## 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 $\mathsf{ker}\ \varphi$.

• 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.