This shows that the eigenvalues are 1, 2, 4 and 4, according to algebraic multiplicity. The eigenspace corresponding to the eigenvalue 1 can be found by solving the equation "Av" = "λ v". It is spanned by the column vector "v" = (−1, 1, 0, 0)T. Similarly, the eigenspace corresponding to the eigenvalue 2 is spanned by "w" = (1, −1, 0, 1)T. Finally, the eigenspace corresponding to the eigenvalue 4 is also one-dimensional (even though this is a double eigenvalue) and is spanned by "x" = (1, 0, −1, 1)T. So, the geometric multiplicity (i.e., the dimension of the eigenspace of the given eigenvalue) of each of the three eigenvalues is one. Therefore, the two eigenvalues equal to 4 correspond to a single Jordan block, and the Jordan normal form of the matrix "A" is the direct sum