I have a question regarding Linear Algebra chapter 5.3: If I am trying to determine if a matrix A= [ 0 1 1                                                                     2 1 2                                                                     3 3 2 ] is diagonizable, and the solution I got for the third row was 1 = 0, does this mean that the matrix is not diagonizable since the eigenvalue = 0 proves that the matrix is inconsistent? Or am I able to still get the basis vector for the eigenspace?

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