A third-order matrix has only two linearly independent eigenvectors, and this matrix has only one eigenvalue of the triple root

A third-order matrix has only two linearly independent eigenvectors, and this matrix has only one eigenvalue of the triple root

Let the triple eigenvalue of the third order square matrix a be c
First, let's see if the only eigenvalue C is 0
1 if C is 0, then AX = CX = 0, then because the matrix has only two linearly independent eigenvectors, that is, the dimension of the solution space is equal to 2, then RKA = n-dim solution space = 3-2 = 1
2 if C is not zero, then the determinant value of a is the third power of C, that is to say, a is nonsingular, so the full rank is 3