here is nxn matrices . (A)=1 and tr(A)=n

find the representative polinomial of A and tell if A is diagonisable.

there is a way which my prof said is wronggg:

i said that this A could be a representation matrices for some transformation

and if (A)=1 then dim ImT=1 too,so dim KerT=n-1

so if we take a basis sfor this kernel and expand it

to .

so the first n-1 columns are zero columns exsept the last n'th column

and from the law that the traces of similar matrices are equal i get

that the diagonal of the representative matrices is all zero except there is 'n' in the corner

and from here its very easy to prove that its diagonisable.

but my prof said that its wrong because even if the representative matrice is diagonisable

it doesnt say anything about A

this is how my prof recommended to start to solve it:

if (A)=1 then we have eigen value 0 with n-1 eigen vectors.

dont know how to continue from here

??