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

1. Aug 10, 2011

### ZachKaiser

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!

Sincerely,
Zach

2. Aug 11, 2011

### HallsofIvy

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. Aug 11, 2011

### ZachKaiser

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