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This question already has an answer here:

Write down all the eigenvalues (along with their multiplicities) of the matrix A = (aij ) ∈ Mn(R) where aij = 1 for all 1 ≤ i, j ≤ n.

My attempt ; first i take A = (aij ) ∈ M2(R) where aij = 1 for all 1 ≤ i, j ≤ 2. here i got Rank A =1 and nullity =1. i got the two eigenvalue ie, λ=1 and λ= 0.....in this pattern i find A = (aij ) ∈ Mn(R) i got two eigenvalue

with λ = 0 with multiplicity n − 1 and λ = 1 with multiplicity 1.

Is my answer is correct or not,,pliz verified my mistakes...

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marked as duplicate by Omnomnomnom linear-algebra Sep 11 '17 at 18:37

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Since $Rank(A) = 1$, $\lambda = 0$ is en eigen value of multiplicity $n-1$

Let's call $\mu$ the remaining eigen value (of multiplicity 1).

You know that $$tr(A) = n=\sum_i \lambda _i=(n-1)*0+\mu$$

So in the end you have two eigen values : $n$ and $0$

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  • $\begingroup$ trace A will be 1 ..because if we A = (aij ) ∈ M3(R),,,then trace of A will be 1 by using row elementary operation,,@stity...im not getting why u take trace A = n... $\endgroup$ – lomber Sep 11 '17 at 17:11
  • $\begingroup$ @lomberlego the trace is the sum of the diagonal. You only have ones on your diagonal so $tr(A)=\sum_1^n 1= n$ $\endgroup$ – stity Sep 12 '17 at 9:31
  • $\begingroup$ thanks a lot @ sity $\endgroup$ – lomber Sep 12 '17 at 10:11

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