Problem on rank and kernel of linear transformation over polynomial vector space

linear algebralinear-transformationsmatricesmatrix-rank

Let $V$ be a vector space of all complex polynomials $p$ with degree $p\leq n$. Let $T:V\rightarrow V$ be the map $T(p(x)) = p'(1)$ then I have to find the dimension of range and kernel of linear transformation $T$.

My approach Considering the usual basis of vector basis $V$ I computed matrix corresponding to linear transformation given below. Order of matrix coming out to be $(n+1)\times (n+1)$.
Its clear that rank of this matrix is $1$ and then dimension of $kernel$ is $n$.

My confusion Here vector space of complex polynomials are given however I have done this by taking usual basis of $V$ as 1, $x$, $x^2$.My approach is correct? Is there any other method to solve this question.

Thank you so much

\begin{pmatrix}0&1&2&~~n\\0&0&0&~~0\\0&0&0&~~0\\\vdots\\0&0&0&\cdots0\end{pmatrix}

Best Answer

$V$ is a vector space over $\color{red}{\mathbb{C}}$ and $\dim(V) =n+1$

$T\in\mathcal{L}(V) $ defined by $$T(p(x)) =p'(1) $$

Claim:

  1. $\textrm{range}(T) =\{a_0:a_0\in\Bbb{C}\}$

  2. $\dim(\textrm{null}(T)) =n$

  1. Take $a_0\in\Bbb{C}$ and then define $p(x) =a_0x$

  2. Use $\textrm{rank- nullity}$ theorem.

Alternative: $\mathcal{P}_n(\Bbb{C})=\{p(x)=a_0+a_1x+\cdots+a_nx^n :a_i\in\Bbb{C}\} $ is the set of all complex polynomials.

Then clearly $\{1, x, x^2, \cdots x^n\}$ is a basis.

One can easily find the matrix associated to the linear map.

Note: $x$ is merely a symbol.If you choose polynomial in $z$ then $\{1, z, z^2, \cdots, z^n\}$ will be a basis.If $\mathcal{P}_n(\Bbb{C})=\{p(z) =a_0+a_1z+\cdots+a_nz^n:a_i\in\Bbb{C}\}$

Then $$p\in\mathcal{P}_n(\Bbb{C})\implies p\in\textrm{span}\{1,z,z^2,\cdots,z^n\}$$ and $\{1,z,z^2,\cdots,z^n\}$ is also LI.

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