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Math 1b Practical — Eigenvalues and eigenvectors, III: Diagonalization
February 14, 2011
If we can express a vector
x
as a linear combination of eigenvectors of an
n
×
n
matrix
A
, then we may compute (±nd a formula for)
A
n
x
and so understand what happens when
n
is large. To this end, it is good thing when we can ±nd a basis for
R
n
consisting of
eigenvectors of
A
.
Important: Note that there are matrices for which no basis of eigenvectors exists. See
the examples below Theorem 3. (There are things one can do to understand
A
n
x
even
when
x
is not in the span of the eigenvectors of
A
, but we will not have time to get into
this in detail in this course.)
Theorem 1.
If
e
1
,
e
2
,...,
e
k
are eigenvectors and
λ
1
,λ
2
,...,λ
k
the corresponding eigen
values of a matrix
A
, and if the eigenvalues
λ
i
are distinct, then the eigenvectors
e
i
are
linearly independent.
Proof:
We proceed by induction on
k
. The assertion is true for
k
= 1 because we have
required that eigenvectors be nonzero vectors. Assume the assertion holds for
k
=
`
.W
e
will show it holds for
k
=
`
+1.
Suppose
c
1
e
1
+
c
2
e
2
+
...
+
c
`
e
`
+
c
`
+1
e
`
+1
=
0
.
(2)
Multiplying by
A
on the left, we get
c
1
A
e
1
+
c
2
A
e
2
+
+
c
`
A
e
`
+
c
`
+1
A
e
`
+1
=
0
,
or
c
1
λ
1
e
1
+
c
2
λ
2
e
2
+
+
c
`
λ
`
e
`
+
c
`
+1
λ
`
+1
e
`
+1
=
0
.
(3)
Multiply (2) by
λ
`
+1
and subtract it from (3) to get
c
1
(
λ
1
−
λ
`
+1
)
e
1
+
c
2
(
λ
2
−
λ
`
+1
)
e
2
+
+
c
`
(
λ
`
−
λ
`
+1
)
e
`
=
0
.
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 Winter '08
 Aschbacher
 Eigenvectors, Vectors

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