Linear Transformations: Polynomial to its second derivatives
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HallsofIvy
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First, a linear transformation on a vector space can be written as a matrix in a given basis. What is the "given basis" for this problem? The "standard basis" for the vector space of polynomials of degree less than or equal to 5 is {1, x, x^2, x^3, x^4, x^5}. Is that what is intended?
Second, to get that matrix representation, you apply the linear transformation to each basis vector, in turn, and write the result as a linear combination of the basis vectors. That gives each column of the matrix. For example the second derivative of both 1 and x is 0 which is just 0 times each basis vector. The first two columns of the matrix are all 0s. The second derivative of x^2 is 2= 2(1)+ 0(x)+ 0(x^2)+ 0(x^3)+ 0(x^4)+ 0(x^5) so the third column is \(\displaystyle \begin{bmatrix}2 \\ 0 \\ 0 \\ 0 \\ 0 \\ 0\end{bmatrix}\). The second derivative of x^3 is 6x so the fourth column is \(\displaystyle \begin{bmatrix}0 \\ 6 \\ 0 \\ 0 \\ 0 \\ 0\end{bmatrix}\). The second derivative of x^4 is 12x^2 so the fifth column is \(\displaystyle \begin{bmatrix}0 \\ 0 \\ 12 \\ 0 \\ 0 \\ 0\end{bmatrix}\). The second derivative of x^5 is 20x^3 so the sixth column is \(\displaystyle \begin{bmatrix}0 \\ 0 \\ 0 \\ 20 \\ 0 \\ 0\end{bmatrix}\).
The matrix representation of this linear transformation, in the standard basis is \(\displaystyle \begin{bmatrix} 0 & 0 & 2 & 0 & 0 & 0 \\ 0 & 0 & 0 & 6 & 0 & 0 \\ 0 & 0 & 0 & 0 & 12 & 0 \\ 0 & 0 & 0 & 0 & 0 & 20 \\ 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \end{bmatrix}\).
Second, to get that matrix representation, you apply the linear transformation to each basis vector, in turn, and write the result as a linear combination of the basis vectors. That gives each column of the matrix. For example the second derivative of both 1 and x is 0 which is just 0 times each basis vector. The first two columns of the matrix are all 0s. The second derivative of x^2 is 2= 2(1)+ 0(x)+ 0(x^2)+ 0(x^3)+ 0(x^4)+ 0(x^5) so the third column is \(\displaystyle \begin{bmatrix}2 \\ 0 \\ 0 \\ 0 \\ 0 \\ 0\end{bmatrix}\). The second derivative of x^3 is 6x so the fourth column is \(\displaystyle \begin{bmatrix}0 \\ 6 \\ 0 \\ 0 \\ 0 \\ 0\end{bmatrix}\). The second derivative of x^4 is 12x^2 so the fifth column is \(\displaystyle \begin{bmatrix}0 \\ 0 \\ 12 \\ 0 \\ 0 \\ 0\end{bmatrix}\). The second derivative of x^5 is 20x^3 so the sixth column is \(\displaystyle \begin{bmatrix}0 \\ 0 \\ 0 \\ 20 \\ 0 \\ 0\end{bmatrix}\).
The matrix representation of this linear transformation, in the standard basis is \(\displaystyle \begin{bmatrix} 0 & 0 & 2 & 0 & 0 & 0 \\ 0 & 0 & 0 & 6 & 0 & 0 \\ 0 & 0 & 0 & 0 & 12 & 0 \\ 0 & 0 & 0 & 0 & 0 & 20 \\ 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \end{bmatrix}\).