Discrete systems of equations227or−2.85078vr1−vr2=0vr1+0.35078vr2=0Letvr1=1 thenvr2= −2.8508.The second eigenvector is found from(A−0.5I)vs=0i.e.−0.35078−112.85078vs1vs2=00or−0.35078vs1−vs2=0vs1+2.85078vs2=0Letvs2=1 thenvs1= −2.8508. Hencevr=1−2.8508,vs=−2.85081One eigenvector represents the stable arm while the other represents the unstablearm. But which, then, represents the stable arm? To establish this, convert thesystem to its canonical form, withzt+1=V−1ut+1. Now take a point on the firsteigenvector, i.e., point (1,−2.8508), thenz0=V−1u0=1−2.8508−2.85081−11−2.8508=10Hence,z1t=2t(1)z2t=(−0.5)t(0)=0Thereforez1t→ +∞ast→ ∞. So the eigenvectorvrmust represent the unstablearm.Now take a point on the eigenvectorvs, i.e., the point (−2.8508,1), thenz0=V−1u0=1−2.8508−2.85081−1−2.85081=01Hence,z1t=2t(0)=0z2t=(−0.5)t(1)
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Trajectory, Trajectory of a projectile, Range of a projectile, Canonical form