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Let V be the vector space of all polynomials of degree ≤ $k$.
Let $D:V→V$ be the linear transformation given by $p(x) → p′(x)$ (the derivative).
Determine the matrix of $D$ with respect to the basis $1, x, . . . , x^k$ and determine the rank and nullity of $D$.

I'm unsure of how to approach this problem. Should I construct a jacobian matrix? If the basis is $1,x,...,x^k$ then all components of the transformation matrix $D$ should be able to be written as $d$1,$d$x,...,$d$$x^k$ but how that would fit with the derivatives is unclear.

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  • $\begingroup$ Hint: We have $\dim V=k+1$ we can identify a polynomial $a_0+a_1x+\ldots +a_kx^k$ with $(a_0,a_1,\ldots,a_k)$. Now the polynomial $x^m$ is represented by $(0,...,0,1,0,...,0)$ where the $1$ stands at position $m+1$. $Dx^m=mx^{m-1}$ would give $(0,...,0,m,0,...0)$ where $m$ is at position $m$. Can you take it from here? $\endgroup$ Feb 16 '18 at 16:16
  • $\begingroup$ Ah, this is helpful. I'll attempt to work this out $\endgroup$ Feb 16 '18 at 16:21
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Your basis is $\{1,x,x^2,\ldots,x^k\}$. Furthermore,

  • $D(1)=0$;
  • $D(x)=1$;
  • $D(x^2)=2x$;
  • $\cdots$
  • $D(x^k)=kx^{k-1}$.

Therefore your matrix is$$\begin{bmatrix}0&1&0&0&\ldots&0\\0&0&2&0&\ldots&0\\0&0&0&3&\ldots&0\\\vdots&\vdots&\vdots&\vdots&\ddots&\vdots\\0&0&0&0&\cdots&k\\0&0&0&0&\ldots&0\end{bmatrix}.$$Can you do the rest?

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  • $\begingroup$ @Justtryingtodomath The rank is $k$, yes, but just saying that $D(1)=0$ is far from enough to justify it. If $D$ was such that $D(1)=0$ and that $D(x^k)=1$ when $k\geqslant1$, then $D(1)$ would still be $0$, but the rank would be $1$, not $k$. $\endgroup$ Feb 16 '18 at 16:58
  • $\begingroup$ I believe so, the rank must Be k+1 since every row is linearly independent $\endgroup$ Feb 16 '18 at 16:59
  • $\begingroup$ @Justtryingtodomath Like I wrote, the rank is $k$, not $k+1$. And the rows are not linearly independent. $\endgroup$ Feb 16 '18 at 17:00
  • $\begingroup$ But the only way to change one row into another in that matrix is to multiply by 0 right? $\endgroup$ Feb 16 '18 at 17:07
  • $\begingroup$ @Justtryingtodomath Look, it's just a matter of noticing that if you delete the first column and the last row, then you'll get a $k\times k$ matrix whose determinant is not $0$. Therefore, the rank is at least $k$. And since it can't be $k+1$… $\endgroup$ Feb 16 '18 at 17:09

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