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Below is a problem from an online class that I am taking (not for credit or anything so I'm not just asking someone to do my homework; it is just bugging me that I cannot figure it out). I am not sure if there is some kind of gap in my knowledge, but I do not know how I would approach a mapping defined by a matrix whose entries are Laurent polynomials. I feel like I might just be able to solve it as a system but I'm not really sure how proceed.

Find the kernel of the linear transformation $T: \mathbb{C}^3(t)\to \mathbb{C}^3(t)$ defined by the matrix:

$\begin{pmatrix} 0 & 1 & t^{-1}\\ 0 & t & 0\\ 0 & 1 & t^{-1}+t \end{pmatrix}$

EDIT (My seemingly incorrect attempt):

  • I am a little confused because the problem said that the mapping acts on the right—doesn't that not work with matrix multiplication? Or am I mixing up the definition of acting on the right? In any case, I did it below with what seems to be the the mapping acting. on the left (?)

Find $\begin{pmatrix}a\\b\\c\end{pmatrix}$ such that

$$\begin{pmatrix} 0 & 1 & t^{-1}\\ 0 & t & 0\\ 0 & 1 & t^{-1}+t \end{pmatrix}\begin{pmatrix}a\\b\\c\end{pmatrix}=\begin{pmatrix}1\\0\\0\end{pmatrix}$$ So $$\left(\begin{array}{c} b+\frac{c}{t}\\ b\,t\\ b+c\,\left(t+\frac{1}{t}\right) \end{array}\right) = \begin{pmatrix}1\\0\\0\end{pmatrix}$$ This is impossible, so the kernel is empty?

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    $\begingroup$ Welcome to Mathematics Stack Exchange! What have you tried so far? Where exactly did you get stuck? A good starting point would be to apply the definition of the kernel. $\endgroup$ Commented Jun 26, 2020 at 18:26
  • $\begingroup$ What is the vector space? $\endgroup$
    – Bernard
    Commented Jun 26, 2020 at 18:29
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    $\begingroup$ If the mapping acts on the right, then you are asking to find the set of $(a,b,c)$ which satisfies $$(a,b,c)\begin{pmatrix} 0 & 1 & t^{-1}\\ 0 & t & 0\\ 0 & 1 & t^{-1}+t \end{pmatrix} = (0,0,0)$$ $\endgroup$ Commented Jun 26, 2020 at 18:53
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    $\begingroup$ Where did your vector $\pmatrix{1\\0\\0}$ come from? It should be the zero vector. $\endgroup$
    – Berci
    Commented Jun 26, 2020 at 19:15
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    $\begingroup$ Well, if it indeed acts on the right, then you have to consider a row vector instead of a column vector, thus it will lead to different solution. Alternatively, you're about to find the kernel of the transpose matrix. $\endgroup$
    – Berci
    Commented Jun 26, 2020 at 20:18

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I'm assuming that $\ \mathbb{C}^3(t)\ $ is the vector space of ordered triples over the field of complex rational functions of the indeterminate $\ t\ $. Then $\ (a(t),b(t),c(t))\ $ is in the kernel of $\ T\ $ if and only if $$ \pmatrix{a(t),b(t),c(t)}\pmatrix{0&1&t^{-1}\\ 0&t&0\\ 0&1&t^{-1}+t}=(0,0,0)\\ \Leftrightarrow\left\{ \begin{matrix}a(t)+tb(t)+c(t)=0\\ \frac{a(t)}{t}+\frac{\left(1+t^2\right)c(t)}{t}=0\end{matrix}\right.\\ \Leftrightarrow\left\{ \begin{matrix}a(t)=-\left(1+t^2\right)c(t)\\ b(t)=tc(t)\end{matrix}\right. $$ Thus, the kernel of the transformation is $$ \left\{\left.f(t)\left(1+t^2,-t,-1\right)\,\right|f(t)\in\mathbb{C}^3(t)\right\}\ . $$

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