Fundamentals of Matrix Algebra, 2011a

Chapter 4
Eigenvalues and Eigenvectors
. Example 84 Find the eigenvalues of A, that is, find λ such that det (A − λI) = 0,
where
[ ] 1 4
A = .
2 3
S (Note that this is the matrix we used at the beginning of this sec-
on.) First, we write out what A − λI is:
[ ] [ ]
1 4 1 0
A − λI = − λ
2 3 0 1
[ ] [ ]
1 4 λ 0
= −
2 3 0 λ
[ ]
1 − λ 4
=
2 3 − λ
Therefore,
det (A − λI) =
∣ 1 − λ 4
2 3 − λ ∣
= (1 − λ)(3 − λ) − 8
= λ 2 − 4λ − 5
Since we want det (A − λI) = 0, we want λ 2 − 4λ − 5 = 0. This is a simple
quadrac equaon that is easy to factor:
λ 2 − 4λ − 5 = 0
(λ − 5)(λ + 1) = 0
λ = −1, 5
According to our above work, det (A − λI) = 0 when λ = −1, 5. Thus, the eigenvalues
of A are −1 and 5. .
Earlier, when looking at the same matrix as used in our example, we wondered if
we could find a vector ⃗x such that A⃗x = 3⃗x. According to this example, the answer is
“No.” With this matrix A, the only values of λ that work are −1 and 5.
Let’s restate the above in a different way: It is pointless to try to find ⃗x where
A⃗x = 3⃗x, for there is no such ⃗x. There are only 2 equaons of this form that have a
soluon, namely
A⃗x = −⃗x and A⃗x = 5⃗x.
As we introduced this secon, we gave a vector ⃗x such that A⃗x = 5⃗x. Is this the
only one? Let’s find out while calling our work an example; this will amount to finding
the eigenvectors of A that correspond to the eigenvector of 5.
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