Is the Proof for a Complex Inner Product Space Correct?

[hsazerty2]

Summary:: Inner Product Spaces, Orthogonality.

Hi there,
This my first thread on this forum :)

I encountered the above problem in Schaum’s Outlines of Linear Algebra 6th Ed (2017, McGraw-Hill) Chapter 7 - Inner Product Spaces, Orthogonality.
Using some particular values for u and v, I proved that a and d must be real positive, and b is the conjugate of c. The solution indicates that a.d-b.c must also be positive, but i can't figure that out.

thanks for your help.

[Moderator's note: Moved from a technical forum and thus no template.]

 

[anuttarasammyak]

---------------
Definition (Inner product) Let V be a vector space over IR.
An inner product ( , ) is a function V × V → IR with the following
properties
1. ∀ u ∈ V , (u, u) ≥ 0, and (u, u) = 0 ⇔ u = 0;
2. ∀ u, v ∈ V , holds (u, v) = (v, u);
3. ∀ u, v, w ∈ V , and ∀ a, b ∈ IR holds
(au + bv, w) = a(u, w) + b(v, w).
------------
Your estimate meets above 1. How about 2. and 3. ?

EDIT
As pointed out in #3 for conjugate inner product
2. (u,v)=(v,u)*

 

[PeroK]

---------------
Definition (Inner product) Let V be a vector space over IR.
An inner product ( , ) is a function V × V → IR with the following
properties
1. ∀ u ∈ V , (u, u) ≥ 0, and (u, u) = 0 ⇔ u = 0;
2. ∀ u, v ∈ V , holds (u, v) = (v, u);
3. ∀ u, v, w ∈ V , and ∀ a, b ∈ IR holds
(au + bv, w) = a(u, w) + b(v, w).
------------
Your estimate meets above 1. How about 2. and 3. ?

We're dealing with a complex inner product here.

 

[wrobel]

first check the condition $(u,v)=\overline{(v,u)}$

Theorem. A function $f(u,v)=u^TA\overline v,\quad A=(\overline {A})^T,\quad u,v\in\mathbb{C}^n$ is positive definite iff the characteristic polynomial of the matrix $A$ has the form
$$\sum_{k=0}^n a_k\lambda^k,\quad a_k\in\mathbb{R},\quad a_k\ne 0,\quad\mathrm{sgn}\, a_{k}=-\mathrm{sgn}\,a_{k+1}$$

 

[SammyS]

Summary:: Inner Product Spaces, Orthogonality.

Hi there,
This my first thread on this forum :)

[Moderator's note: Moved from a technical forum and thus no template.]

Hello, @hsazerty2 !

 

[hsazerty2]

@wrobel, Thanks for the reply, but i need to find a solution that does not involve the notion of characteristic polynomial or eigenvalues.
Is the below proof right ?

if f(u,v) is a scalar product, then f(u,u) must be positive for all u in C2, which means that the above matrix A must be positive definite, and for that its determinant should be positive. and i already proved that the above matrix is hermitian. So all the conditions are : a and d real positive, b is the conjugate of c, and the determinant is positive. Conversely, if all the above conditions are satisfied, then the matrix is hermitian and definite positive, so f(u,v) is a scalar product.
Right ?