MHB Is this a valid proof for the invertibility of $\Phi$?

[rputra]

I am working on this problem: Show that $\Phi =I +\Delta^2 + \Delta^3$ is invertible, if $\Delta$ is a derivative linear mapping from $P_n$ to itself, $I$ is the identity mapping, where $P_n$ is vector space of one-variable polynomials with complex coefficients.

I would like to solve this problem by using the theorem that says if (a) the dimension of $\Phi$'s domain equals to the dimension of the range, and if (b) $\Phi$ is one-to-one, then $\Phi$ is invertible.

The (a) is obvious, $dim (P_n) = n+1$. For the (b), I would like to find out the kernel of $\Phi$, and the to show that $Ker (\Phi) = \{0\}$ therefore proving the $\Phi$ is indeed one-to-one. But unfortunately I do not know how to get the $Ker(\Phi)$. My understanding is that $\Delta^3$ and $\Delta^2$ are respectively the third second and the second derivative of a polynomial, does this mean I have to resort to differential equation?

Any hint or guidance would be very much appreciated. Thank you for your time and effort.

 

[Euge]

Hi Tarrant

You haven't actually proved (a), since you haven't proven that the image of $\Phi$ has dimension $n + 1$. Also, since $P_n$ is finite dimensional, it suffices to prove either (a) or (b). This is because a linear transformation on a finite dimensional vector space is one-to-one if and only if it is onto.

Since $\{1,z,z^2,\ldots, z^n\}$ is a basis for $P_n$, it suffices to consider the rank (or nullity) of the matrix $[\Phi]$ of $\Phi$ relative to this basis. Since $\Phi(z^m) = z^m + m(m - 1)z^{m-2} + m(m-1)(m-2)z^{m-3}$ for all $m \ge 3$, it follows that $[\Phi]$ is upper triangular with ones along the diagonal. Therefore, $[\Phi]$ has full rank $n + 1$, and $\Phi$ is invertible.

 

[rputra]

Hi Tarrant

You haven't actually proved (a), since you haven't proven that the image of $\Phi$ has dimension $n + 1$. Also, since $P_n$ is finite dimensional, it suffices to prove either (a) or (b). This is because a linear transformation on a finite dimensional vector space is one-to-one if and only if it is onto.

Since $\{1,z,z^2,\ldots, z^n\}$ is a basis for $P_n$, it suffices to consider the rank (or nullity) of the matrix $[\Phi]$ of $\Phi$ relative to this basis. Since $\Phi(z^m) = z^m + m(m - 1)z^{m-2} + m(m-1)(m-2)z^{m-3}$ for all $m \ge 3$, it follows that $[\Phi]$ is upper triangular with ones along the diagonal. Therefore, $[\Phi]$ has full rank $n + 1$, and $\Phi$ is invertible.

Thank you for your swift response. But unfortunately we are still in the young early weeks of this semester therefore we have not gone through the concepts of upper triangle, rank or nullity. The most we have covered are only bases, matrix associated with linear mapping, etc. Is there anyway you can modify your guidance along these lines?

Again thank you very much for your time and effort, am looking forward to hear from you again.

 

[Opalg]

Here's a more elementary way to prove (a), using the basis $\{1,z,z^2,\ldots,z^n\}$ of $P_n.$ First, can you see that $1$ must be in the range of $\Phi$? Then what about $z,\,z^2,\,\ldots$? Use induction if necessary to show that $z^k$ is in the range of $\Phi$ for all $k \leqslant n.$

 

[rputra]

I am working on this problem: Show that $\Phi =I +\Delta^2 + \Delta^3$ is invertible, if $\Delta$ is a derivative linear mapping from $P_n$ to itself, $I$ is the identity mapping, where $P_n$ is vector space of one-variable polynomials with complex coefficients.

I would like to solve this problem by using the theorem that says if (a) the dimension of $\Phi$'s domain equals to the dimension of the range, and if (b) $\Phi$ is one-to-one, then $\Phi$ is invertible.

The (a) is obvious, $dim (P_n) = n+1$. For the (b), I would like to find out the kernel of $\Phi$, and the to show that $Ker (\Phi) = \{0\}$ therefore proving the $\Phi$ is indeed one-to-one. But unfortunately I do not know how to get the $Ker(\Phi)$. My understanding is that $\Delta^3$ and $\Delta^2$ are respectively the third second and the second derivative of a polynomial, does this mean I have to resort to differential equation?

Any hint or guidance would be very much appreciated. Thank you for your time and effort.

The above is the original posting that started this thread. However, in hindsight, I think I misquoted the theorem. (For that I sincerely apologize!) The complete theorem from Serge Lange's textbook should be as follow:

Let $L: V \longrightarrow W$ be a linear map. Assume that $dim (V) = dim (W)$. If $Ker (L) = \{0\}$ or if $Im(L) = W$, then $L$ is bijective.

Here the theorem says that "if the dimension of the domain equals the dimension of the co-domain..." and not the dimension of range. I suspect this nuance makes lots of difference. Therefore here are my question: Can I still use this theorem to solve the problem? Or perhaps there is other simpler, easier-to-understand approach? Keep in mind that we are still in the early young period of this semester, therefore we have not gone through any sophisticated theorems yet.

Thank you again for your time and effort.

 

[rputra]

Problem: Show that $\Phi =I +\Delta^2 + \Delta^3$ is invertible, if $\Delta$ is a derivative linear mapping from $P_n$ to itself, $I$ is the identity mapping, where $P_n$ is vector space of one-variable polynomials with complex coefficients.

Do you think this following solution has merit? First of all, I would like to solve this problem with the theorem that I quoted earlier from Serge Lange's textbook:

Let $L: V \longrightarrow W$ be a linear map. Assume that $dim (V) = dim (W)$. If $Ker (L) = \{0\}$ or if $Im(L) = W$, then $L$ is bijective.

(a) It is obvious that the dimension of domain equals to the dimension of co-domain, as $dim(P_n) = (n+1) = dim(P_n).$
(b) Consider $\rho^n \in P_n,$ therefore
$$\begin{align}
\Delta (\rho^n) &= n\rho^{n-1}\\
\Delta^2 (\rho^n) &= n(n-1)\rho^{n-2}\\
\Delta^3 (\rho^n) &= n(n-1)(n-2)\rho^{n-3}\\
\end{align}$$
and
$$\Phi(\rho^n) = \rho^{n} + n(n-1)\rho^{n-2} + n(n-1)(n-2)\rho^{n-3}. $$

(c) In trying to find the $Ker (\Phi)$, I observe the followings:
*In order to make the last term zero, anyone of the $\rho^n, n, (n-1), (n-2)$ has to be zero.
*For the second term, anyone of the $\rho^n, n, (n-1)$ has to be zero.
*For the first term, $\rho^n$ has to be zero.
*In order to make all terms zero, $\rho^n$ has to be zero.
Hence, $Ker(\Phi) = \{0\}$, implying that $\Phi$ is one-to-one. Therefore $\Phi$ is invertible based on the theorem above.

Please let me know if the solution is acceptable. Thank you for your time and effort.