row-reducing a general matrix helps compute the eigenvalues: if $M$ is an arbitrary matrix, then row-reduction gives

$$EM = U$$
where $E$ is a product of elementary matrices and $U$ is upper-triangular. Now you know the eigenvalues of $EM$, but this doesn't tell you anything about the eigenvalues of $M$.
If a matrix can be row reduced into a triangular form, then you have the eigenvalues already.

$BUT$ every matrix can't be reduced into a triangular form. 

They can be factorized into two matrices, a lower triangular and upper triangular (by LU factorization, you might be aware of it already). But, the diagonals of those matrices won't give you eigenvalues.

$Ax = \lambda x$ OR characteristic equation det|A -$\lambda$I| = 0 must be solved to find the eigenvalues

Also, there are other methods to find the complete eigensystem for a matrix.