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Chapter 5. Eigenvalues and Eigenvectors
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Some texts define
Let’s look at some examples of eigenvalues and eigenvectors. If
eigenvectors as we
have, except that 0 is a ∈ F, then aI has only one eigenvalue, namely, a, and every vector is
an eigenvector for this eigenvalue.
2
declared not to be an For a more complicated example, consider the operator T ∈L(F )
eigenvector. With the defined by
definition used here,
the set of eigenvectors 5.4 T(w, z) = (−z, w).
corresponding to a
fixed eigenvalue is a If F = R, then this operator has a nice geometric interpretation: T is
2
◦
subspace. just a counterclockwise rotation by 90 about the origin in R .An
operator has an eigenvalue if and only if there exists a nonzero vector
in its domain that gets sent by the operator to a scalar multiple of itself.
2
The rotation of a nonzero vector in R obviously never equals a scalar
multiple of itself. Conclusion: if F = R, the operator T defined by 5.4
has no eigenvalues. However, if F = C, the story changes. To find
eigenvalues of T, we must find the scalars λ such that
T(w, z) = λ(w, z)
has some solution other than w = z = 0. For T defined by 5.4, the
equation above is equivalent to the simultaneous equations

5.5 −z = λw, w = λz.

Substituting the value for w given by the second equation into the first
equation gives
2
−z = λ z.
Now z cannot equal 0 (otherwise 5.5 implies that w = 0; we are looking
for solutions to 5.5 where (w, z) is not the 0 vector), so the equation
above leads to the equation

2
−1 = λ .
The solutions to this equation are λ = i or λ =−i. You should be
able to verify easily that i and −i are eigenvalues of T. Indeed, the
eigenvectors corresponding to the eigenvalue i are the vectors of the
form (w, −wi), with w ∈ C, and the eigenvectors corresponding to the
eigenvalue −i are the vectors of the form (w, wi), with w ∈ C.
Now we show that nonzero eigenvectors corresponding to distinct
eigenvalues are linearly independent.