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Suppose we are given $\mathbf A = \begin{pmatrix} 0&1&0\\ -4&4&0\\ -2&1&2 \end{pmatrix}$

Its JNF is $\mathbf J = \begin{pmatrix} 2&1&0\\ 0&2&0\\ 0&0&2 \end{pmatrix}$

I don't understand the following statement

If we are in the basis $\{e_1, e_2, e_3\}$ wrt which $\mathbf A$ is in JNF, then the matrix tells us: $\mathbf A e_1 = 2e_1, \mathbf A e_2 = 2e_2 + e_1, \mathbf Ae_3 = 2e_3.$

I understand that if $\{e_1, e_2, e_3\}$ is the standard basis then this is true. But it is not necessary that in the standard bases $\mathbf A$ looks like $\mathbf J,$ so why the matrix tells as that?

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  • $\begingroup$ I understand that $A$ is the linear application, not the matrix in one particular basis. $\endgroup$ – Miguel Aug 1 at 15:59
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If you consider $A$ as information of a change of basis then this can be described as $$a_1=-4e_2-2e_3,$$ $$a_2=e_1+4e_2+e_3,$$ $$a_3=2e_3.$$ after the calculation of the eigen-vectors and the generalized ones of $A$ you get three vectors $$v_1=-e_1-2e_2,$$ $$v_2=e_1+2e_2+e_3,$$ $$v_3=e_2,$$ which form a matrix $S$ such that $S^{-1}AS=J$.

Then by solving for $e_1,e_2,e_3$ and subbing you are going to get $$a_1=2v_1,$$ $$a_2=v_1+2v_2,$$ $$a_3=2v_2.$$

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