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I have a $3 \times 3$ matrix $A$ for which I need to determine the characteristic polynomial.

Suppose I had $\det(A)=f$. If I performed row operations on $A$ I could manipulate $f$ accordingly to find the determinant for the new matrix. So for example if I multiplied row $1$ by $k$ and swapped row $2$ and row $3$ the determinant for the new matrix would be given by $-1 \times k \times f$.

My question: Could I simplify $A$ by performing row operations on it, then use the simplified $A$ in $(\lambda I_{3}-A)$, calculate $\det(\lambda I_{3}-A)$ and finally manipulate $\det(\lambda I_{3} -A)$ according to the row operations that I did on the original $A$. So if I swapped row $1$ and $3$ to get the simplified $A$, I would multiply $\det(\lambda I_{3}-A)$ by $-1$ in the end to get the final determinant.

Would this change the characteristic polynomial? If yes, why?

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  • $\begingroup$ Could you put up an example? (Preferably one where this algorithm speeds up the calculation, but if you don't have one just one so I can see what exactly the steps are) $\endgroup$ – alphacapture Oct 20 '17 at 2:34
  • $\begingroup$ $$ A= \begin{bmatrix} x && y && z \\ p && q && r \\ f && g && h \end{bmatrix}$$ I want the eigenvalues for $A$. I swap row 1 and row 3, and multiply row 2 by 2. I will call the result $B$. $$B = \begin{bmatrix} f && g && h \\ 2p &&2 q &&2 r \\ x && y && z\end{bmatrix}$$ My question is, would $det(\lambda I-B)$ be equal to $-1 \times 2 \times det(\lambda I-A)$? $\endgroup$ – user140161 Oct 20 '17 at 2:54
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No, elementary matrix operations don't preserve the eigenvalues in general. As an example, consider $A = I$, where $I$ is the $3 \times 3$ identity matrix. $$ \det{(\lambda I - A)} = \det{(\lambda I - I)} = (\lambda-1)^3. $$ Now, lets construct $B$ by swapping rows 1 and 3 while multiplying row 2 by 2 (as suggested in your comment), i.e. $$ B = \begin{pmatrix} 0 & 0 & 1\\ 0 & 2 & 0\\ 1 & 0 & 0 \end{pmatrix}, $$ and $$ \det{(\lambda I - B)} = (x-2)(x^2 - 1) \neq -2 (\lambda - 1)^3 = -2 \det{(\lambda I - A)}. $$

However, we could construct $B_\lambda$ by starting with the matrix $\lambda I - A$, swapping rows 1 and 3 and multiplying row 2 by 2, i.e. $$ B_\lambda = \begin{pmatrix} 0 & 0 & \lambda-1\\ 0 & 2\lambda-2 & 0\\ \lambda-1 & 0 & 0 \end{pmatrix}. $$ Then $$ \det{B_\lambda} = -2(\lambda-1)^3 = -2 \det{(\lambda I - A)} $$ as expected.

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