# 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

Staff Emeritus
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