So V is the set of polynomials of degree 2 or less over the real numbers.

In order to represent a linear transformation as a matrix you have to first have a basis and the standard basis for this space is \(\displaystyle \{1, x, x^2\}\). Apply the linear transformation to each vector in the basis in turn and write the result in terms of the basis. The coefficients in that linear combination form one column of the basis.

Here, T(f(x))= xf'+ f(2)x+ f(3) so T(1)= x(0)+ 1x+ 1= 1+ x. Since that is \(\displaystyle 1(1+ 1(x)+ 0(x^2)\), the first column of the matrix is \(\displaystyle \begin{bmatrix}1 \\ 1 \\ 0\end{bmatrix}\).

\(\displaystyle T(x)= x(1)+ 2x+ 3= 3x+ 3= 3(1)+ 3(x)+ 0(x^2)\) and the second column of the matrix is \(\displaystyle \begin{bmatrix}3 \\ 3 \\ 0\end{bmatrix}\).

Finally, \(\displaystyle T(x^2)= x(2x)+ 4x+ 9= 9(1)+ 4(x)+ 2(x^2)\) so the third column of the matrix is \(\displaystyle \begin{bmatrix}9 \\ 4\\ 2\end{bmatrix}\).

The matrix representation of this linear transformation is \(\displaystyle \begin{bmatrix}1 & 3 & 9 \\ 1 & 3 & 4 \\ 0 & 0 & 2\end{bmatrix}\).

But you shouldn't **have to** write a linear transformation as a matrix to find eigenvalues- eigenvalue problems show up in many fields of mathematics, often in "function spaces" of infinite dimension where the transformations **cannot** be written as matrices.

Here, any "vector" is of the form \(\displaystyle f(x)= ax^2+ bx+ c\) and so \(\displaystyle f'= 2ax+ b\), f(2)= 4a+ 2b+ c, and f(3)= 9a+ 3b+ c.

\(\displaystyle T(f)= (2ax+ b)x+ (4a+ 2b+ c)x+ (9a+ 3b+ c)\)\(\displaystyle = 2ax^2+ (4a+ 3b+ c)x+ (9a+ 3b+ c)= \lambda f\)\(\displaystyle = a\lambda x^2+ b\lambda x+ c\lambda\)

so, comparing "like" coefficients, we must have \(\displaystyle a\lambda= 2a\), \(\displaystyle b\lambda= 4a+ 3b+ c\), and \(\displaystyle c\lambda= 9a+ 3b+ c\).

The first equation, \(\displaystyle a\lambda= 2a\) is the same as \(\displaystyle \lambda= 2\) so that either a= 0 or the eigenvalue is 2.

Set a= 0 to find other eigenvalues.