The question is as follows : A linear operator T on a complex vector space V has a characteristic polynomial $x^3(x-5)^2$ and minimal polynomial $x^2(x-5)^2$ .
Now, i need to prove that the operator induced by T on the quotient space $ V/\ker[T-5I]$ has all eigen values = $0$.
$ Attempt $: The matrix from the given information can be deduced as follows :
\begin{pmatrix}
0 &1 &0 &0 &0\\
0 &0 &0 &0 &0\\
0 &0 &0 &0 &0\\
0 &0 &0 &5 &0\\
0 &0 &0 &0 &5
\end{pmatrix}
$$W = \ker(T-5I) = \operatorname{span} [(0,0,0,1,0)^T, (0,0,0,0,1)^T ].$$
The basis of the quotient is given by: $[ W+(1,0,0,0,0)^T, W+(0,1,0,0,0)^T , W+(0,0,1,0,0)^T] $.
Now, I don't understand specifically much by the phrase operator induced by T on the quotient space spanned by the above basis , but , i think it means T acting on a vector in the quotient space i.e. $T [ \alpha_1 (W + [1,0,0,0,0]^T ) + \alpha_2 (W + [0,1,0,0,0]^T + \alpha_3 (W + [0,0,1,0,0]^T ]$
$$
\begin{align}
& = T [W + \alpha_1 [1,0,0,0,0]^T + \alpha_2 [0,1,0,0,0]^T + \alpha_3 [0,0,1,0,0]^T ] \\
& = Tw + \alpha_1 .0 + \alpha_2 [1,0,0,0,0]^T + \alpha_3 .0
\end{align}
$$
$w$ is a vector which belongs to $W$
$$= T [ \alpha_4 [(0,0,0,1,0)^T + \alpha_5 (0,0,0,0,1)^T ]
= 5\alpha_4 [(0,0,0,1,0)^T + 5\alpha_5 (0,0,0,0,1)^T$$
$$=(0,0,0,5\alpha_4, 5\alpha_5)^T$$
and the eigenvalues don't come as $0$ as desired in the question statement. Where am i going wrong in understanding the question?
Thanks.
Best Answer
The quotient vector space has dimension $3$, and is spanned by the classes of the first three basis vectors. This you have computed above. In other words, the linear map on $W$ with respect to this basis has the matrix $$ \begin{pmatrix} 0 & 1 & 0 \cr 0 & 0 & 0\cr 0 & 0 & 0 \end{pmatrix}. $$ This matrix has only $0$ as an eigenvalue, with multiplicity $3$.