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In all the part of this problem, consider a matrix A=[abcd] with the eigenvaluesλ1andλ2.

a. show thatλ12+λ22=a2+d2+2bc

b. show thatλ12+λ22≤a2+d2+c2+d2

c. for which matrices A Does the equalityλ12+λ22=a2+d2+c2+d2

Short Answer

Expert verified

The part of this problem, consider a matrix.

det(A-λl)=0a-λbcd-λ

a)=a2+d2+2bc

b)λ12+λ22=a2+d2+2bc≤a2+b2+c2+d2

c)b2+c2=2bc⇔b2=2bc+c2=0⇔(b-c)2=0⇔b=c

Step by step solution

01

definition of matrix

A function is defined as a relationship between a set of inputs that each have one output.

Given,

A=abcddet(A-λl)=0a-λbcd-λ

Multiplication:

a-λd-λ-bc=0λ2-a+dλ+ad-bc=0λ2=a+d±(a+d)2-4ad-bc2

a. Given,λ12+λ22=a2+d2+2bc

we compute,

λ12+λ22=2(a+d)2+2(a+d)2-8(ad-bc)4=(a+d)2-2(ad-bc)=a2+d2≥2bc

02

square the value

b. given, λ12+λ22≤a2+b2+c2+d2

apply the b-c2≥0,b2+c2-2bc≥0,

⇒b2+c2≥2bc

Now,

using part a, we get theλ12+λ22=a2+d2+2bc≤a2+b2+c2+d2

c.Given, λ12+λ22=a2+b2+c2+d2

The equality holds if and only

ifb2+c2=2bc⇔b2-2bc+c2=0⇔b-c2=0⇔b=c

hence,

a)=a2+b2+2bc

b)λ12+λ22=a2+d2+2bc≤a2+b2+c2+d2

c)b2+c2=2bc⇔b2-2bc+c2=0⇔b-c2=0⇔b=c

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