Consider the differential operator $T:P3(R)→P3(R)$ given by
$T(p(x))=2p′(x)−2p′′(x)+2p′′′(x)$
Find an ordered basis $F$ for $P3(R)$ such that $T$ acts like a shift operator with respect to $F$, i.e.
$M^{F}_{F} (P)$ = \begin{pmatrix}0&0&0&0\\1&0&0&0\\0&1&0&0\\0&0&1&0\end{pmatrix}
Basis $F$ = {$ . . . $}
Enter your basis as polynomials separated by comma: e.g. $x2,2x+1,2$
So, I have a few similar questions like that and don't know how to solve them can someone help me with it.
On
Interpreting the columns of the desired matrix form as the images of the basis vectors, you need to find a linearly-independent set of four polynomials $p_k$ such that $T(p_1)=p_2$, $T(p_2)=p_3$, $T(p_3)=p_4$ and $T(p_4)=0$.
Clearly $p_4\in\ker(T)$, so you could start there. The kernel of $T$ consists of the constant polynomials, so you might as well choose $p_4=1$. Now you have to find a polynomial $p_3$ such that $T(p_3)=p_4$ and $p_3$ is not a constant polynomial. Observe that $T$ reduces the degree of a nonconstant polynomial by $1$, which means that $p_3=ax+b$ for some unknown coefficients $a$ and $b$. So, solve $T(ax+b)=1$ for $a$ and $b$ to get a $p_3$. Proceed in this same vein for the other two polynomials. At each step you’ll generate a polynomial of degree one greater than the previous one, so you’re assured that they’ll be linearly-independent. Note that the solutions to the equations might not be unique, so you’ll just need to pick whichever seems most convenient.
Since $T$ reduces the degree of the polynomial by one, there’s an easier way to proceed: pick a cubic polynomial $p$ and repeatedly apply $T$ to it until you get to zero. This sequence of polynomials will be your basis. The simplest choice is $x^3$, of course.
I assume $R$ is the field of real numbers, or at least a ring with unity, and $P^3(R)$ is the vectorspace of polynomials with coefficients in $R$ of degree atmost 3.
You can identify $P^3(R)$ with $R^4$ via $$p(x) = ax^3 + bx^2 +cx +d \mapsto \begin{pmatrix} a \\ b \\c \\ d \end{pmatrix}$$ Under this identification the standard basis of $R^4$ corresponds to the polynomials $x^3, x^2, x ,1$. Problems like this can usually be treated by considering an isomorphism of this kind and reducing it to linear algebra on $R^n$ for some $n$.
In the ordered basis $F = (x^3, x^2, x, 1)$ the map $p(x) \mapsto p'(x)$ can be identified with the matrix $$D = \begin{pmatrix} 0 & 0 & 0 & 0 \\ 3 & 0 & 0 & 0 \\ 0 & 2 & 0 & 0 \\ 0 & 0 & 1 & 0 \end{pmatrix}$$.
Your operator $T$ is equal to $$T = 2 D - 2D^2 + 2 D^3.$$ Now you just need to find a basis in which $T$ is of the desired form.
Can you handle it from here on?
Edit
Plugging in the numbers we get the matrix $$T = \begin{pmatrix} 0 & 0 & 0 & 0 \\ 6 & 0 & 0 & 0 \\ -12 & 4 & 0 & 0 \\ 12 & -4 & 2 & 0 \end{pmatrix}$$
Let $F=(f_1,f_2,f_3,f_4)$ be a basis such that $T$ is as requested. Given the form of $T$ in this basis we know that under $T$ $f_1$ gets mapped to $f_2$, $f_2$ gets mapped to $f_3$, $f_3$ to $f_4$ and $f_4$ itself is in the kernel of $T$ (i.e $Tf_4 = 0$). This translates to a set of equations:
We start by choosing $f_1 = (1,1,1,1)^T$, as there are no constraints on $f_1$ (except that it has to be linearly independent of the other basis elements of course). This yields $$f_2 = Tf_1 = \begin{pmatrix} 0 \\ 6 \\ -8 \\ 10\end{pmatrix}, ~~f_3 = Tf_2 = \begin{pmatrix} 0 \\ 0 \\ 24 \\ -40\end{pmatrix} ~~ \text{and} ~~f_4 = Tf_3 = \begin{pmatrix} 0 \\ 0 \\ 0 \\ 48\end{pmatrix}$$
Now we have to check whether the fourth condition holds. The kernel of $T$ is the linear span of the vector $(0, 0 , 0, 1)^T$, which means that $f_4$ has to be of the form $f_4 = (0, 0 , 0, \lambda)^T$ for some $\lambda \in R$. Clearly $f_4 \in \mathrm{ker}(T)$ and since $f_1$ up to $f_4$ are linearly independent we have found our basis. All that is left to do, is translating these vectors back to polynomials e.g. $f_1 = (1,1,1,1)^T$ corresponds to $x^3 + x^2 +x + 1$. Note that you can start with any polynomial of degree 3 you like.