Find bases for the image and kernel of a given linear transformation

Let V \subseteq \mathbb{Q}[x] be the 6 dimensional vector space over \mathbb{Q} consisting of the polynomials having degree at most 5. Let \varphi : V \rightarrow V be the map given by \varphi(p) = x^2p^{\prime\prime} - 6xp^\prime + 12p, where p^\prime and p^{\prime\prime} denote the first and second derivative of p with respect to x. (See this previous exercise.)

  1. Prove that \varphi is a linear transformation.
  2. We showed previously that the set \{1,x,x^2,x^3,x^4,x^5\} is a basis of V. Find bases for the image and kernel of \varphi with respect to this basis.

We begin with a lemma.

Lemma: Let R be a commutative unital ring. Then D : R[x] \rightarrow R[x] given by D(p) = p^\prime is a module homomorphism. Proof: Let p(x), q(x) \in R[x] and let r \in R. Then (p+rq)^\prime(x) = \sum (i+1)(p_i+rq_i)x^i = \sum(i+1)p_ix^i + r\sum (i+1)q_ix^i = p^\prime(x) + rq^\prime(x). Thus D(p+rq) = D(p) + rD(q), and so D is a module homomorphism. \square

Thus it is clear that \varphi is a linear transformation.

Note the following:

  1. \varphi(1) = 12
  2. \varphi(x) = 6x
  3. \varphi(x^2) = 2x^2
  4. \varphi(x^3) = 0
  5. \varphi(x^4) = 0
  6. \varphi(x^5) = 2x^5

Thus we see that the matrix of \varphi is

A = \left[ \begin{array}{cccccc} 12 & 0 & 0 & 0 & 0 & 0 \\ 0 & 6 & 0 & 0 & 0 & 0 \\ 0 & 0 & 2 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 2 \end{array} \right].

The reduced row echelon form of A is the matrix

A^\prime = \left[ \begin{array}{cccccc} 1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \end{array} \right].

Since only the 1st, 2nd, 3rd, and 6th columns of A^\prime are pivotal, we see that the set \{12, 6x, 2x^2, 2x^5 \} forms a basis for \mathsf{im}\ \varphi. Next, the solutions of A^\prime X = 0 have the form X(x_4,x_5) = \left[ \begin{array}{c} 0 \\ 0 \\ 0 \\ x_4 \\ x_5 \\ 0 \end{array} \right]. Letting (x_4,x_5) \in \{(1,0),(0,1)\}, we see that \{(0,0,0,1,0,0),(0,0,0,0,1,0)\} is a basis for \mathsf{ker}\ \varphi.

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Comments

  • mcoulont  On October 7, 2011 at 11:38 am

    Are you sure of phi(x^5) ?

    • nbloomf  On October 10, 2011 at 8:30 am

      No; but it should be fixed now.

      Thanks!

      • deedeerr  On November 29, 2011 at 7:08 pm

        It’s not fixed yet. Still says phi(x^5) = 12x^5 instead of phi(x^5) = 2x^5

        • nbloomf  On December 3, 2011 at 12:42 am

          Alright- it should be fixed for real this time.

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