Prove that the operator induced by $T$ on the quotient space $ V/\ker[T-5I]$ has all eigenvalues $=0$ 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.
 A: 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 $3$. 
A: Your definitions of quotient operator and its eigenvalue are wrong, hence the confusion.
First of all, look at what you think of as the quotient operator over $W$ $$ T(W+α_1[1,0,0,0,0]^T+α_2[0,1,0,0,0]^T+α_3[0,0,1,0,0]^T) $$ - let's call this $T(W+v)$,
being equal to this
$$Tw+α_1.0+α_2[1,0,0,0,0]T+α_3.0 = Tw + Tv\quad(1)$$
You have defined a map from an (affine) subset to a single vector: 
$$T(v+W) = Tv+Tw,\space w\in W$$ which means each set $v+W$ could be mapped to multiple outputs corresponding to multiple $w$ by the same map $T$. Thus this definition doesn't make sense unless you specify how to pick $w$ through $T$. 
The better definition of quotient operator is:
$$T|_W: V/W\to V/W$$ $$T|_W(v+W) = Tv+W$$
which maps from an affine subset to another affine subset in the same vector space (hence the name operator). You can verify that this mapping actually makes sense.
I hope you are familiar with algebraic operations on quotient space: $$(v+W)+(u+W)=v+u+W$$
and $$k(v+W) =kv +W $$
Note that this means the additive identity on $V/W$ is $W$
And then we can define eigenvalue of quotient operator $T|_W$:  
$\lambda$ is an eigenvalue of $T|_W$ $\iff$
$\exists v \notin W: T|_W(v+W)=\lambda v+W $
Using these definitions on $(1)$ you will arrive at the desired result. 
