$A=$$\large \bigl(\begin{smallmatrix} 4 &-3 & -3\\ 3 &-2 &-3 \\ -1 &1 & 2 \end{smallmatrix}\bigr)$
$A^{2}=$$\large \bigl(\begin{smallmatrix} 4 &-3 & -3\\ 3 &-2 &-3 \\ -1 &1 & 2 \end{smallmatrix}\bigr)$$\large \bigl(\begin{smallmatrix} 4 &-3 & -3\\ 3 &-2 &-3 \\ -1 &1 & 2 \end{smallmatrix}\bigr)$
$=$ $\large \bigl(\begin{smallmatrix} 10 &-9 & -9\\ 9 &-8 &-9 \\ -3 & 3 & 4 \end{smallmatrix}\bigr)$
So , $A^{2}\neq A$ So , It is not idempotent matrix .
Now Find the Eigen values:-
$\begin{vmatrix} 4-\lambda &-3 &-3 \\ 3&-2-\lambda &-3 \\ -1&1 & 2-\lambda \end{vmatrix}$ $=0$
$(4\lambda ^{2}-\lambda ^{3}+\lambda -4+9-9\lambda -3+3\lambda )=0$
$(2-\lambda ^{3}+4\lambda ^{2}-5\lambda )=0$
$(\lambda ^{3}-4\lambda ^{2}+5\lambda -2 )=0$
$\lambda ^{2}(\lambda -1)-3\lambda (\lambda -1)+2(\lambda -1)=0$
$(\lambda -1)(\lambda ^{2}-3\lambda +2)=0$
$(\lambda -1)(\lambda ^{2}-2\lambda-\lambda +2)=0$
$(\lambda -1)(\lambda -1)(\lambda -2)=0$
So the eigen values are $2,1,1$ .
Now we know that ,
A matrix is nilopotent if and only if all its eigen value is zero .
Link:- https://math.stackexchange.com/questions/1265679/why-are-eigenvalues-of-nilpotent-matrices-equal-to-zero
Now the eigen values are not zero so it is not nilopotent as well.
So option $B ,C$ are gone .
Now we need to compute eigen vector for the eigen values .
Lets try with $\lambda=1$;
$(A-\lambda I)X=0$
$\begin{pmatrix} 3 &-3 &-3 \\ 3&-3 &-3 \\ -1& 1& 1 \end{pmatrix}$$X=0$
$\begin{pmatrix} 3 &-3 &-3 \\ 0&0 &0 \\ 0& 0& 0 \end{pmatrix}$$\begin{pmatrix} x\\ y\\ z \end{pmatrix}$ $=0$
So $x-y-z=0$
Free variables are y,z .
So , $y=k1,z=k2$
So , $x=k1+k2$
So , $X$=$\begin{pmatrix} k1+k2\\ k1\\ k2 \end{pmatrix}$
$X=$ $k1\begin{pmatrix} 1\\ 1\\ 0 \end{pmatrix}$ $+$ $k2\begin{pmatrix} 1\\ 0\\ 1 \end{pmatrix}$
Hence the basis for 1-eigen space $B1=\left \{ v1,v2 \right \}$, where $v1=\begin{pmatrix} 1\\ 1\\ 0 \end{pmatrix}$ , $v2=\begin{pmatrix} 1\\ 0\\ 1 \end{pmatrix}$
Now compute eigen vector space for $\lambda=2$,
$(A-\lambda I)X=0$
$\begin{pmatrix} 2 &-3 & -3\\ 3 & -4 &-3 \\ -1&1 & 0 \end{pmatrix}X=0$
$\begin{pmatrix} 2 &-3 & -3\\ -1 & 1 &0 \\ -1&1 & 0 \end{pmatrix}X=0$
$\begin{pmatrix} 2 &-3 & -3\\ -1 & 1 &0 \\ 0&0 & 0 \end{pmatrix}X=0$
$\begin{pmatrix} 2 &-3 & -3\\ 0 & -1 &-3 \\ 0&0 & 0 \end{pmatrix}X=0$
$2x-3y-3z=0$
$y+3z$=0
$z$ be the free variable .
So ,$z=k$
$y=-3k$
$x=-3k$
So , the eigen vector is $X=\begin{pmatrix} -3k\\ -3k\\ k \end{pmatrix}$
$X=k\begin{pmatrix} -3\\ -3\\ 1 \end{pmatrix}$
So, $v3=\begin{pmatrix} -3\\ -3\\ 1 \end{pmatrix}$
So the eigen vector basis $v1,v2,v3$ are linearly independent.$v1,v2$ forms the basis for the eigen value 1 and $v3$ is not contained in 1-eigenspace because its eigen value is 2 . So Diagonilazation theorem says
$A=CDC^{-1}$ where C=$\begin{pmatrix} 1 &1 & -3\\ 1 & 0 &-3 \\ 0&1 & 1 \end{pmatrix}$ ,D= $\begin{pmatrix} 1 &0 & 0\\ 0 & 1 &0 \\ 0& 0& 2 \end{pmatrix}$
So the given matrix is diagonalizable.
Correct answer is (A) .
