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Given a linear transformation $T: \mathbb{R[Y]_{\leq2}} \rightarrow \mathbb{R[Y]_{\leq 2}}$ so that

$T(1+2Y) = 3 + 3Y + 2Y $

$T(1+Y) = 1 + Y + Y^2,$

$T(Y^2) = -2 -2Y - Y^2$

and i want to find kernel and image of T.

I know that kernel of $T$ are all vectors $\{v_i \in V | T(v_i) = 0\}$

Now this is confusing for me because i don't know anything about $T$ , how do i know which vector $T$ will map to zero?

I don't need full solution , only explaination about how do i know which vectors $T$ will map to zero?

$T(\begin{bmatrix}1 & 1 & 0 \\ 2 & 1& 0 \\ 0 & 0 & 1\end{bmatrix} = \begin{bmatrix}3 & 1& -2 \\ 3 & 1& -2 \\ 2 & 1& -1 \end{bmatrix}$

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Actually, you know everything about $T$.

In order to compute the kernel of $T$, solve the equation$$T\bigl(\alpha(1+2Y)+\beta(1+Y)+\gamma Y^2\bigr)=0.$$And the image of $T$ is the set$$\left\{\alpha(3+3Y+2Y^2)+\beta(1+Y+Y^2)+\gamma(-2-2Y-Y^2)\,\middle|\,\alpha,\beta,\gamma\in\mathbb R\right\}.$$Note that here I wrote $3+3Y+2Y^2$ instead of $3+3Y+2Y$; my guess is that you made a typo here.

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