Find the kernel of this linear transformation

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?

• 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. Commented Jun 26, 2020 at 18:26
• What is the vector space? Commented Jun 26, 2020 at 18:29
• If the mapping acts on the right, then you are asking to find the set of $(a,b,c)$ which satisfies $$(a,b,c)\begin{pmatrix} 0 & 1 & t^{-1}\\ 0 & t & 0\\ 0 & 1 & t^{-1}+t \end{pmatrix} = (0,0,0)$$ Commented Jun 26, 2020 at 18:53
• Where did your vector $\pmatrix{1\\0\\0}$ come from? It should be the zero vector. Commented Jun 26, 2020 at 19:15
• 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. Commented Jun 26, 2020 at 20:18

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 $$\left\{\left.f(t)\left(1+t^2,-t,-1\right)\,\right|f(t)\in\mathbb{C}^3(t)\right\}\ .$$