edited by
1,645 views

1 Answer

Best answer
3 votes
3 votes

It is not true that when u have distinct eigen values then u will have that much linearly independent vectors , take a 2*2 identity matrix , it has same eigen values but it has 2 different linearly independent vectors (1,0) and (0,1) .

How many linearly independent vectors we will have also depends on the rank of matrix , so if u have 2 distinct eigen values then there may be a case that u have repeated similar values which may be producing distinct linearly independent vectors ,therefore , u can only say that it will have atleast 2 distinct linearly independent vectors, option A is not always right .


Take A =

1 0 0

0 2 1

0 1 2
has eigenvalues 1 , 1 and 3.
The eigenvector for lambda=3 is [0 1 1] , and the repeated eigen value
lambda=1 has linearly independent eigen vectors [1 0 0] and [0 -1 1].

selected by

Related questions

2 votes
2 votes
0 answers
2
1 votes
1 votes
2 answers
4
Nishant Arora asked Dec 15, 2016
692 views
can we use some identity to find value of ab+by+ya???without finding its eigen values and then multiplying..earlier i have read it somewhere but i forget it.