I have this matrix:
$$A_t=\begin{pmatrix}\phantom{-}2+t&\phantom{-}4&\phantom{-}2+t&\phantom{-}2+t\\\phantom{-}t-2&\phantom{-}0& -6+t& -2-t\\-t+2&-4&-t+2&-2-t\\\phantom{-t+}0&\phantom{-}0&\phantom{-t+}0&\phantom{-t+}2t\end{pmatrix}$$
I must determine all $t\in\mathbb{R}$ for which $A_t$ is diagonalizable.
My solution:
PAt = (2t - X) det =
2+t-X 4 2+t
-2+t -X -6+t
2-t -4 2-t-X
And through Gauss I achieve this form:
(2t-X) det =
4-X 0 4-X
0 -4-X -4-X
2-t -4 2-t-X
= (2t -X) (4-X) ((-4-X)(2-t-X)-(16+4X)-(-4-X)(2-t))
= (2t - X) (4-X) (-4-X) ((2-t-X) - (2-t) +4)
= (2t-X)(4-X)(-4-X)(2-t-X-2+t+4)
= (2t-X)(4-X)(-4-X)(4-X)
Kernel =
-2+t 4 2+t 2+t
0 -8 -8 -4-2t
0 0 0 2t-4
0 0 0 0
Kernel =
-2+t 4 2+t 2+t
-8 -8 -8
2t-4
Kernel =
-2+t 0 -2+t 0
-8 -8 -8
2t-4
So the basis is:
1 1 -1 0
for t != 2
for t = 2 the basis is:
1 0 0 0
0 1 -1 0
0 0 1 -1
So my question is, are my thoughts right ? Are they complete this way ? If not and it's not much which is left, could you complete it for me or at least give me a hint ? Or are there better ways to solve this, since our professor didn't really teach us how to solve this kind of tasks with parameters. Thank you.