Diagonalizability of an operator

Question

Given $T:Vto V$ a linear operator, such that $m_{T}=(x-c_{1})^{r_{1}}...(x-c_{k})^{r_{k}}, ineq jto c_{i}neq c_{j}$, we'll denote for each $i$, $E_{i}$ is the projection onto $Ker((T-c_{i})^{r_{i}})$, with respect to the aforementioned decomposition of $V$ and $T$.
First question was to prove $T=TE_{1}+TE_{2}+...+TE_{k}$ which is quite easy.
Now I have to prove that the linear operator $D=c_{1}E_{1}+...+c_{k}E_{k}$ is diagonalizable, but I can't seem to come up with anything smart to say about this operator.

Vincent · Answer

Let's denote the space onto which $E_i$ is the projection by $V_i$, that's a bit shorter.
Step 1) Pick an arbitrary basis for each $V_i$
Step 2) Show that the union of these bases is a basis of all of $V$
Step 3) Write down the matrix representation of $D$ with respect to this basis
Generally speaking: when working with a very concrete matrix it is easy to see whether or not it is diagonalizable. And in this case it is particularly easy because, as you might have guessed, the matrix constructed in step 3 is already diagonal.
Step 2 is the most interesting one. The underlying result is that $V = bigoplus V_i$. This is the only way in which the operator $T$ still plays some (background) role in the second question.

Diagonalizability of an operator

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