different independent eigen vectors

Question

In A = (a_ij)_nxn where a_ij = 1    ∀ i,j then number of different independent Eigen Vectors of A are  _________ . 

(a)  1

(b)  n-1

(c)  2

(d)  n

Answer 1

Ans should be n.

As there will be (n-1) eigen value will be 0 and 1 value will be real .
for eigen value 0 there will be n-1 independent solution.
for real eigen value there will be 1 independent solution .

So total independent solution = (n-1)+1= n

Comments

yup answer will be "n" not 2

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Given question is asking for the number of different independent Eigen Vectors of A 

and you are telling the total independent solution not different independent solution.

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This should be the best solution

Answer 2

Answer will be 2

one eigen value is real no.

others are 0

So no. of different independent eigen vector also be 2 ( one for real no, other for 0)

because it will depend on eigen value

Comments

n×n matrix have n eigen values, out of which n-1 eigen values are 0, and one is real. 

No of eigen vectors? 

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updated

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I guess answer should be n instead of '2'.

try to solve for Eigen Value 0 - U will get n-1  constants in Eigen Vector.

What does this Vector signify ?  How many linealy Independent Eigen Vector it may give ?

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what should be ans - 2 or n  ??

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What is your answer?

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@Srestha, @Praveen Sir, Please tell me how-

 A[n×n] matrix where A[i][j]= 1, have n eigen values, out of which n-1 eigen values are 0, and one is real.  ??

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So waht is the answer 2 or N?

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@srestha I also agree on no. of the different independent eigenvector is 2.

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http://www.globalspec.com/reference/69712/203279/7-9-eigenvectors-for-repeated-eigenvalues

Follow this post it clearly says after finding out the eigen values if we have repeated eigen values then we put the repeated eigen value in the matrix, then find the rank. After that number of linearly independant vectors = order of matrix - rank of matrix(after putting value of the repeated eigen value under consideration). So in our case we get two eigen values, 0 and a non zero. For 0 the repetition is (n-1) so we put zero as eigen value in the matrix and we get rank of matrix 1 and so number of linearly independant vectors = n-1. For the non zero value we get rank as n-1, so number of linearly independant vectors= n-(n-1)=1.

SO TOTAL NUMBER OF DIFF LINEARLY INDEPENDANT VECTORS=N-1 + 1= N

PLZ CORRECT ME IF I AM WRONG

Answer 3

Answer here is n. Bcoz we get 2 eigen vales 0 and another is a real number. For real number one eigenvactor.

And for n-1 eigen values(0) place it in

A-lambda I = 0. We get rank of system as 1. Hence n-1 linearly independt solutions possible(eigenvectors).

Which gives us n eigen vectors.

D) is answer.