Question : Find the eigen values and eigen vector of the matrix $\displaystyle
\begin{bmatrix}
1 & -6 & -4 \\
0 & 4 & 2 \\
0 & -6 & -3
\end{bmatrix}
$
Get them by solving $\displaystyle -\lambda (\lambda - 1)^2 = 0$ for $\displaystyle \lambda$.
If you cannot solve this equation then you are strongly advised to go back and thoroughly revise the prerequisite mathematics (such as solving polynomial equations) that these questions assume you are competent in.
What work is it that you have shown? The first thing you did was simply ask for the eigenvalues and eigenvectors. When you were challenged to show your work said you had got to $\displaystyle A - \lambda I = (1- \lambda) (- \lambda + \lambda^2) = - \lambda + 2 \lambda^2 - \lambda^3$ and were stuck.
Are you saying that you did not know that you set that equal to 0 and solve: $\displaystyle - \lambda + 2 \lambda^2 - \lambda^3= 0$?
Then, when Tonio pointed out that this factored as $\displaystyle -\lambda(1- \lambda)^2$, your response was "could you tell what would be the values of $\displaystyle \lambda$".
Are you saying you do not know how to solve $\displaystyle -\lambda(1- \lambda)^2= 0$?
Then, finally, you just state "my assumptions is that $\displaystyle \lambda$= -1, 0, 1 But i dont know if that is correct or not ".
Well, you can see whether they satisfy the equation of not by putting them into the equation: if $\displaystyle \lambda= -1$ then $\displaystyle -\lambda+ 2\lambda^2- \lambda^3= -(-1)+ 2(-1)^2-(-1)^3= 1+ 2+ 1= 4$. No, that does NOT satisfy the equation.
Or you could see if they are eigenvalues by putting them into the original eigenvalue problem. Does there exist non-trivial [x y z] such that $\displaystyle \begin{bmatrix}1 & -6 & -4 \\ 0 & 4 & 2 \\ 0 & -6 & -3\end{bmatrix}\begin{bmatrix}x \\ y \\ z\end{bmatrix}= -1\begin{bmatrix}x \\ y \\ z\end{bmatrix}$?
Those reduce to the three equations x- 6y- 4z= -x, 4y+ 2z= -y, and -6y- 3z= -z. The last two equations are 5y+ 2z= 0 and -6y+ 2z= 0. Subtracting the second of those equations from the first, 11y= 0 so y must be 0. Then -6(0)+ 2z= 2z= 0 so z must be 0. Finally, x- 6(0)- 4(0)= -x becomes 2x= 0 so x=0. [x y z] must be [0 0 0], the trivial solution. There is no non-trivial solution so -1 is NOT an eigenvalue.
Finally, you say
"If u look above that is my work and from that i can conclude that the values could be -1,0,1
I havealready provided my work above what else do u want? "
Other than your assertion that the characteristic polynomial is $\displaystyle - \lambda + 2 \lambda^2 - \lambda^3$ you have shown no work!
You did not show that you know you must set that equal to 0 to get the characteristic equation, $\displaystyle - \lambda + 2 \lambda^2 - \lambda^3= 0
$. You did not show that you could factor that to get $\displaystyle -\lambda(1- \lambda)^2= 0$. And even after being told that you could factor it in that way, you did not say how you "can conclude that the values could be -1,0,1 " which, you have been told twice now, is wrong.
What are the solutions to $\displaystyle -\lambda+ 2\lambda^2- \lambda^3= -\lambda(1- \lambda)^2= 0$?
If you HAVE to ask this then you REALLY need to go back ASAP to your high school maths, since this is 8th-9th stuff at most, and please do not take offense from this: many of us can sometimes forget very basic material, but if you're trying to deal with linear algebra then you can NOT allow yourself to have these kind of gaps in your mathematical background, lest you'll suffer a lot when things become a little more hard, and believe me: they will.
Tonio