How can I find a basis for the span of some eigenvectors?

In summary, the individual is seeking a basis for the span of eigenvectors that are not currently known, without having to calculate them directly. The matrix in question is positive semi-definite with real and distinct eigenvalues, and has linearly independent eigenvectors. An example is provided to illustrate the question, and a potential solution involving finding the orthogonal complement is discussed. However, the individual notes that this approach may not work in all cases.
  • #1
ZachKaiser
2
0
Hello all. This is my first post here. Hope someone can help. Thank you guys in advance.

Here is the question:

I have a n-by-n matrix A, whose eigenvalues are all real, distinct. And the matrix is positive semi-definite. It has linearly independent eigenvectors V_1...V_n. Now I have known part of them, let's say V_1...V_m. How can I get a basis for span{V_(m+1)...V_n} without calculating V_(m+1)...V_n (because n may be large and calculating all the eigenvectors is unfeasible)?

To better illustrate the question, here is a working example. Let's say

A=[1 1 -1;
0 2 1;
0 0 3;]

whose eigenvalues and eigenvectors are:
lamda_1=1, V_1=[1 0 0]'
lamda_2=2, V_2=[1 1 0]'
lamda_3=3, V_3=[0 1 1]'

If I only know lamda_1 and V_1 now, how can I get a basis for span{V_2,V_3} without calculating V_2 and V_3?

Thanks again and I appreciate your help!


Zach
 
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  • #2
Assuming that you know that the matrix has three independent eigenvectors, two of them lie in the space orthogonal to the one you have. Here, your v1 is [1, 0, 0], then the "orthogonal complement" consists of all [x, y, z] such that [x, y, z][1, 0, 0]= x= 0. That is [0, y, z].
 
  • #3
Thank you HallsofIvy. But unfortunately, the eigenvectors are not necessarily orthogonal (orthogonal only for symmetric matrices). So your idea seems not correct.

In the example I gave earlier, if you just find a basis for the orthogonal complement of V1, e.g. [0 1 0] and [0 0 1], they are not the basis for span{V2,V3}. Simply because span{[0 1 0],[0 0 1]} is not an invariant subspace for A, the "orthogonal" property in one subspace is not preserved after you multiply it by A.

Thanks anyway
 

1. How do I determine if a set of vectors forms a basis for the span of eigenvectors?

To determine if a set of vectors forms a basis for the span of eigenvectors, you can check if the vectors are linearly independent and if they span the entire vector space. This can be done by using methods such as Gaussian elimination or calculating the determinant of the matrix formed by the vectors.

2. Can I find a basis for the span of eigenvectors if I have repeated eigenvalues?

Yes, it is still possible to find a basis for the span of eigenvectors even if there are repeated eigenvalues. In this case, you would need to find generalized eigenvectors, which are vectors that satisfy the equation (A-λI)^kx=0, where A is the original matrix and λ is the repeated eigenvalue.

3. What is the importance of finding a basis for the span of eigenvectors?

Finding a basis for the span of eigenvectors is important because it allows us to easily understand and manipulate the behavior of a matrix. Eigenvectors and their corresponding eigenvalues provide information about the transformations of a matrix, and having a basis for their span allows us to easily perform calculations and make predictions.

4. Is it possible to have more than one basis for the span of eigenvectors?

Yes, it is possible to have more than one basis for the span of eigenvectors. This is because there may be multiple sets of linearly independent eigenvectors that span the same vector space. However, the number of eigenvectors in each basis will be equal to the dimension of the vector space.

5. How can I efficiently find a basis for the span of eigenvectors?

One efficient way to find a basis for the span of eigenvectors is to use the diagonalization method. This involves finding the eigenvalues and eigenvectors of a matrix and then using them to create a diagonal matrix. The eigenvectors of the original matrix form a basis for the span of eigenvectors, and the diagonal matrix can be easily manipulated for calculations.

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