Page 419 - 35Linear Algebra

G.12 Diagonalization 419

Note that

~e 2 X .
~x = ~e 1
Also lets call

~e 2 V .
~v := ~e 1
Now the matrix M is the matrix of some linear transformation L in the basis
of the engineers. Lets convert it to the basis of the nutritionists:

s s s
~ ~ ~e 2 P ~e 1
L~x = L f 1 f 2 = L ~e 1 = MP .
f f ~e 2 f
Note here that the linear transformation on acts on vectors -- these are the
objects we have written with a~ sign on top of them. It does not act on columns
of numbers!
We can easily compute MP and find
1
1 1 − 2 0 1
MP = λ 1 = 3 .
2 −1 −1 − 5
λ λ
Note that P −1 MP is the matrix of L in the nutritionists basis, but we don’t
need this quantity right now.
Thus the last task is to solve the system, lets solve for sugar and fruit.
We need to solve

s 0 1 s 27
MP = 3 = .
f − 5 f 0
λ
This is solved immediately by forward substitution (the nutritionists basis
is nice since it directly gives f):
f = 27 and s = 45λ .

2 × 2 Example

Lets diagonalize the matrix M from a previous example

Eigenvalues and Eigenvectors: 2 × 2 Example

4 2
M =
1 3
We found the eigenvalues and eigenvectors of M, our solution was

2 1
λ 1 = 5, v 1 = and λ 2 = 2, v 2 = .
1 −1
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