Exercise 3: Predicting Long-Term Behavior From Eigenvalue Magnitudes — Possible Solution
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GIVEN
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L1 = 0.5 (eigenvector direction u1)
L2 = 1.5 (eigenvector direction u2)

WHICH DIRECTION DOMINATES
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Per this chapter's own repeated-application finding, applying a
transformation many times to a generic starting vector drives the
result increasingly toward the eigenvector belonging to the
LARGEST-MAGNITUDE eigenvalue. Comparing |L1| = 0.5 and |L2| = 1.5,
L2 is the larger of the two, so after many repeated applications the
result will end up dominated by the u2 direction - any starting
contribution along u1 becomes comparatively negligible.

WHY u1'S CONTRIBUTION FADES OUT
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Each repeated application multiplies the component of the vector
lying along u1 by another factor of L1 = 0.5. Repeatedly multiplying
by a number less than 1 in magnitude shrinks that component toward
zero the more times the transformation is applied - so the u1 part of
the vector effectively disappears over many iterations, even though it
never becomes exactly zero.

WHY u2'S CONTRIBUTION GROWS
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The component along u2 gets multiplied by L2 = 1.5 at every step.
Repeatedly multiplying by a number greater than 1 in magnitude grows
that component larger and larger without bound, exactly the behavior
this chapter's own worked example demonstrated with L = 3 (roughly
tripling in magnitude at each step).

OVERALL MAGNITUDE: GROWS WITHOUT BOUND
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Since the dominant eigenvalue's magnitude (1.5) is greater than 1,
the overall vector's magnitude grows without bound as the
transformation is applied more and more times - this is the unstable
case from this chapter's own stability-analysis application, not the
stable, shrinking-toward-zero case (which would require every
eigenvalue's magnitude to be below 1).

WHY THIS WORKS AS AN ANSWER
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The prediction is built directly from this chapter's own repeated-
application finding (dominance follows the largest-magnitude
eigenvalue), explains both the fading and the growing contributions
in terms of what repeated multiplication by a number below vs. above
1 actually does, and correctly classifies the overall system as
unstable using this chapter's own stability-analysis criterion rather
than guessing.