How do I find a canonical basis for A?
I know how to find the JFC but I don't know what's a canonical basis for A. Can someone explain?
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$\begingroup$$A$ is the matrix that represents a linear transformation in the standard basis: $\{(1,0,0)^T,(0,1,0)^T,(0,0,1)^T\}$.
The canonical basis for the same linear transformation is the basis in which the transformation is expressed by a matrix in the Jordan canonical form. This basis is made by the proper and generalized eigenvectors of the tranformation.
In your case these vectors are: $\{(-1,0,1)^T,(1,0,0)^T,(2,1,0)^T\}$and we have:
$$A= \begin{pmatrix} 1&2&-1\\ 0&2&0\\ 1&-2&3 \end{pmatrix}= \begin{pmatrix} -1&1&2\\ 0&0&1\\ 1&0&0 \end{pmatrix} \begin{pmatrix} 2&1&0\\ 0&2&0\\ 0&0&2 \end{pmatrix} \begin{pmatrix} 0&0&1\\ 1&-2&1\\ 0&1&0 \end{pmatrix} =SJS^{-1} $$
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