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Mathematics: Gate 2016 EE set-2
Shubhanshu
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Linear Algebra
May 19, 2017
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A 3X3 matrix P is such that, P^3 = P. Then the eigenvalues of P are:
1) 1, 1, -1
2) 1, 0.5 + j0.866, 0.5 - j0.866
3) 1, -0.5 + j0.866, -0.5 - j0.866
4) 0, 1, -1
gate2016-ee-2
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Shubhanshu
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May 19, 2017
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Best answer
Suppose $x$ is an eigenvector of $P$, and let $\lambda$ be corresponding eigenvalue, then
$$Px=\lambda x$$
Also
$$P^3x=\lambda^3 x$$.
We are given $P^3=P$, which means
$$P^3x=Px \Rightarrow \lambda^3 x = \lambda x \Rightarrow (\lambda^3-\lambda)x=0$$
Since $x$ is non-zero, $\lambda^3-\lambda = 0$. Solving it, we get $\lambda={0,1,-1}$
Happy Mittal
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May 24, 2017
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option 4??
joshi_nitish
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May 19, 2017
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Shubhanshu
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May 20, 2017
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Yes but how?
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joshi_nitish
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May 20, 2017
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every square matrix saisfies its own characterstic equation...therefore,
λ^3-λ=0
λ=0,1,-1..
but since i am deriving characterstic equation back from given matrix condition, i am not sure about it.
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then why not option 1 ???
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joshi_nitish
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@shubhanshu because roots of λ^3-λ=0 will be 0,1,-1...
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