I tried doing this with regular method and computations became very complex .Is there any shorter way , Am i missing something ?
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Hint: Since all columns (or alternatively all rows) all equal, the rank of the matrix is 1

Droplet
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Yes rank is 1 , so 0 must be an eigen value what about others ? – Sophie Clad Jan 11 '15 at 08:50
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http://math.stackexchange.com/questions/146927/relation-between-rank-and-number-of-non-zero-eigenvalues-of-a-matrix Maybe this question can help – Droplet Jan 11 '15 at 08:54
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So here M can have atmost 2 distinct eigen values , but how o che k whether it isone or two – Sophie Clad Jan 11 '15 at 08:58
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You can use the fact that the trace (here 4) is also the sum of the eigenvalues – Droplet Jan 11 '15 at 09:04
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1Three eigen values here are zero which equals nullity so other will be 4 .Is that right ? – Sophie Clad Jan 11 '15 at 09:07
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Yes, that's right. Congrats! – Droplet Jan 11 '15 at 09:12
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It can be done by using the characteristic equation:
Let us assume that the characteristic equaton of the matrix A is:
$$ \lambda^4 + a_1\lambda^3 + a_2\lambda^2 + a_3 \lambda + a_4 = 0$$
If you check then,
$$A^2 = 4 A$$
Using the fact that a matrix satisfies its own characteristic equation we get:
$$A^4 = 4 A^3 $$
Again writing it in terms of the Eigen values we get:
$$ \lambda^4 - 4 \lambda^3 = 0$$
Solve this and you will get two distinct Eigen values: 4 and 0.

FAM
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