Question

2. Given two arbitrary vectors |phi_1 angle and |phi_2 angle belonging to the inner product space mathcal{H}, the Cauchy-Schwartz inequality states that |langlephi_1|phi_2 angle|^2 leq langlephi_1|phi_1 anglelanglephi_2|phi_2 angle. (1) The purpose of this problem is to use the properties of inner product to prove this inequality. To proceed with the proof consider the vector |Psi angle defined as: |Psi angle = |phi_1 angle + lambda|phi_2 angle where lambda is a complex number that can be written as lambda = a + ib. (a) Write an expression for the inequality langlePsi|Psi angle geq 0 as a function of lambda i.e. f(lambda). Then, re-write this expression as a function of a and b i.e. f(a, b). [2] (b) Show that the value of lambda that minimises langlePsi|Psi angle is lambda_{min} = -frac{langlephi_2|phi_1 angle}{langlephi_2|phi_2 angle} (2) Hint:Compute the derivates of the function f(a, b) obtained in (a) with respect to a and b. Solve these equations to obtain a_{min} and b_{min} and then compute lambda_{min}. [2]

          2. Given two arbitrary vectors |phi_1
angle and |phi_2
angle belonging to the inner product space mathcal{H}, the Cauchy-Schwartz inequality states that
|langlephi_1|phi_2
angle|^2 leq langlephi_1|phi_1
anglelanglephi_2|phi_2
angle. (1)
The purpose of this problem is to use the properties of inner product to prove this inequality. To proceed with the proof consider the vector |Psi
angle defined as:
|Psi
angle = |phi_1
angle + lambda|phi_2
angle
where lambda is a complex number that can be written as lambda = a + ib.
(a) Write an expression for the inequality langlePsi|Psi
angle geq 0 as a function of lambda i.e. f(lambda). Then, re-write this expression as a function of a and b i.e. f(a, b). [2]
(b) Show that the value of lambda that minimises langlePsi|Psi
angle is
lambda_{min} = -frac{langlephi_2|phi_1
angle}{langlephi_2|phi_2
angle} (2)
Hint:Compute the derivates of the function f(a, b) obtained in (a) with respect to a and b. Solve these equations to obtain a_{min} and b_{min} and then compute lambda_{min}. [2]

$2. Given two arbitrary vectors |phi1 angle and |phi2 angle belonging to the inner product space mathcalH, the Cauchy-Schwartz inequality states that |langlephi1|phi2 angle|^2 leq langlephi1|phi1 anglelanglephi2|phi2 angle. (1) The purpose of this problem is to use the properties of inner product to prove this inequality. To proceed with the proof consider the vector |Psi angle defined as: |Psi angle = |phi1 angle + lambda|phi2 angle where lambda is a complex number that can be written as lambda = a + ib. (a) Write an expression for the inequality langlePsi|Psi angle geq 0 as a function of lambda i.e. f(lambda). Then, re-write this expression as a function of a and b i.e. f(a, b). [2] (b) Show that the value of lambda that minimises langlePsi|Psi angle is lambdamin = -fraclanglephi2|phi1 anglelanglephi2|phi2 angle (2) Hint:Compute the derivates of the function f(a, b) obtained in (a) with respect to a and b. Solve these equations to obtain amin and bmin and then compute lambdamin. [2]$

Added by Sergio I.

Question

Please give Ace some feedback