# Eigen vectors

• Aug 7th 2008, 03:03 PM
JCIR
Eigen vectors
I need to know how to compute T with respect to the basis b. and i need a way to know where b is a basis consisting of egeinvectors of T.

Let v= p_1(R), T(a+bx)= (6a-6b)+(12a-11b)x and the basis B={3+4x,2+3x}

Let V= R^3, T =$\displaystyle \begin{pmatrix}\ a \\ b \\ c\end{pmatrix}\$ = $\displaystyle \begin{pmatrix}\ 3a + 2b - 2c\\ -4a-3b+2c\\-c\end{pmatrix}\$

The basis B= { $\displaystyle \begin{pmatrix}\ 0 \\ 1 \\ 1\end{pmatrix}\$ $\displaystyle \begin{pmatrix}\ 1 \\ -1 \\ 0\end{pmatrix}\$ $\displaystyle \begin{pmatrix}\ 1 \\ 0 \\ 2 \end{pmatrix}\$ }

general steps will defenitely help.
• Aug 10th 2008, 07:42 PM
elizsimca
To determine if a set is a basis for a subspace, you must show that the set meets two out of the three conditions:

1. S is linearly independent
2. S is a spanning set for V
3. S contains exactly k vectors, where k is the dimension of the space you are working in.

To show that the basis consists of the eigenvectors of T, you need to show that the matrix-vector product of the standard matrix of T and each vector in B is equal to the product of a scalar and the vector.

Although, I think it would follow directly from proving that the assumed eigenvectors in the set that you have correspond to distinct eigenvalues, because if an n x n matrix consists of n distinct eigenvalues, and you have separate, linearly independent vectors.

To determine if the vectors in the set are eigenvectors, you must show that T*v where T is the standard matrix of the linear operator and v is a vector in the basis is equal to some scalar times v, for each vector in the set. The scalars would be your eigenvalues.

Sorry, this may be jumbled...making sense at all?