Given are linear transformation $T$, $U$:$R_2[x]\to R_2[x]$, where $T$ and $U$ are:

$(Tp)(x) = xp'(x) - p(0)$

$(Up)(x) = xp(1)$

Find $ker(U\circ T)$ and $ker(T\circ U)$

Just by doing the composition of those transformations I got: $ker(U\circ T) = (x^2+2,x+1)$ and $ker(T\circ U) = (x^2-1, x-1)$, but if I try by writing out the matrices for $U$ and $T$ and multiplying them, I don't get the same result. Can someone clarify please.


1 Answer 1


Note that, if $p(x)\in\mathbb R_2[x]$, then\begin{align}U\bigl(T\bigl(p(x)\bigr)\bigr)&=U\bigl(xp'(x)-p(0)\bigr)\\&=x\bigl(p'(1)-p(0)\bigr).\end{align}Therefore,\begin{align}\ker(U\circ T)&=\{p(x)\in\mathbb R_2[x]\mid p(0)=p'(1)\}\\&=\{a+2b+ax+bx^2\mid a,b\in\mathbb R\}\\&=(x+1,x^2+2)\end{align}(as you got).

Now, if I try to determine $\ker(U\circ T)$ using matrices, then I consider the matrices $M_T$ and $M_U$ of $T$ and $U$ respectively with respect to that basis $\{1,x,x^2\}$; these are:$$M_T=\begin{bmatrix}-1&0&0\\0&1&0\\0&0&2\end{bmatrix}\text{ and }M_U=\begin{bmatrix}0&0&0\\1&1&1\\0&0&0\end{bmatrix}.$$But$$M_U.M_T=\begin{bmatrix}0&0&0\\-1&1&2\\0&0&0\end{bmatrix}$$and$$\begin{bmatrix}0&0&0\\-1&1&2\\0&0&0\end{bmatrix}.\begin{bmatrix}a\\b\\c\end{bmatrix}=\begin{bmatrix}0\\-a+b+2c\\0\end{bmatrix}$$and therefore the kernel consists of those polynomials $a+bx+cx^2$ such that $a=b+2c$. But that's what I had got before.

Can you do the same thing with $T\circ U$?

  • $\begingroup$ Thanks a lot, I see where I made a mistake, insted of multiplying by a vector $ \begin{bmatrix}a\\b\\c\end{bmatrix} $ a multiplied by $ \begin{bmatrix}x^2\\x\\1\end{bmatrix}$. $\endgroup$ Dec 24, 2019 at 19:27
  • $\begingroup$ I'm glad I could help. $\endgroup$ Dec 24, 2019 at 20:01

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