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Answer should be A. But they gave D. Their Explanation: Corresponding to each distinct eigen value, we have atleast one independent eigen vector. 

in Linear Algebra by Active (2.1k points)
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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].

by Loyal (6.3k points)
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How did you get two different eigen vector for  repeated $\lambda$=1 ?
One might get [0 1 -1 ] or [ 0 -1 1 ] for $\lambda $ = 1 and also for repeated  $\lambda $ = 1 . Correct me if I'm wrong

looks right to me...
Are you supporting what i said ? if not pls correct me @Arpit Tripathi
@pC supporting...
i also got same vectors which u mentioned above

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