TUTORIAL 8 MTH3201 LINEAR ALGEBRA

𝐴 =−1 38 4

1(a)

Eigen value = λ To find λ, λ I − A = 0

λ = ?

To find bases for eigen space

λ I − A 𝑥 = 0 𝑥 = ?

λ I − A = λ1 00 1

−−1 38 4

=λ + 1 −3−8 λ − 4

= λ + 1 λ − 4 − 24 = λ2 − 3λ −28

λ I − A = 0, λ2 − 3λ −28=0, λ = 7, −4

when λ = 7, when λ = −4,

λ I − A 𝑥 = 0 λ + 1 −3−8 λ − 4

𝑥1

𝑥2=

00

8 −3

−8 3

𝑥1

𝑥2=

00

8 −3−8 3

00

1 −3/80 0

00

𝑥1 −3

8𝑥2 = 0

𝐿𝑒𝑡 𝑥2 = 𝑡

𝑥1 =3

8𝑡 ∴ 𝑥 =

t

838

λ I − A 𝑥 = 0 λ + 1 −3−8 λ − 4

𝑥1

𝑥2=

00

−3 −3−8 −8

𝑥1

𝑥2=

00

−3 −3−8 −8

00

1 10 0

00

𝑥1 + 𝑥2 = 0

𝐿𝑒𝑡 𝑥2 = 𝑡 𝑥1 = −𝑡

∴ 𝑥 = 𝑡−11

𝐸𝑅𝑂 𝐸𝑅𝑂

∴ Hence,38

form a basis corresponding to λ = 7

−11

form a basis corresponding to λ = −4

1(b) λ = 1,6,7

bases =−15/16−1/2

1

,01

−3,

021

1(c) λ = 3,9, −6

bases =1

−11

,1

−21

,−221

1(e) λ = 2,2,3

bases =−210

,−201

,010

1(f) λ = 2, −2, −4

bases =−110

,013

,101

𝐴 =2 −64 −8

, λ I − A = λ1 00 1

−2 −64 −8

=λ − 2 6−4 λ + 8

= λ − 2 λ + 8 + 24 = λ2 + 6λ +8 λ I − A = 0, λ2 + 6λ +8=0, (λ+2)(λ+4)=0, λ = −2, −4

1(d)

when λ = −2, when λ = −4, −4 6−4 6

𝑥1

𝑥2=

00

−4 6−4 6

00

1 −3/20 0

00

𝑥1 −3

2𝑥2 = 0

𝐿𝑒𝑡 𝑥2 = 𝑡

𝑥1 =3

2𝑡 ∴ 𝑥 =

t

232

−6 6−4 4

𝑥1

𝑥2=

00

−6 6−4 4

00

1 −10 0

00

𝑥1 − 𝑥2 = 0

𝐿𝑒𝑡 𝑥2 = 𝑡 𝑥1 = 𝑡

∴ 𝑥 = 𝑡11

𝐸𝑅𝑂 𝐸𝑅𝑂

∴32

form a basis corresponding to λ = −2 ∴ 1

1 form a basis corresponding to λ = −4

2(a) 𝐴 =

1 −1 −11 2 −20 −1 0

det 𝜆𝐼 − 𝐴 =𝜆 − 1 1 1−1 𝜆 − 2 20 1 𝜆

= −1 2 𝜆 − 1 + 1 + 𝜆 𝜆 − 1 𝜆 − 2 + 1

𝜆𝐼 − 𝐴 = 𝜆3 − 3𝜆2 + 𝜆 + 1 = 𝜆 − 1 𝜆2 − 2𝜆 − 1

Characteristic equation of A: 𝜆𝐼 − 𝐴 = 0

𝜆 − 1 𝜆2 − 2𝜆 − 1 = 0 𝜆 = 1, 𝜆 = 1 ± 2

∴ Eigen values of 𝐴11 : 𝜆 = 111, 𝜆 = 1 ± 211

2(b)

∴ Eigen values of 𝐴11 :

𝜆 = 211, 𝜆 = 111𝜆 = −1 11

𝜆 = 2,1, −1

3(a) 𝐴 =

3 −1 11 0 1

−5 2 −3

det 𝜆𝐼 − 𝐴 =𝜆 − 3 1 −1−1 𝜆 −15 −2 𝜆 + 3

𝜆𝐼 − 𝐴 = 𝜆3 − 5𝜆 + 2 = 𝜆 − 2 𝜆2 + 2𝜆 − 1

Characteristic equation of A: 𝜆𝐼 − 𝐴 = 0

𝜆 − 2 𝜆2 + 2𝜆 − 1 = 0

𝜆 = 2, 𝜆 = −1 ± 2 ∴ λ ≠ 0, A is invertible

3(b)

𝜆 = 6, 𝜆 =−9 ± 57

2

∴ λ ≠ 0, A is invertible

det 𝜆𝐼 − 𝐴 =𝜆 − 5 −2 −6

0 𝜆 + 8 1−1 2 𝜆

4(a) 𝐴 =1 0 2

−3 0 −20 1 5

det 𝜆𝐼 − 𝐴 =𝜆 − 1 0 −2

3 𝜆 20 −1 𝜆 − 5

𝜆𝐼 − 𝐴 = 𝜆3 − 6𝜆2 + 7𝜆 + 4 = 𝜆 − 4 𝜆2 − 2𝜆 − 1

Characteristic equation of A: 𝜆𝐼 − 𝐴 = 0

𝜆 − 4 𝜆2 − 2𝜆 − 1 = 0 𝜆 = 4, 𝜆 = 1 ± 2

∴ det A = 4 1 + 2 1 − 2 = −4

∴ trace A = 4 + 1 + 2 + 1 − 2 = 6

4(b) 𝜆 = 2, 𝜆 = −2 ± 2 3

∴ det A = 2 −2 + 2 3 −2 − 2 3 = −16

∴ trace A = 2 + −2 + 2 3 + −2 − 2 3 = −2

4(c) 𝜆 = 0, 𝜆 = −1 ± 2

∴ det A = 0

∴ trace A = −2

4(d) 𝜆 = 1, 𝜆 = 6, 𝜆 = 6

∴ det A = 36

∴ trace A = 13

5(a)

10 6 22−1 −1 −2−4 −2 −9

000

ERO

𝐴 =−9 −6 −221 2 24 2 10

, det 𝜆𝐼 − 𝐴 =𝜆 + 9 6 22−1 𝜆 − 2 −2−4 −2 𝜆 − 10

𝜆𝐼 − 𝐴 = 𝜆3 − 3𝜆2 + 2𝜆 = 𝜆 𝜆 − 2 𝜆 − 1 𝜆 = 1,2,0

For 𝜆 = 1

1 0 5/20 1 −1/20 0 0

000

𝐿𝑒𝑡 𝑥3 = 𝑡

𝑥 = 𝑡/2−512

To find bases for eigen space

λ I − A 𝑥 = 0 𝑥 = ?

Basis for eigenspace corresponding to λ = 1 is 𝑝 1 =−512

11 6 22−1 0 −2−4 −2 −8

000

ERO

For 𝜆 = 2

1 0 20 1 00 0 0

000

𝐿𝑒𝑡 𝑥3 = 𝑡

𝑥 = 𝑡−201

To find bases for eigen space

λ I − A 𝑥 = 0 𝑥 = ?

Basis for eigenspace corresponding to λ = 2 is 𝑝 2 =−201

9 6 22

−1 −2 −2−4 −2 −10

000

ERO

For 𝜆 = 0

1 0 8/30 1 −1/30 0 0

000

𝐿𝑒𝑡 𝑥3 = 𝑡

𝑥 = 𝑡/3−813

Basis for eigenspace corresponding to λ = 0 is 𝑝 3 =−813

𝐶𝑜𝑚𝑏𝑖𝑛𝑒 𝑝 1, 𝑝 2 𝑎𝑛𝑑 𝑝 3

𝑃 =−5 −2 −81 0 12 1 3

𝑃−1𝐴 𝑃 = 𝐷 =

𝜆1 0 00 𝜆2 00 0 𝜆3

=1 0 00 2 00 0 0

5(b)

0 0 01 0 0

−4 3 −3 000

ERO

𝜆 = 2,2,5

For 𝜆 = 2

1 0 00 1 −10 0 0

000

𝐿𝑒𝑡 𝑥3 = 𝑡

𝑥 = 𝑡011

𝑝 1 =011

3 0 01 3 0

−4 3 0 000

ERO

For 𝜆 = 5

1 0 00 1 00 0 0

000

𝐿𝑒𝑡 𝑥3 = 𝑡

𝑥 = 𝑡001

𝑝 2 =001

∴ Since A is 3X3 matrix, there are only two basis vector. Thus, A is not diagonalizable

5(c) 𝜆 = 1, −1,2

5(d) 𝜆 = 1,4,9

𝑃 =0 2 −1

−1 −1 −11 −1 2

, 𝑃−1𝐴 𝑃 = 𝐷 =

𝜆1 0 00 𝜆2 00 0 𝜆3

=1 0 00 −1 00 0 2

𝑃 =−1 0 0−1 1 −11 0 1

, 𝑃−1𝐴 𝑃 = 𝐷 =

𝜆1 0 00 𝜆2 00 0 𝜆3

=1 0 00 4 00 0 9

6. 𝐴 =4 −31 0

,

𝑃−1𝐴 𝑃 = 𝐷 =1 00 3

, from here, we know that λ1 = 1 and λ2=3

−3 3−1 1

00

ERO

For 𝜆 = 1

𝐿𝑒𝑡 𝑥2 = 𝑡 𝑥 = 𝑡

11

𝑝 1 =11

𝜆𝐼 − 𝐴 =𝜆 − 4 3−1 𝜆

1 −10 0

00

−1 3−1 3

00

ERO

𝐿𝑒𝑡 𝑥2 = 𝑡 𝑥 = 𝑡

31

𝑝 2 =31

1 −30 0

00

For 𝜆 = 3

𝑃 =1 31 1

𝑃 𝐼 𝐼 𝑃−1

1 31 1

1 00 1

1 00 1

−1/2 3/21/2 −1/2

ERO

𝑃−1 =−1/2 3/21/2 −1/2

𝐷 = 𝑃−1𝐴 𝑃

𝐷𝑛 = 𝑃−1𝐴𝑃 𝑛

𝐷𝑛 = 𝑃−1𝐴𝑃 ∙ 𝑃−1𝐴𝑃 ∙ 𝑃−1𝐴𝑃 ∙ 𝑃−1𝐴𝑃… ∙ 𝑃−1𝐴𝑃

𝐷𝑛 = 𝑃−1𝐴𝑛𝑃

𝑃𝐷𝑛 = 𝑃𝑃−1𝐴𝑛𝑃

𝑃𝐷𝑛𝑃−1 = 𝑃𝑃−1𝐴𝑛

𝑃𝐷𝑛𝑃−1 = 𝐴𝑛

𝐴𝑛 = 𝑃𝐷𝑛𝑃−1

𝐴𝑛 =1 31 1

1 00 3

𝑛 −1/2 3/21/2 −1/2

𝐴𝑛 =1 3 ∙ 3𝑛

1 3𝑛

1

2−1 31 −1

𝐴𝑛 =1

2−1 + 3 ∙ 3𝑛 3 − 3 ∙ 3𝑛

−1 + 3𝑛 3 − 3𝑛

𝐴𝑛 =1

23 00 3

−4 −31 0

+1

2∙ 3𝑛 4 −3

1 0

𝐴𝑛 =1

23𝐼 − 𝐴 +

3𝑛

2𝐴 − 𝐼

𝐴𝑛 =3𝑛 − 1

2𝐴 +

3 − 3𝑛

2𝐼

∗ 𝑝𝑟𝑜𝑣𝑒𝑑

7(a) 𝐴 =

1 1 0−1 −2 −13 1 −2

λ = 0, −1, −2

𝑃 =1 1 −1

−1 −2 31 1 1

𝑃 𝐼 𝐼 𝑃−1

𝑃−1 =1 1 −1

−1 −2 31 1 1

ERO

𝐴𝑛 = 𝑃𝐷𝑛𝑃−1

𝐴10 = 𝑃𝐷10𝑃−1

=1 1 −1

−1 −2 31 1 1

010 0 00 −1 10 0

0 0 −2 10

1 1 −1−1 −2 31 1 1

𝐴10 =510 −1 −511

−1532 2 1534−514 −1 513

7(b) 𝐴 =

2 0 16 4 −32 0 3

λ = 4,4,1

𝑃 =1 0 −10 1 32 0 1

𝑃 𝐼 𝐼 𝑃−1

𝑃−1 =1/3 0 1/32 1 −1

−2/3 0 1/3

ERO

𝐴𝑛 = 𝑃𝐷𝑛𝑃−1

𝐴10 = 𝑃𝐷10𝑃−1

=1 0 −10 1 32 0 1

410 0 00 410 00 0 110

1/3 0 1/32 1 −1

−2/3 0 1/3

𝐴10 =349526 0 3495252097150 1048576 −1048575699050 0 699051

Use calculator

8(a) 𝐴 =1 −3

−3 9

λ = 0,10

−1 33 −9

00

ERO

For 𝜆 = 0

1 −30 0

00

9 33 1

00

ERO

1 1/30 0

00

For 𝜆 = 10

𝐿𝑒𝑡 𝑥2 = 𝑡

𝑥 = 𝑡31

𝑝 1 =31

𝐿𝑒𝑡 𝑥2 = 𝑡

𝑥 = 𝑡/3−13

𝑝 2 =−13

Step 1

8(a)

𝑝 1 =31

𝑝 2 =−13

Step 2 𝑞 𝑛 =

𝑣 𝑛𝑣 𝑛

Gram-Schmidt process

𝑞 1 =𝑣 1𝑣 1

=3,1

10

=3

10,

1

10

=

3

101

10

𝑞 2 =𝑣 2𝑣 2

=−1,3

10

=−1

10,

3

10

=

−1

103

10

𝑃 =

3

101

10

−1

103

10

𝐷 = 𝑃−1𝐴 𝑃 =0 00 10

Step 3