Fundamentals of Matrix Algebra, 2011a

4.1 Eigenvalues and Eigenvectors
We solve (A − (1)I)⃗x = ⃗0 for ⃗x by row reducing the appropriate matrix:
[ −4 0
] 0
5 0 0
−→
rref
[ ] 1 0 0
.
0 0 0
We’ve seen a matrix like this before, 9 but we may need a bit of a refreshing. Our
first row tells us that x 1 = 0, and we see that no rows/equaons involve x 2 . We conclude
that x 2 is free. Therefore, our soluon, in vector form, is
⃗x = x 2
[ 0
1
]
.
We pick x 2 = 1, and find
⃗x 2 =
[ 0
1
]
.
To summarize, we have:
eigenvalue λ 1 = −3 with eigenvector ⃗x 1 =
[ ] −5
4
and
eigenvalue λ 2 = 1 with eigenvector ⃗x 2 =
[ 0
1
]
.
.
So far, our examples have involved 2×2 matrices. Let’s do an example with a 3×3
matrix.
. Example 88 .Find the eigenvalues of A, and for each eigenvalue, give one eigenvector,
where
⎡
⎤
−7 −2 10
A = ⎣ −3 2 3 ⎦ .
−6 −2 9
S We first compute the characterisc polynomial, set it equal to 0,
then solve for λ. A warning: this process is rather long. We’ll use cofactor expansion
along the first row; don’t get bogged down with the arithmec that comes from each
step; just try to get the basic idea of what was done from step to step.
9 See page 31. Our future need of knowing how to handle this situaon is foretold in footnote 5.
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