3

Let:

$$A=\begin{pmatrix} 3 & 0 & -2 & -3\\ 4 & -8 & 14 & -15\\ 2 & -4 & 7 & -7\\ 0 & 2 & -4 & 3 \end{pmatrix}$$

Find a change of basis matrix $P$ such that $PAP^{-1}$ is in Jordan canonical form.

After finding the characteristic polynomial I can find the Jordan form as it has eigenvalues $1$ with multiplicity $3$ and $2$ with multiplicity $1$. However, what I am struggling with is finding the characteristic polynomial in a fast and effective way rather than finding the $\mathrm{det}(A-\lambda I)$ as it would take a lot of time and error could happen easily. It was described here, but I am wondering if this method is unique or does the exchanging of rows and dividing them depend on the elements of the matrix.

Thanks in advance!

HMGB1
  • 31
  • Thanks Moo! I edited the matrix! – HMGB1 Aug 11 '21 at 23:35
  • If you multiply a row or a column by a constant, the determinant is also multiplied by the same constant. Exchanging two rows or two columns flips the sign of the determinant, but otherwise leaves it unchanged. In the various answers, these changes were carefully tracked and offset by other changes. As long as you do that, the methods listed can be applied to any matrix. Note that the selected answer does require two submatrices to commute. If they don't, the method fails. The answer handled this requirement by manipulating a submatrix to be a multiple of the identity. – Paul Sinclair Aug 12 '21 at 15:53

0 Answers0