A linear operator $T$ on a complex vector space $V$ has the characteristics polynomial $x^3(x-5)^2$ and the minimal polynomial $x^2(x-5)$ .Choose all correct options.
$(a)$ The Jordan Form of $T$ is uniquely determined by the given information.
$(b)$ There are exactly two Jordan blocks in the Jordan decomposition of $T$.
$(c)$ The operator induced by $T$ on the quotient space $V/Ker(T-5I)$ is nilpotent , where $I$ is the Identity operator.
$(d)$ The operator induced by $T$ on the quotient space $V/Ker(T)$ is a scalar multiple of the Identity operator.
My attempt
$(a)$ This is the uniquely determined Jordan Form.
\begin{bmatrix}0&1&&&\\0&0&&&\\&&0&&\\&&&5&\\&&&&5\end{bmatrix}
$(b)$ There are $4$ Jordan Blocks.
$(c)$ Let $B=\{v_1,v_2,v_3,v_4,v_5\}$ be the ordered basis with respect to which the matrix representation of $T$ is the above Jordan Canonical Form . The Vector Space $V$ has the standard inner product.
I know that
"Every subspace of a finite dimensional Inner Product Space has a unique orthogonal complement."
and
"The quotient space corresponding to a subspace $W$ of a finite dimensional vector space $V$ is isomorphic to the complementary subspace of $W$.
i.e $V/W \simeq W'$ where $W\oplus W'=V$
Keeping these theorems in mind,
We have, $T-5I=\begin{bmatrix}-5&1&&&\\0&-5&&&\\&&-5&&\\&&&0&\\&&&&0\end{bmatrix}$
Let $(x_1,x_2,x_3,x_4,x_5)^T \in Ker(T-5I)$ .This gives
$-5x_3=0, -5x_2=0, -5x_1+x_2=0$ giving $x_1=x_2=x_3=0$
Hence $\{Ker(T-5I)\}= \{(0,0,0,x_4,x_5)^T: x_4,x_5 \in \mathbb{C} \}$
Hence $\{Ker(T-5I)\}^{\perp}= \{(x_1,x_2,x_3,0,0)^T: x_1,x_2,x_3 \in \mathbb{C} \}$
So, $T$ operating on $V/Ker(T-5I)$ is same as $T$ operating on $\{Ker(T-5I)\}^{\perp}$
Also , $\{Ker (T-5I)\}^{\perp}=L(\{v_1,v_2,v_3\})$
Hence
$T:V/Ker(T-5I) \to V/Ker(T-5I)$
is given by $Tv_1=0, Tv_2=v_1, Tv_3=0$ as obtained fron the Jordan Canonical Form. Thus the matrix of $T$ relative to the ordered basis $\{v_1,v_2,v_3\}$ is
$\begin{bmatrix}0&1&0\\0&0&0\\0&0&0\end{bmatrix} $
which is a nilpotent matrix .
For $(d)$
$Ker(T)=\{(x_1,0,x_3,0,0)^T : x_1,x_3 \in \mathbb{C}\}$
Hence
$\{Ker(T)\}^{\perp}=\{(0,x_2,0,x_4,x_5)^T : x_2,x_4,x_5 \in \mathbb{C}\}$
Hence by the same reason
$T: V/Ker(T) \to L(\{v_1,v_4,v_5\})$ is given by
$Tv_2=v_1,Tv_4=5v_4,Tv_5=5v_5$
Thus the matrix of this induced operator
$\begin{bmatrix}1&0&0\\0&5&0\\0&0&5\end{bmatrix} $
which is not scalar matrix.
Thanks a lot for your patience in seeing my work. Do you see any discrepancy there.?
If you have any other metods, please share. Thanks for your time.