Exercise 3.7.4: let a = 2 1 0 0 2 0 0 0 2 .

a. what are the eigenvalues?

b. what is/are the defect(s) of the eigenvalue(s)?

c. find the general solution of ~x 0 = a~x in two di erent ways and verify you get the same answer.

With

we have

so has one eigenvalue, , with multiplicity 3.

In order for to not be defective, we need the dimension of the eigenspace to match the multiplicity of the repeated eigenvalue 2. But has nullspace of dimension 2, since

That is, we can only obtain 2 eigenvectors,

and there is no other. We needed 3 in order to complete the basis of eigenvectors.

we know that

applying the law of sines

°

so

∠j=18°

∠k=29°

∠l=180-(18+29)=133°

the answer part a) is

the measure of angle k is  ≈29°

part b)  the measure of angle k could also be

(180-29)=151°

so

∠j=18°

∠k=151°

∠l=180-(18+151)=11°

the answer part b) is

the measure of angle k could also be 151°

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answer: 1000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000

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