Compute the minimal polynomials for each of the following operators. Determine which of the following operators is diagonalizable.(adsbygoogle = window.adsbygoogle || []).push({});

a) [itex]T : P_2(\mathbb{C}) \to P_2(\mathbb{C})[/itex], where:

[tex](Tf)(x) = -xf''(x) + (i + 1)f'(x) - 2if(x)[/tex].

b) Let [itex]V = M_{k \times k}(\mathbb{R})[/itex].

[tex]T : V \to V[/itex] by [itex]T(A) = -2A^t[/tex].

For (a), I think the notation (Tf)(x) is confusing me a little. Do they mean T(f(x))? If f = ax² + bx + c, am I right in saying that:

(Tf)(x) = x²(-2ia) + x(2i)(a - b) + (b + i(b - 2c))?

For (b), I start by finding the characteristic polynomial of T. Let B be the matrix representation of T with respect to some ordered basis. Then, the characteristic polynomial of T is:

g(t) = det(B - tI)

g(t) = det(BI + B(½tI))

g(t) = det(B((1 + ½t)I))

g(t) = det((1 + ½t)BI)

g(t) = det((1 + ½t)(-2I))

g(t) = det(-(2 + t)I)

g(t) = (-1)ⁿ(2 + t)ⁿ, where n = dim(V)

I'm not sure whether n = k or n = k². Now, either way, the minimal polynomial of T is the same as the minimal polynomial of B, which will be some power of (2 + t). Let's try the first power, so the minimal polynomial is:

p(t) = 2 + t

Then:

p(B) = 2I + B = 2I + BI = 2I - 2I = 0, so the first power seems right.

Now, for some [itex]A \in V, A \neq A^t[/itex], we have:

[tex]0 = p(B)(A) = (2I + B)A = 2A + BA = 2A - 2A^t = 2(A - A^t) \neq 0,[/tex]

a contradiction. Where did I go wrong?

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# Homework Help: Minimal Polynomials

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