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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: $$D: P_3(\mathbb{R}) \to P_3(\mathbb{R})$$ $$p(x) \to p'(x)$$


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?

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  • $\begingroup$ We do not need to find the expression of $D$ w.r.t. a basis, it is "clear" that the constant polynomials are mapped to zero - and only such polynomials, so the kernel is the subspace of constant polynomial, and the image are the polynomials of degree $\le 1$. For a connection to the question, yes, the specified systems generate indeed these spaces. $\endgroup$ – dan_fulea May 31 '18 at 15:46
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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.

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  • $\begingroup$ That makes sense, but isn't the $Im(L) = span\text{{1,2X}}$ by your method? $\endgroup$ – Simbörg May 31 '18 at 15:44
  • $\begingroup$ That is equivalent. By defintion, $\mathrm{span}\{1,2X\}$ consists of the vector space of all linear combinations of the two vectors $1$ and $2X$. This means that by performing elementary row operations on the two vectors, the vector space spanned by these vectors remains the same. The convention is to write the span as simple, or as straightforward as possible. Both my way to write the span and yours would be fine. $\endgroup$ – Václav Mordvinov May 31 '18 at 15:47
  • $\begingroup$ Thank you, it helped! $\endgroup$ – Simbörg May 31 '18 at 15:48
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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 $\operatorname{range}D = \operatorname{span}(1,2x)$.

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