# Basis for JNF and matrix relationship for this basis

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?

• I understand that $A$ is the linear application, not the matrix in one particular basis. – Miguel Aug 1 at 15:59

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 $$a_1=2v_1,$$ $$a_2=v_1+2v_2,$$ $$a_3=2v_2.$$