Row reduction and the characteristic polynomial of a matrix Can you row reduce the matrix before computing $\det(\lambda I-A)$? Will this still give an equivalent characteristic polynomial?
 A: A typical presentation of elementary row operations sets out three kinds:
(1) Multiply a row by a nonzero scalar.
(2) Add a multiple of one row to another.
(3) Swap two rows.
The effects on the determinant of a (square) matrix when these are applied are easily determined.  (1) multiplies the determinant by the same scalar used to multiply the row.  (2) leaves the determinant unchanged.  (3) changes the sign of the determinant.
However even if the cumulative effects of a series of row operations were contrived to leave the determinant of $A$ unchanged, this does not imply that the characteristic polynomial is preserved.  For the characteristic polynomial to remain unchanged, we would need all the elementary symmetric invariants of characteristic roots (the coefficients of the characteristic polynomial, effectively) to stay the same.
For simplicity let's consider only the trace of $A$, the sum of characteristic roots, which determines the coefficient of $\lambda^{n-1}$, which is also the sum of the diagonal entries of $A$.  Operation (1) adds $(r-1)a_{ii}$ to the trace, when the ith row is multiplied by $r$.  Operation (2) adds $r a_{ij}$ to the trace, when $r$ times the ith row is added to the jth row.  Operation (3) adds $a_{ij}+a_{ji}-a_{ii}-a_{jj}$ to the trace when the ith and jth rows are swapped.  All these are fairly unpredicatable effects on the trace, and hence on the characteristic polynomial.
Considering the case of a $2\times 2$ matrix, we see that the reduced row-echelon form of a matrix $A$ is has characteristic polynomial either $\lambda^2$, $\lambda(\lambda-1)$, or $(\lambda-1)^2$.  Reconstructing even the characteristic polynomial of $A$ from the characteristic polynomial of its reduced row-echelon form seems unwieldy.
On the other hand it does make sense to consider computing $\det(\lambda I - A)$ by applying elementary row operations to row reduce $\lambda I - A$.  However since the matrix entries are polynomials, say from $\mathbb{R}[\lambda]$, the ring operations are not as easy to carry out as in the case of row reducing real matrix $A$.
For example consider the matrix $A$ and its reduced row echelon form $R$:
$$ A = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}, \; R = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} $$
The characteristic polynomial for $A$ is $\lambda^2 - 1$, while the characteristic polynomial for $R$ is $(\lambda - 1)^2$.  Only a single elementary row operation, swapping the two rows, was required to change $A$ into $R$.
However the polynomial matrix:
$$ \lambda I - A = \begin{pmatrix} \lambda & -1 \\ -1 & \lambda \end{pmatrix} $$
may be reduced by a sequence of three elementary row operations (one of each kind!) to upper triangular form:
$$ \begin{pmatrix} 1 & -\lambda \\ 0 & \lambda^2 - 1 \end{pmatrix} $$
whose determinant is evidently $\lambda^2 - 1$.  Thus elementary row operations applied to $\lambda I - A$ can provide us the characteristic polynomial of $A$.
A: No, you can't row reduce in advance. You will get a different characteristic polynomial if you do that.
For example,matrix A=$\left[
\begin{array} {lcr}
-1 & 1 & 0 \\
-4 & 3 & 0 \\
1 & 0 & 2 \\
\end{array}
\right] $ has a characteristic polynomial (2-$\lambda$)(1-$\lambda$)$^2$ 
But the reduced matrix A'= $\left[
\begin{array} {lcr}
-1 & 1 & 0 \\
0 & -1 & 0 \\
0 & 0 & 2 \\
\end{array}
\right]$ has a different characteristic polynomial (2-$\lambda$)(1+$\lambda$) $^2$ 
