A is a square matrix over F field
if k is the eigen value of A
prove that k is eigen value of A^t too
and has the same eigen vectors
eigen vectors are the solution space P(A)
is found by solving (A-kI)x=0
dim P(A)=dim n -dim (ro(a))
these are the laws i maneged to come up with to solve it
Just because two different matrices have an eigenvalue in common doesn't mean that the eigenvector is the same in both cases. Just try it on some small examples to see that the vector can change.
By the way..... spell check really is a wonderful thing. It really helps us answer your questions if they are written correctly and neatly.
Well, you can't. As you have been told that is not true.
For example, if has eigenvalues 1 and 2 with corresponding eigenvectors and , respectively.
Its transpose is which also has eigenvalues 1 and 2 but with corresponding eigenvectors and , respectively. Those eigenvectors are not multiples of the eigenvectors of A.