TITLE: Why cant I row reduce and get the same Eigen values?
QUESTION [8 upvotes]: A example is:
$$A=
 \begin{bmatrix}
    3 & 2   \\
    1 & 4   
  \end{bmatrix} \leftrightarrow
 -\begin{bmatrix}
    1 & 4   \\
    3 & 2   
  \end{bmatrix} \leftrightarrow
 -\begin{bmatrix}
    1 & 4   \\
    0 & -10   
  \end{bmatrix} = H
$$
The row operations I used are $$(1)   R_1 \leftrightarrow R_2$$
$$(2) -3R_1 + R_2 \to R_2$$
The original matrix 
$$|A - \lambda I| = (\lambda-2)(\lambda-5) $$
The reduced matrix:
$$|H - \lambda I| = -(1-\lambda)(-10-\lambda) $$
Why are they different?
For a couple other questions this was working 
EDIT: A matrix that gave the same eigen value as the one reduced:
$$A=
 \begin{bmatrix}
    2 & 0 & 1  \\
    6 & 4 & -3 \\
    2 & 0 & 3  
  \end{bmatrix} \leftrightarrow  \begin{bmatrix}
    1 & 0 & 1  \\
    9 & 4 & -3 \\
    -1 & 0 & 3  
  \end{bmatrix} \leftrightarrow \begin{bmatrix}
    1 & 0 & 0  \\
    9 & 4 & -12 \\
    -1 & 0 & 4 
  \end{bmatrix} = H
$$
Column operations were $$-C_3 + C_1 \to C_1$$
$$-C_1 + C_3 \to C_3$$
The eigen values are the same for A and H

REPLY [7 votes]: There's no reason this will work because row operations don't preserve similarity and so might change the eigenvalues. More explicitly, by performing row operations you move from a matrix $A$ to a matrix $PA$ where $P$ is invertible and in general $PA$ is not similar to $A$. 
To see this even more explicitly, assume that $A$ is diagonal with entries $(1,0)$ and multiply the first row by $2$. Clearly the eigenvalues of the resulting matrix will be $(2,0)$ while those of the original matrix were $(1,0)$.

REPLY [6 votes]: The eigenvalues of a matrix are the solutions to the polynomial
$$
\det(A-\lambda I)=0
$$
not the solution of 
$$
\det(SA-\lambda I)=0
$$
where $S$ is the product of a bunch of elementary matrices. When you row reduce, you are multiplying on the left (usually) by an invertible matrix. The polynomials above will not in general have their roots coincide.