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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!

Theorem
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  • Try to compute the eigenspaces corresponding to these eigenvalues. Their dimensions will be dependent on $a$, and so... – Andrei Kh Sep 04 '16 at 19:01
  • So I need to check when $3-\text{r}(2I-A)=1$ and $3-\text{r}(10I-A)=2$? That happens iff $T$ is diagonalizable, right? $A$ is obviously $[T]_B$ – Theorem Sep 04 '16 at 19:06
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    Exactly (this text should remove the 15 char restriction on comments) – Andrei Kh Sep 04 '16 at 19:17
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    I don't think you did the change of basis correctly. For example, you have $T((2,0,0)) = (a-9,a+1,0)$, but this vector is in the standard basis. You need to convert it into the $B$- basis (which amounts to dividing the first and second components by $2$). It would be easier to just work in the given basis $((1,-1,0), (1,1,0), (1,1,1))$. I think you will see that the characteristic polynomial will depend on $a$. – angryavian Sep 04 '16 at 19:27
  • And do you think it's the most efficient way? We need to solve 3 systems of equations. – Theorem Sep 05 '16 at 14:54
  • @Theoniix Apologies, I was mistaken; I think working in the given basis would be harder than your approach. Just continue with what you have done but make sure to convert back into the $B$-basis as I mentioned. – angryavian Sep 05 '16 at 22:22

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