# Shortcuts for computing the eigenvalues of a linear transformation

How would you calculate the eigenvalues of the following matrix?

$A = \begin{pmatrix} -3 & 1 & -1 \\ -7 & 5 & -1\\ -6 & 6 & -2\end{pmatrix}$  $\ \ \ \ \ $$\chi_A(\lambda) = \det(A-\lambda I)= \begin{vmatrix} -3-\lambda & 1 & -1 \\ -7 & 5-\lambda & -1\\ -6 & 6 & -2-\lambda \end{vmatrix}=\ldots =0 I'd really like to avoid using the rule of Sarrus. This will just lead to a huge list of multiplications and finally I may even have to guess the roots of the polynomial (maybe with Vieta's Theorem) and factor them out via polynomial division. - This whole process is tedious and prone to errors so I'd like to take some shortcuts whenever I can. Here are some shortcuts I already know, which may be used in conjunction: • Multiplying two rows by (-1) (this will not change the determinant) • Developing the determinant via a row or column that has a lot of zeros in it (ideally just one factor) to get out linear factors of the polynomial. • Transforming the matrix via gaussian elimination to a matrix which has more zeros in one column or row (ideally: transform it to a lower/upper triangular matrix). • Compute the determinant via the theorem for block-diagonal matrices. However, none of these shortcuts are useful for calculating the eigenvalues of A. How would you approach the computation of the determinant of the matrix of above? What other shortcuts do you have to share? • For dense matrix expanding of determinant is only method, I suppose. – Michael Galuza Jul 24 '15 at 15:25 ## 1 Answer Here's a start: Add column 1 to column 2 (this doesn't change the determinant); then take out a factor from column 2; after that, add column 2 to column 3:$$ \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$\$

• 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. – ndrizza Jul 25 '15 at 15:56
• 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. ;-) – Hans Lundmark Jul 25 '15 at 17:25