# Apparently, matrices of linear transformation are my weakness

• May 13th 2010, 01:22 AM
kaelbu
Apparently, matrices of linear transformation are my weakness
I have another eigenvector problem, where the transformation is given by T(f(x))= xf'(x) + f(2)x + f(3) . I haven't slightest clue how to represent this transformation with a matrix, and as such am completely stumped.
Thanks!
• May 13th 2010, 04:54 AM
HallsofIvy
Quote:

Originally Posted by kaelbu
I have another eigenvector problem, where the transformation is given by T(f(x))= xf'(x) + f(2)x + f(3) . I haven't slightest clue how to represent this transformation with a matrix, and as such am completely stumped.
Thanks!

You haven't said what the vector space is- what kinds of functions f(x) can be. The most general would be "the space of all differentiable functions" but that is an infinite dimensional vector space and T cannot be represented by a matrix.

It is impossible to represent T as a matrix until we know what the vector space is! Please state the entire problem.
• May 13th 2010, 07:08 AM
kaelbu
Oh! Sorry! V= P2(R)
• May 14th 2010, 01:49 AM
HallsofIvy
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.