I am given $T:\mathbb{R}^3\rightarrow\mathbb{R}^3$ such that
$T(1,-1,0)=(a-4,a+6,0),\ T(1,1,0)=(-5,-5,0),\ T(1,1,1)=(2,2,2)$
I need to find for which $a$'s $T$ is diagonalizable (and for which it isn't)
So I took a basis $B=((2,0,0),(0,2,0),(0,0,1))$ and I found relatively easily
$[T]_B=\begin{bmatrix} a-9 & -a-1 &7 \\ a+1&-a-11 & 7\\ 0& 0& 2 \end{bmatrix}$
after calculating the char. polynomial I got $p(x)=(x-2)(x+10)^2$. It kinda makes sence from the given $T(1,1,1)=(2,2,2)$ but it doesn't depend on $a$ so I can't use my direction which was find when all the eigenvalues are different, and then treat each left case. I guess I need to connect it to linear independence of the corrosponding eigenvectors, but how?
I'm very new to the subject, so I'm still very lost and need some assistence. Thank to everyone who tries in advance!