Page 420 - 35Linear Algebra

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So we can diagonalize this matrix using the formula D = P −1 MP where P =
(v 1 , v 2 ). This means

2 1 −1 1 −1 −1
P = and P = −
1 −1 3 −1 2
The inverse comes from the formula for inverses of 2 × 2 matrices:
−1

a b 1 d −b
= , so long as ad − bc 6= 0.
c d ad − bc −c a
So we get:

1 −1 −1 4 2 2 1 5 0
D = − =
3 −1 2 1 3 1 −1 0 2
But this does not really give any intuition into why this happens. Let look

x
at what happens when we apply this matrix D = P −1 MP to a vector v = .
y

x
Notice that applying P translates v = into xv 1 + yv 2 .
y

x 2x + y
P −1 MP = P −1 M
y x − y

2x y
−1
= P M +
x −y

2 1
−1
= P x M + y M
1 −1
= P −1 [x Mv 1 + y Mv 2 ]
Remember that we know what M does to v 1 and v 2 , so we get
P −1 [x Mv 1 + y Mv 2 ] = P −1 [xλ 1 v 1 + yλ 2 v 2 ]
= 5x P −1 v 1 + 2y P −1 v 2

1 0
= 5x + 2y
0 1

5x
=
2y

1 0
Notice that multiplying by P −1 converts v 1 and v 2 back in to and
0 1
respectively. This shows us why D = P −1 MP should be the diagonal matrix:

λ 1 0 5 0
D = =
0 λ 2 0 2

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