text: # Homework Help: Minimal Polynomials
  
  1. Oct 31, 2004
  
  ### AKG
  
  Compute the minimal polynomials for each of the following operators. Determine which of the following operators is diagonalizable.
  
  a) $T : P_2(\mathbb{C}) \to P_2(\mathbb{C})$, where:
  
  $$(Tf)(x) = -xf''(x) + (i + 1)f'(x) - 2if(x)$$.
  
  b) Let $V = M_{k \times k}(\mathbb{R})$.
  
  $$T : V \to V[/itex] by $T(A) = -2A^t$$. 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$, we have:
  
  $$0 = p(B)(A) = (2I + B)A = 2A + BA = 2A - 2A^t = 2(A - A^t) \neq 0,$$
  
  a contradiction. Where did I go wrong?
  
  2. Oct 31, 2004
  
  ### Hurkyl
  
  Staff Emeritus
  What is $P_2(\mathbb{C})$? I'm going to assume it means complex quadratic polynomials.
  
  Notice the domain and range of T: it takes a polynomial and spits out another polynomial. The parentheses are correct: you compute Tf first. Tf is a polynomial, so you can evaluate it at x: (Tf)(x).
  
  Your problem for (b) is that you got your identy matrices confused. In:
  
  g(t) = det(B - tI)
  
  I is the identity operator on V, a k^4 dimensional space. However, you treated it as if it was the identity matrix in V, a k^2 dimensional space, when you went to the next line.
  
  (BTW, did you remember to check that T was linear before you assumed it was a matrix?)
  
  I think diagonalizing first is a better approach for this one. I thought it was fairly easy to find k^2 linearly independent eigenvectors of T. Then, once you have the eigenvalues, you can write down the minimum polynomial.
  
  Last edited: Oct 31, 2004
  3. Oct 31, 2004
  
  ### AKG
  
  I think you mean k²-dimensional.
  I checked that T was linear, so there should be a matrix representation of it. You make it seem that the matrix should be a (k² x k²)-matrix. If B is this matrix, then:
  
  T(v) = Bv for all v in V. But this means that we're multiplying a (k² x k²)-matrix, B, by a (k x k)-matrix v, which isn't possible.
  
  Another thing: assuming all else was right, and I found g(t) correctly (with n = k² according to you), then the rest should still hold, and I should still get that contradiction.
  
  4. Oct 31, 2004
  
  ### AKG
  
  Basically, T is linear, so it should have a matrix representation, and if v is in V, then T(v) = Bv, so B must be an (k x k)-matrix. However, when we usually have operators over n dimensional spaces, their matrices are (n x n)-matrices, and since we have a k²-dimensional space here, we would expect a (k² x k²)-matrix. Now, is there any reason why the fact that we don't have such a matrix problematic? Do we really get any contradictions, or is it just unusual? As far as I can tell, it isn't really a problem, what do you think?
  
  5. Oct 31, 2004
  
  ### Hurkyl
  
  Staff Emeritus
  First thing you have to notice for (b):
  
  V, the vector space of all kxk matrices over R, is a k^2 dimensional vector space.
  
  T is a linear operator on V, so it does have a matrix representation, which you can get by selecting a basis for V...
  
  However, just like for any other vector space, you also have to write the elements of V in terms of the basis vectors -- so if you're writing T as a matrix, you have to write elements of V as k^2-tuples, and T would indeed be written as a k^2 x k^2 matrix.
  
  6. Oct 31, 2004
  
  ### AKG
  
  Oh, perfect. Thanks!
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