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It is widely known that if a matrix is given in upper triangular form, then one can just read off the eigenvalues (and their algebraic multiplicity) on the main diagonal of the matrix.

My question is: what if I get a non-upper triangular matrix to start, and I then put it into row-echelon form - not the row-reduced echelon form with all 1's in the pivot variables. Can I spot any of the eigenvalues of the original matrix from this upper triangular matrix?

Thanks,

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$\begin{bmatrix}1&-1\\1&3\end{bmatrix}$ has eigenvalues $\lambda_1 = 2$ and $\lambda_2 = 2$.

$\begin{bmatrix}1&-1\\0&4\end{bmatrix}$ has eigenvalues $\lambda_1 = 4$ and $\lambda_2 = 1$

On the other hand, note that $2\times 2 = 4 \times 1$.

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  • $\begingroup$ Interesting observation you make here, @MichaelBiro. I see that the determinant was invariant under an elementary row operation - and also equal to the product of the eigenvalues in both the non-upper triangular and upper triangular cases (the upper triangular case is widely known). Is the determinant always going to be equal to the product of the eigenvalues of a square matrix, even if the matrix is not given in upper-triangular form to start with? $\endgroup$ – User001 Dec 3 '14 at 2:23
  • $\begingroup$ So that...for a larger matrix, once I compute its characteristic polynomial, solve for its eigenvalues (if it has any at all), then we sort of get the determinant for free? Though, this sounds too good to be true. $\endgroup$ – User001 Dec 3 '14 at 2:25
  • $\begingroup$ Yes, the determinant is the product of the eigenvalues, and is preserved by Gaussian elimination (possibly with a change of parity). Computing eigenvalues has its own difficulties, especially if you want their exact values... $\endgroup$ – Michael Biro Dec 3 '14 at 2:35
  • $\begingroup$ Ok, got it. Thanks so much, @MichaelBiro. $\endgroup$ – User001 Dec 3 '14 at 2:40
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In general, the eigenvalue preserving transformations are precisely the similarity transformations: $A$ and $B$ have the same eigenvalues precisely when $B=P A P^{-1}$ for some invertible matrix $P$. It is routine to check that elementary row operations are not similarity transformations. For instance, if $EA$ is $A$ after a row operation, then $(EA)^2 \neq EA^2$, in general, yet $(PAP^{-1})^2 = P A^2 P^{-1}$ independent of $P$.

Alternately one can simply do a direct calculation on a $2 \times 2$ example, as in Michael's answer.

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  • $\begingroup$ Ok, thanks a lot, Ian. Have a great night, guys. $\endgroup$ – User001 Dec 3 '14 at 2:42

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