The number of linearly independent eigenvectors for the eigenvalue A lS equal to nujty(A AI) Hence, 2 [l . . . . . . ifA : (U 2) ,then there is one eigenvalue A : 2 and nIhty(A 7 2]] : I Number I , and the number of linearly Independent eigenvectors fer A : 2 IsI Number I , and the number at linearly independent eigenvecters for A : 2 is I Number . ifB : (i i) then there is enly one eigenvalue A : 2 and nuit 7 21') : I Number I Because the matrix B does not have enough linearly independent eigenvectors, it lS called a defective matrix, and cannot be diagonalised