Find the matrix for linear transformation \(T : \mathbb{P}_n \to \mathbb{P}_n\), \[T f(x) =f(x) -2f'(x)+3f''(x)\]

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Find the matrix for linear transformation \(T : \mathbb{P}_n \to \mathbb{P}_n\), \[T f(x) =f(x) -2f'(x)+3f''(x)\]

Mathematics
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\(\mathbb{P}_n\) is the set of polynomials of degree not greater than \(n\)
\[f(x)=a_0=a_1x+a_2x^2+...a_nx^n=\\\begin{bmatrix} a_0 &a_1 &a_2 &... &a_n \end{bmatrix} \begin{bmatrix} 1\\ x\\ x^2\\ x^3\\ ...\\ x^n \end{bmatrix}\\f'=0+a_1 +2a_2x+3a_3x^2+...=\\\begin{bmatrix} a_0 &a_1 &a_2 &... &a_n \end{bmatrix} \begin{bmatrix} 0\\ 1\\ 2x\\ 3x^2\\ ...\\n x^n-1 \end{bmatrix}\]
\[f''=0+0+2a_2+6a_3x+12a_4x^2+...=\\\begin{bmatrix} a_0 &a_1 &a_2 &... &a_n \end{bmatrix} \begin{bmatrix} 0\\ 0\\ 2\\ 6x\\12x^2\\ ...\\n(n-1)x^{n-2} \end{bmatrix}\]

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almost solved!
how do we get the transformation matrix ?
doing \(f(x) -2f'(x)+3f''(x)\) gives a vector, not matrix right ?
for f\[\begin{bmatrix} 1\\ x\\ x^2\\ x^3\\ x^4\\ x^5\\ ... \end{bmatrix}\] for f'\[\begin{bmatrix} 0\\ 1\\ 2x\\ 3 x^2\\ ... \end{bmatrix}\] for f'' \[\begin{bmatrix} 0\\0\\2\\ 6x\\ 12 x^2\\ ... \end{bmatrix}\] find f-2f'+3f'' vector and multiply by \[\begin{bmatrix} a_0 &a_1 &a_2 &... &a_n \end{bmatrix}\] it gives you matrix
hmm \(f-2f'+3f'\)' gives \(n+1\) x \(1\) vector premultiplying \(\begin{bmatrix} a_0 &a_1 &a_2 &... &a_n \end{bmatrix}\) gives 1x1 matrix (scalar)
this the way to wirte matrix form \[\begin{bmatrix} 0\\ 1\\ 2x\\ 3x^2\\... \end{bmatrix}=\begin{bmatrix} 0 &0 &0 &0 &... \\ 1&0 &0 &0 &... \\ 0&2 &0 &0 &... \\ 0&0 &3 &0 &... \\ 0 &0 &0 &4 &... \end{bmatrix}\begin{bmatrix} 1\\ x\\ x^2\\ x^3\\ x^4\\ ... \end{bmatrix}\]
that was f' we do like this for f'' then you have sum of some matrix to find the transformation
I think that should work because now the vector has correct exponents ordering
'.'
matrix for \(Tf(x)=f(x)\) is \[\begin{bmatrix} 1&0&0&0&\cdots&0&0\\ 0&1&0&0&\cdots&0&0\\ 0&0&1&0&\cdots&0&0\\ 0&0&0&1&\cdots&0&0\\ \vdots&\vdots&\vdots&\vdots&\ddots&0&0\\ 0&0&0&0&\cdots&1&0\\ 0&0&0&0&\cdots&0&1\\ \end{bmatrix}\] matrix for \(Tf(x)=f'(x)\) is \[\begin{bmatrix} 0&1&0&0&\cdots&0&0\\ 0&0&2&0&\cdots&0&0\\ 0&0&0&3&\cdots&0&0\\ 0&0&0&0&\cdots&0&0\\ \vdots&\vdots&\vdots&\vdots&\ddots&n-1&0\\ 0&0&0&0&\cdots&0&n\\ 0&0&0&0&\cdots&0&0\\ \end{bmatrix}\] matrix for \(Tf(x)=f''(x)\) is \[\begin{bmatrix} 0&0&2\cdot 1&0&\cdots&0&0\\ 0&0&0&3\cdot 2 &\cdots&0&0\\ 0&0&0&0&\cdots&0&0\\ 0&0&0&0&\cdots&(n-1)(n-2)&0\\ \vdots&\vdots&\vdots&\vdots&\ddots&0&n(n-1)\\ 0&0&0&0&\cdots&0&0\\ 0&0&0&0&\cdots&0&0\\ \end{bmatrix}\]
i don't really see a more simpler way to express the final matrix hmm
for the overall transformation, I think il end up with matrix for \(Tf(x)=f(x)\color{blue}{-2\cdot}f'(x)+\color{red}{3\cdot }f''(x)\) is \[\begin{bmatrix} 1&\color{blue}{-2\cdot}1&\color{red}{3\cdot }2\cdot 1&0&\cdots&0&0\\ 0&1&\color{blue}{-2\cdot}2&\color{red}{3\cdot }3\cdot 2 &\cdots&0&0\\ 0&0&1&\color{blue}{-2\cdot}3&\cdots&0&0\\ 0&0&0&1&\cdots&\color{red}{3\cdot }(n-1)(n-2)&0\\ \vdots&\vdots&\vdots&\vdots&\ddots&\color{blue}{-2\cdot}(n-1)&\color{red}{3\cdot }n(n-1)\\ 0&0&0&0&\cdots&1&\color{blue}{-2\cdot}n\\ 0&0&0&0&\cdots&0&1\\ \end{bmatrix}\]
You can represent the entries of the derivative matrix using the Kronecker delta and the binomial coefficients by starting at the top left entry being indexed with 0,0 instead of 1,1 (which is more natural because \(x^0\) is also the first entry of the vector multiplying it, not \(x^1\) ): \[\Large D_{nm}^{(k)} = \delta_{n,(m-k)} \binom{n+k}{n} \] So for instance, k=0 corresponds to the identity matrix, \[\Large D_{nm}^{(0)} = \delta_{nm} \] Or let's look at this entry on the second derivative which represents the map from \(x^4\) to \(12 x^2\): \[\Large D_{24}^{(2)} = \delta_{22} \binom{4}{2} =12\] Cool, also I found this other matrix the other day too that might be related through the Taylor series since it represents shifting the function f(x) to f(x+a), that is if f(x) can be represented by the same vector you're using. Part of A for just \(f(x)= px^2+qx+r\) $$A= \begin{bmatrix} 1 & a & a^2 \\ 0 & 1 & 2a \\ 0 & 0 & 1 \end{bmatrix}$$ $$A_{ij}=\binom{i}{j}a^{j-i}$$ anywho I thought you might think this is interesting.
So using this matrix D I put we can write: \[Ty = y-2y'+3y''\]\[Ty=D^{(0)}y-2D^{(1)}y +3D^{(2)}y\]\[T=I-2D^{(1)}+3D^{(2)}\] So I guess that's one way to write T down, I don't know if that's quite what you like or not but I think it's a little better.
wow this has yo be some really advanced math
to*
That looks neat, pretty sure you meant \[\Large D_{nm}^{(k)} = \delta_{n,(m-k)} \binom{n+k}{n}\color{red}{k!}\]
\[Ty = y-2y'+3y''\]\[Ty=D^{(0)}y-2D^{(1)}y +3D^{(2)}y\]\[T=I-2D^{(1)}+3D^{(2)}\] This is indeed compact, was just wondering if we can do any operations with the kronecker delta symbol and maybe further simplify or something..
Ahhh yeah I somehow left that out haha, oh well you got it. As it stands, I don't think there's a whole lot more we can do. Each one of those D matrices is going to be its own separate band at least, so together it will form one fairly predictable banded matrix.
Something kind of interesting I'm seeing is, each of these D matrices are linearly independent of each other and all these D matrices form a vector space.

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