Kernel and image of linear operator - matrix representation

Question

Let $P_3(\mathbb{R})$ be a real vector space with the basis $V = (1,x,x^2)$.

Find the kernel and the image of the image of the linear operator: $$D: P_3(\mathbb{R}) \to P_3(\mathbb{R})$$ $$p(x) \to p'(x)$$

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With respect to the basis $V$ we know that: $$D(1) = 0 + 0x + 0x^2$$ $$ D(x) = 1 + 0x + 0x^2$$ $$ D(x^2) = 0 +2x + 0x^2$$

So the matrix representation would be: $$ _V[D]_V = \begin{pmatrix} 0 & 1 & 0 \\ 0 & 0 & 2 \\ 0 & 0 & 0 \end{pmatrix}$$

Thus we know that $x_1 =$ free variable, $x_2 = 0$ and $x_3 = 0$. The null space of the matrix representation is $\begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}$

The image of the matrix representation is $(\begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix},\begin{pmatrix} 0 \\ 2 \\ 0 \end{pmatrix})$

My question is, how do I find the kernel and the image of the linear operator, when I know the Null space and the image of the corresponding matrix representation?

Answer 1

The matrix is just a representation of the linear operator with respect to some basis. In this case you took the 'standard' basis $\beta=\{1,X,X^2\}$ of $\mathbb{R}[X]_{\leq2}$, so we have $\ker(L)=\mathrm{span}\{1\}$ and $\mathrm{Im}(L)=\mathrm{span}\{1,X\}$, by transforming the kernel and image of the matrix representation back to the polynomial vector space. Formally this is done by an inverse coordinate transformation.

Answer 2

If i have understood your query correctly then you can use the fact that $\mathcal{P}_3(\mathbf{F})$ and $\mathbf{F}^3$ are isomorphic to each other denote $\mathcal{P}_3(\mathbf{F})\cong\mathbf{F}^3$ then assuming that you know that $$\operatorname{null}[D]_V = \operatorname{span}\left(\begin{pmatrix}1\\0\\0\end{pmatrix}\right)\text{ }\text{ and }\text{ }\operatorname{range}[D]_V = \operatorname{span}\left(\begin{pmatrix}1\\0\\0\end{pmatrix},\begin{pmatrix}0\\2\\0\end{pmatrix}\right)$$

and the fact that $V = 1,x,x^2$ is the basis in question it follows that the corresponding spaces in $\mathcal{P_3}(\mathbf{F})$ are $\operatorname{null}D = \operatorname{span}(1)$ and $\operatorname{range}D = \operatorname{span}(1,2x)$.