Compute the matrix of a given linear transformation

Let V \subseteq \mathbb{Q}[x] be the set of all polynomials with rational coefficients having degree at most 5. Let \varphi : V \rightarrow V be the linear transformation given by \varphi(p) = p^\prime (the usual derivative of p with respect to x). Compute the matrix of \varphi with respect to the bases B = \{x^t \ |\ 0 \leq t \leq 5\} and E = \{\sum_{i=0}^t x^i \ |\ 0 \leq t \leq 5\}. That is, compute M^B_B(\varphi) and M^E_E(\varphi).


First we compute M^B_B(\varphi). Note the following:

  1. \varphi(1) = 0
  2. \varphi(x) = (1)1
  3. \varphi(x^2) = (2)x
  4. \varphi(x^3) = (3)x^2
  5. \varphi(x^4) = (4)x^3
  6. \varphi(x^5) = (5)x^4

Thus, we have

M^B_B(\varphi) = \begin{bmatrix} 0 & 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 2 & 0 & 0 & 0 \\ 0 & 0 & 0 & 3 & 0 & 0 \\ 0 & 0 & 0 & 0 & 4 & 0 \\ 0 & 0 & 0 & 0 & 0 & 5 \\ 0 & 0 & 0 & 0 & 0 & 0 \end{bmatrix}

Now we compute M^E_E(\varphi).

  1. \varphi(1) = 0
  2. \varphi(x+1) = 1(1)
  3. \varphi(x^2+x+1) = 2x+1 = 2(x+1) - 1(1)
  4. \varphi(x^3+x^2+x+1) = 3x^2+2x+1 = 3(x^2+x+1) - 1(x+1) - 1(1)
  5. \varphi(x^4+x^3+x^2+x+1) = 4x^3+3x^2+2x+1 = 4(x^3+x^2+x+1) - 1(x^2+x+1) - 1(x+1) - 1(1)
  6. \varphi(x^5+x^4+x^3+x^2+x+1) = 5x^4 + 4x^3 + 3x^2 + 2x + 1 = 5(x^4+x^3+x^2+x+1) - 1(x^3+x^2+x+1) - 1(x^2+x+1) - 1(x+1) - 1(1).

Thus we have

M^E_E(\varphi) = \begin{bmatrix} 0 & 1 & \text{-}1 & \text{-}1 & \text{-}1 & \text{-}1 \\ 0 & 0 & 2 & \text{-}1 & \text{-}1 & \text{-}1 \\ 0 & 0 & 0 & 3 & \text{-}1 & \text{-}1 \\ 0 & 0 & 0 & 0 & 4 & \text{-}1 \\ 0 & 0 & 0 & 0 & 0 & 5 \\ 0 & 0 & 0 & 0 & 0 & 0 \end{bmatrix}

Just for fun, we can also compute M^B_E(\varphi) and M^E_B(\varphi). Note first that

  1. \varphi(1) = 0
  2. \varphi(x) = 1(1)
  3. \varphi(x^2) = 2x = 2(x+1) - 2(1)
  4. \varphi(x^3) = 3x^2 = 3(x^2+x+1) - 3(x+1)
  5. \varphi(x^4) = 4x^3 = 4(x^3+x^2+x+1) - 4(x^2+x+1)
  6. \varphi(x^5) = 5x^4 = 5(x^4+x^3+x^2+x+1) - 5(x^3+x^2+x+1)

so that

M^E_B(\varphi) = \begin{bmatrix} 0 & 1 & \text{-}2 & 0 & 0 & 0 \\ 0 & 0 & 2 & \text{-}3 & 0 & 0 \\ 0 & 0 & 0 & 3 & \text{-}4 & 0 \\ 0 & 0 & 0 & 0 & 4 & \text{-}5 \\ 0 & 0 & 0 & 0 & 0 & 5 \\ 0 & 0 & 0 & 0 & 0 & 0 \end{bmatrix}.

Finally,

  1. \varphi(1) = 0
  2. \varphi(x+1) = 1 = 1(1)
  3. \varphi(x^2+x+1) = 2x+1 = 2(x) + 1(1)
  4. \varphi(x^3+x^2+x+1) = 3x^2+2x+1 = 3(x^2) + 2(x) + 1(1)
  5. \varphi(x^4+x^3+x^2+x+1) = 4x^3+3x^2+2x+1 = 4(x^3) + 3(x^2) + 2(x) + 1(1)
  6. \varphi(x^5+x^4+x^3+x^2+x+1) = 5x^4+4x^3+3x^2+2x+1 = 5(x^4) + 4(x^3) + 3(x^2) + 2(x) + 1(1)

Thus

M^B_E(\varphi) = \begin{bmatrix} 0 & 1 & 1 & 1 & 1 & 1 \\ 0 & 0 & 2 & 2 & 2 & 2 \\ 0 & 0 & 0 & 3 & 3 & 3 \\ 0 & 0 & 0 & 0 & 4 & 4 \\ 0 & 0 & 0 & 0 & 0 & 5 \\ 0 & 0 & 0 & 0 & 0 & 0 \end{bmatrix}.
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Comments

  • mcoulont  On October 5, 2011 at 4:22 am

    If I don’t mistake, a row is missing.

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