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How would you calculate the eigenvalues of the following matrix?

$A = \begin{pmatrix} -3 & 1 & -1 \\ -7 & 5 & -1\\ -6 & 6 & -2\end{pmatrix}$ $ $ $\ \ \ \ \ $$\chi_A(\lambda) = \det(A-\lambda I)= \begin{vmatrix} -3-\lambda & 1 & -1 \\ -7 & 5-\lambda & -1\\ -6 & 6 & -2-\lambda \end{vmatrix}=\ldots =0$

I'd really like to avoid using the rule of Sarrus. This will just lead to a huge list of multiplications and finally I may even have to guess the roots of the polynomial (maybe with Vieta's Theorem) and factor them out via polynomial division. - This whole process is tedious and prone to errors so I'd like to take some shortcuts whenever I can.

Here are some shortcuts I already know, which may be used in conjunction:

  • Multiplying two rows by $(-1)$ (this will not change the determinant)
  • Developing the determinant via a row or column that has a lot of zeros in it (ideally just one factor) to get out linear factors of the polynomial.
  • Transforming the matrix via gaussian elimination to a matrix which has more zeros in one column or row (ideally: transform it to a lower/upper triangular matrix).
  • Compute the determinant via the theorem for block-diagonal matrices.

However, none of these shortcuts are useful for calculating the eigenvalues of $A$. How would you approach the computation of the determinant of the matrix of above? What other shortcuts do you have to share?

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  • $\begingroup$ For dense matrix expanding of determinant is only method, I suppose. $\endgroup$ – Michael Galuza Jul 24 '15 at 15:25
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Here's a start:

Add column 1 to column 2 (this doesn't change the determinant); then take out a factor from column 2; after that, add column 2 to column 3: $$ \chi_A(\lambda) = \begin{vmatrix} -3-\lambda & 1 & -1 \\ -7 & 5-\lambda & -1 \\ -6 & 6 & -2-\lambda\end{vmatrix} = \begin{vmatrix} -3-\lambda & -2-\lambda & -1 \\ -7 & -2-\lambda & -1 \\ -6 & 0 & -2-\lambda\end{vmatrix} = -(2+\lambda) \begin{vmatrix} -3-\lambda & 1 & -1 \\ -7 & 1 & -1 \\ -6 & 0 & -2-\lambda\end{vmatrix} = -(2+\lambda) \begin{vmatrix} -3-\lambda & 1 & 0 \\ -7 & 1 & 0 \\ -6 & 0 & -2-\lambda\end{vmatrix} = \dotsb $$

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  • $\begingroup$ Thanks! I didn't think of taking out a factor of a column! That's a good idea! I guess there's really no simpler way than doing it that way when the matrix is dense. $\endgroup$ – ndrizza Jul 25 '15 at 15:56
  • $\begingroup$ In general, you may not be able to avoid multiplying everything out. For tricks like this to work, one needs to be a bit lucky. Or the exercise needs to be rigged. ;-) $\endgroup$ – Hans Lundmark Jul 25 '15 at 17:25

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