# 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. [duplicate]

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...

## marked as duplicate by Omnomnomnom linear-algebra StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Sep 11 '17 at 18:37

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 $0$
• @lomberlego the trace is the sum of the diagonal. You only have ones on your diagonal so $tr(A)=\sum_1^n 1= n$ – stity Sep 12 '17 at 9:31