## Description

this question deals with the “Laguerre Polynomials,” which are orthogonal with respect to the inner product hf, gi = Z ∞ 0 f(x)g(x) e −x dx. Use the fact that ∀ integers p, q ≥ 0 hx p , xq i = (p + q)! (“p plus q factorial”), to derive the first 4 (order 0, 1, 2, 3) orthonormal Laguerre Polynomials starting from elements of the standard polynomial basis {1, x, x2 , x3}.

Let T ∈ L(C 7 ) be defined by T(z1, z2, z3, z4, z5, z6, z7) = (πz1+z2+z3+z4, πz2+z3+z4, πz3+z4, πz4, √ 7z5+z6+z7, √ 7z6+z7, √ 7z7) Let Bs(C 7 ) = {e1, e2, e3, e4, e5, e6, e7} be the standard basis of C 7 (a) (25 pts.) Find M(T, Bs(C 7

(b) (25 pts.) Find the eigenvalues {λk}k=1,...,? (c) For each eigenvalue, λk: i. (30 pts.) Find the eigenspace E(λk, T) ii. (30 pts.) Find the generalized eigenspace G(λk, T)

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## Explanation & Answer

View attached explanation and answer. Let me know if you have any questions.Here are my answers to your assignment.Both documents are the same - one is in Ms. Word format, and the other in PDF - for your convenience.

∞

1. 〈𝑓, 𝑔〉 = ∫0 𝑓(𝑥)𝑔(𝑥)𝑒 −𝑥 𝑑𝑥 , ∀ 𝑖𝑛𝑒𝑡𝑒𝑔𝑟 𝑝, 𝑞 ≥ 0 〈𝑥 𝑝 , 𝑥 𝑞 〉 = (𝑝 + 𝑞)!

𝑠𝑡𝑎𝑛𝑑𝑎𝑟𝑑 𝑝𝑜𝑙𝑦𝑛𝑜𝑚𝑖𝑎𝑙 𝑏𝑎𝑠𝑖𝑠 {1, 𝑥, 𝑥 2 , 𝑥 3 }

𝐿𝑒𝑡 𝑢0 = 1

𝑢1 = 𝑥 −

〈𝑥, 1〉 ∙ 1

〈𝑥 1 , 𝑥 0 〉

(1 + 0)!

=

𝑥

−

=𝑥−

= 𝑥−1

|1|2

〈𝑥 0 , 𝑥 0 〉

(0 + 0)!

〈𝑥 2 , 1〉

〈𝑥 2 , 𝑥 − 1〉

𝑢2 = 𝑥 −

∙1−

∙ (𝑥 − 1)

|1|2

|𝑥 − 1|2

2

a. 〈𝑥 2 , 1〉 = (2 + 0)! = 2

b. |1|2 = 1

∞

∞

∞

c. 〈𝑥 2 , 𝑥 − 1〉 = ∫0 (𝑥 2 )(𝑥 − 1)𝑒 −𝑥 𝑑𝑥 = ∫0 𝑥 3 𝑒 −𝑥 𝑑𝑥 − ∫0 𝑥 2 𝑒 −𝑥 𝑑𝑥 = 6 − 2 = 4

∞

∞

d. |𝑥 − 1|2 = 〈𝑥 − 1, 𝑥 − 1〉 = ∫0 (𝑥 − 1)2 𝑒 −𝑥 𝑑𝑥 = ∫0 (𝑥 2 − 2𝑥 + 1)𝑒 −𝑥 𝑑𝑥

∞

∞

∞

= ∫0 𝑥 2 𝑒 −𝑥 𝑑𝑥 − 2 ∫0 𝑥𝑒 −𝑥 𝑑𝑥 + ∫0 𝑒 −𝑥 𝑑𝑥 = 2 − 2(1) + 1 = 1

2

4

𝑢2 = 𝑥 2 − ∙ 1 − ∙ (𝑥 − 1) = 𝑥 2 − 2 − 4𝑥 + 4 = 𝑥 2 − 4𝑥 + 2

1

1

𝑢3 = 𝑥 3 −

a.

b.

c.

d.

e.

〈𝑥 3 , 1〉

〈𝑥 3 , 𝑥 − 1〉

〈𝑥 3 , 𝑥 2 − 4𝑥 + 2〉

(𝑥

∙

1

−

∙

−

1)

−

∙ (𝑥 2 − 4𝑥 + 2)

|1|2

|𝑥 − 1|2

|𝑥 2 − 4𝑥 + 2|2

〈𝑥 3 , 1〉 = (3 + 0)! = 3! = 6

|1|2 = 1

∞

∞

∞

〈𝑥 3 , 𝑥 − 1〉 = ∫0 (𝑥 3 )(𝑥 − 1)𝑒 −𝑥 𝑑𝑥 = ∫0 𝑥 4 𝑒 −𝑥 𝑑𝑥 − ∫0 𝑥 3 𝑒 −𝑥 𝑑𝑥 = 24 − 6 = 18

|𝑥 − 1|2 = 1

∞

∞

∞

〈𝑥 3 , 𝑥 2 − 4𝑥 + 2〉 = ∫0 (𝑥 3 )(𝑥 2 − 4𝑥 + 2)𝑒 −𝑥 𝑑𝑥 = ∫0 𝑥 5 𝑒 −𝑥 𝑑𝑥 − 4 ∫0 𝑥 4 𝑒 −𝑥 𝑑𝑥 +

∞

2 ∫0 𝑥 3 𝑒 −𝑥 𝑑𝑥 = 120 − 4(24) + 2(6) = 120 − 96 + 12 = 36

f.

∞

∞

〈𝑥 2 − 4𝑥 + 2, 𝑥 2 − 4𝑥 + 2〉 = ∫0 (𝑥 2 − 4𝑥 + 2)(𝑥 2 − 4𝑥 + 2)𝑒 −𝑥 𝑑𝑥 = ∫0 𝑥 4 𝑒 −𝑥 𝑑𝑥 −

∞

∞

∞

∞

8 ∫0 𝑥 3 𝑒 −𝑥 𝑑𝑥 + 20 ∫0 𝑥 2 𝑒 −𝑥 𝑑𝑥 − 16 ∫0 𝑥𝑒 −𝑥 𝑑𝑥 + 4 ∫0 𝑒 −𝑥 𝑑𝑥 = 24 − 8(6) + 20(2) −

16(1) + 4(1) = 4

6

18

36

𝑢3 = 𝑥 3 − ∙ 1 −

∙ (𝑥 − 1) −

∙ (𝑥 2 − 4𝑥 + 2) = 𝑥 3 − 9𝑥 2 + 18𝑥 − 6

1

1

4

(−1)0

𝑢0 = (1)(1) = 1

0!

(−1)1

𝐿1 =

− 𝑢1 = (−1)(𝑥 − 1) = 1 − 𝑥

1!

→

(−1)2

1

𝑥2

𝐿2 =

𝑢2 = ( ) (𝑥 2 − 4𝑥 + 2) =

− 2𝑥 + 1

2!

2

2

(−1)3

1 3

𝑥 3 3𝑥 2

2

(𝑥

− 9𝑥 + 18𝑥 − 6) = − +

− 3𝑥 + 1

{ 𝐿3 = 3! 𝑢3 = − 6

6

2

𝐿0 =

∴ 𝐿 {1, 1 − 𝑥,

𝑥2

𝑥 3 3𝑥 2

− 2𝑥 + 1, − +

− 3𝑥 + 1}

2

6

2

2. Given that ℬ𝑠 (ℂ7 ) = {𝑒1 , 𝑒2 , 𝑒3 , 𝑒4 , 𝑒5 , 𝑒6 , 𝑒7 } is std basis of ℂ7

a. 𝐹𝑜𝑟 ℳ(𝑇, ℬ𝑠 (𝐶 7 )):

𝑇(𝑒1 ) = (4, 0, 0, 0, 0, 0, 0)

𝑇(𝑒2 ) = (1, 4, 0, 0, 0, 0, 0)

𝑇(𝑒3 ) = (1, 1, 4, 0, 0, 0, 0)

𝑇(𝑒4 ) = (1, 1, 1, 4, 0, 0, 0)

𝑇(𝑒5 ) = (0, 0, 0, 0, 3, 0, 0)

𝑇(𝑒6 ) = (0, 0, 0, 0, 1, 3, 0)

𝑇(𝑒7 ) = (0, 0, 0, 0,1, 1, 3)

The solution will be:

4

0

0

ℳ(𝑇) = 0

0

0

[0

1

4

0

0

0

0

0

1

1

4

0

0

0

0

1

1

1

4

0

0

0

0

0

0

0

3

0

0

0

0

0

0

1

3

0

b. {𝜆𝑘 }𝑘=1 = 4, 4, 4, 4, 3, 3, 3

0

0

0

0

1

1

3]

c. For each eigenvalue, 𝜆𝑘 :

i.

Find the eigenspace

For 𝜆 = 4, solution set of [(𝐴 − 4𝐼)𝑥 = 0]

So:

0

0

0

0

0

0

[0

1

0

0

0

0

0

0

𝑥1

1 0

0

0

0

𝑥

1 0

0

0

0

2

𝑥3

0

1 0

0

0

0 0

0

0 ∙ 𝑥4 = 0

𝑥5

0

0 −1 1

1

𝑥6

0

0 0 −1 1

−1] [𝑥7 ] [0]

0 0

0

1

1

0

0

0

0

0

𝑥3 = �...