Page 246 - 35Linear Algebra

246 Diagonalization

and

1 2 1 2
0
0
(w , w ) = (e 1 + 2e 2 , 2e 1 + e 2 ) = (e 1 , e 1 ) ⇒ Q = .
2
1
2 1 2 1
Hence
 
1 0 1

1 1 −2 1 2 3 1 1 2
0
−1
M = Q MP = −  1 1 0  = .
3 −2 1 2 1 3 1 2 1
0 1 1
Notice that the change of basis matrices P and Q are both square and invertible.
Also, since we really wanted Q −1 , it is more efficient to try and write (e 1 , e 2 ) in
0
0
terms of (w , w ) which would yield directly Q −1 . Alternatively, one can check that
1 2
0
MP = QM .
13.3 Changing to a Basis of Eigenvectors
If we are changing to a basis of eigenvectors, then there are various simplifi-
cations:

• Since L : V → V , most likely you already know the matrix M of L
using the same input basis as output basis S = (u 1 , . . . , u n ) (say).

0
• In the new basis of eigenvectors S (v 1 , . . . , v n ), the matrix D of L is
diagonal because Lv i = λ i v i and so

 
λ 1 0 · · · 0
 0 0 
λ 2
 
L(v 1 ), L(v 2 ), . . . , L(v n ) = (v 1 , v 2 , . . . , v n )  . . .  .
 . . . . . . 
0 0 · · · λ n
0
• If P is the change of basis matrix from S to S , the diagonal matrix of
eigenvalues D and the original matrix are related by

D = P −1 MP

This motivates the following definition:

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