# Linear algebra – Jordan Matrix switch

Matrices $$B in mathbf {C} ^ {n times n}$$ move with a given matrix $$A in mathbf {C} ^ {n times n}$$ in Jordan it is known that the normal form is constructed from Toeplitz's superior triangular matrices. I have seen convincing evidence, but I wanted to derive this fact by my own method (in Dirac notation):

1. Consider the base of generalized eigenbasis (Jordan base) for A and its adjunct $$A ^ daga$$ belonging to self-worth $$lambda$$:

$$| R_i rangle in ker (A- lambda) ^ i- ker (A- lambda) ^ {i-1}, | L_j rangle in ker (A ^ dagger- lambda ^ *) ^ i- ker (A ^ dagger- lambda ^ *) ^ {i-1}$$

giving rise to the Jordanian chains.

1. Squeeze $$B$$ in this base (right and left): $$B = sum_ {ij} B_ {ij} | R_i rangle langle L_j |$$

2. Apply $$A$$ on the left resp. Right:

$$AB = sum_ {ij} B_ {ij} A | R_i rangle langle L_j | = sum_ {ij} B_ {ij} lambda | R_i rangle langle L_j | + sum_ {ij} B_ {ij} | R_ {i-1} rangle langle L_j |$$

$$BA = sum_ {ij} B_ {ij} | R_i rangle langle L_j | A = sum_ {ij} B_ {ij} lambda | R_i rangle langle L_j | + sum_ {ij} B_ {ij} | R_i rangle langle L_ {j-1} |$$

1. As $$| R_i rangle langle L_j |$$ It is a basis for the space of matrices, equality $$AB = BA$$ is equivalent to the equality of the respective coefficients separately, in particular:

$$sum_ {ij} B_ {ij} | R_ {i-1} rangle langle L_j | = sum_ {ij} B_ {ij} | R_i rangle langle L_ {j-1} |$$

1. Redefining the summation rates via $$k: = i-1$$ Y $$l: = j-1$$ yields

$$sum_ {kj} B_ {k + 1, j} | R_k rangle langle L_j | = sum_ {il} B_ {i, l + 1} | R_i rangle langle L_l |$$ and therefore $$B_ {i + 1, j} = B_ {i, j + 1}$$

But the Toeplitz matrices are characterized by $$B_ {i + 1, j} = B_ {i, j-1}$$! What's wrong here?