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272 CHAPTER 9 Eigenvalues, Diagonalization, and Special Matrices

⎛ ⎞
e 1
n n
⎜ ⎟
t
e 2 2
⎜ ⎟
E E = e 1 e 2 ··· e n ⎜ . ⎟ = e j e j = |e j | .
.
j=1 j=1
⎝ . ⎠
e n
Therefore the conclusion of Lemma 9.1 can be written
n n
i=1 j=1 a ij e i e j
λ = n .
|e j | 2
j=1
Proof of Lemma 9.1 Since AE = λE, then
t t
E AE = λE E,
yielding the conclusion of the lemma.
When we discuss diagonalization, we will need to know if the eigenvectors of a matrix
are linearly independent. The following theorem answers this question for the special case
that the n eigenvalues of A are distinct (the characteristic polynomial has no repeated
roots).

THEOREM 9.2
Suppose the n × n matrix A has n distinct eigenvalues. Then A has n linearly independent
eigenvectors.
To illustrate, in Example 9.4, A was 2×2 and had two distinct eigenvalues. The eigenvectors
produced for each eigenvalue were linearly independent.
Proof We will show by induction that any k distinct eigenvalues have associated with them k
linearly independent eigenvectors. For k = 1 there is nothing to show. Thus suppose k ≥ 2 and
the conclusion of the theorem is valid for any k − 1 distinct eigenvalues. This means that any
k − 1 distinct eigenvalues have associated with them k − 1 distinct eigenvectors. Suppose A has
k distinct eigenvalues λ 1 ,··· ,λ k with corresponding eigenvectors V 1 ,··· ,V k . We want to show
that these eigenvectors are linearly independent.
If they were linearly dependent, then there would be numbers c 1 ,··· ,c k , not all zero, such
that
c 1 V 1 + c 2 V 2 + ··· + c k V k = O.

By relabeling if necessary, we may assume for convenience that c 1 = 0. Multiply this equation
by λ 1 I n − A:
O =(λ 1 I n − A)(c 1 V 1 + c 2 V 2 + ··· + c k V k )

=c 1 (λ 1 I n − A)V 1 + c 2 (λ 1 I n − A)V 2
+ ··· + c k (λ 1 I n − A)V k
=c 1 (λ 1 V 1 − λ 1 V 1 ) + c 2 (λ 1 V 2 − λ 2 V 2 )
+ ··· + c k (λ 1 V k − λ k V k )
=c 2 (λ 1 − λ 2 )V 1 + ··· + c k (λ 1 − λ k )V k .
Now V 2 ,··· ,V k are linearly independent by the inductive hypothesis, so these coefficients are all
zero. But λ 1 = λ j for j =2,··· ,k by the assumptions that the eigenvalues are distinct. Therefore
c 2 = ··· = c k = 0.

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October 14, 2010 14:49 THM/NEIL Page-272 27410_09_ch09_p267-294