Prove Positive Definiteness On Tensor Product
Hey everyone! Today, we're diving deep into the fascinating world of tensor products and inner products, specifically focusing on proving the positive definiteness of the inner product defined on the tensor product of two complex inner product spaces. This might sound like a mouthful, but trust me, we'll break it down into manageable pieces and make it super clear.
Introduction to Tensor Products and Inner Products
Before we jump into the proof, let's quickly recap what tensor products and inner products are. Think of tensor products as a way to combine two vector spaces into a larger one. If you have two vectors, say v from a vector space V and w from a vector space W, their tensor product, denoted as v ⊗ w, lives in the tensor product space V ⊗ W. This new space is spanned by elements of this form, and it captures the essence of combining V and W in a specific way. In simpler terms, guys, it's like multiplying two vectors in a way that respects their individual vector space structures.
Now, inner products provide a way to measure the “angle” and “length” of vectors in a vector space. They're a generalization of the dot product you might have encountered in basic linear algebra. An inner product, denoted as ⟨⋅,⋅⟩, takes two vectors as input and returns a scalar (a complex number in our case). It satisfies certain properties like linearity, conjugate symmetry, and, most importantly for our discussion today, positive definiteness. The positive definiteness property ensures that the inner product of a vector with itself is always non-negative, and it's zero only if the vector is the zero vector. This property is crucial for defining notions like orthogonality and norms (lengths) in the vector space.
Setting the Stage: Defining the Inner Product on the Tensor Product
Okay, so we have two complex inner product spaces, (V, ⟨⋅,⋅⟩₁) and (W, ⟨⋅,⋅⟩₂) . Our goal is to define an inner product on their tensor product V ⊗ W and then prove that it's positive definite. The natural way to define this inner product, which we'll denote as ⟨⋅,⋅⟩, is by first defining it on elementary tensors (tensors of the form v ⊗ w) and then extending it linearly to the entire tensor product space. Specifically, for v₁, v₂ ∈ V and w₁, w₂ ∈ W, we define:
⟨v₁ ⊗ w₁, v₂ ⊗ w₂⟩ = ⟨v₁, v₂⟩₁ ⟨w₁, w₂⟩₂
In essence, the inner product of two elementary tensors is the product of their respective inner products in V and W. This definition makes intuitive sense because it respects the tensor product structure. To extend this definition to all of V ⊗ W, we use the fact that any tensor in V ⊗ W can be written as a linear combination of elementary tensors. So, if we have two tensors x = Σᵢ aᵢ(vᵢ ⊗ wᵢ) and y = Σⱼ bⱼ(uⱼ ⊗ zⱼ), where aᵢ and bⱼ are scalars, then the inner product is given by:
⟨x, y⟩ = ⟨Σᵢ aᵢ(vᵢ ⊗ wᵢ), Σⱼ bⱼ(uⱼ ⊗ zⱼ)⟩ = Σᵢ Σⱼ aᵢ b̄ⱼ ⟨vᵢ, uⱼ⟩₁ ⟨wᵢ, zⱼ⟩₂
Where b̄ⱼ denotes the complex conjugate of bⱼ. This formula might look a bit intimidating, but it's simply a consequence of the linearity and conjugate symmetry properties of inner products. Now that we have a concrete definition of the inner product on V ⊗ W, let's get to the heart of the matter: proving its positive definiteness.
The Proof: Positive Definiteness
Alright, let's get down to business and prove that the inner product we defined on V ⊗ W is indeed positive definite. To show this, we need to demonstrate two things:
⟨x, x⟩ ≥ 0for allx ∈ V ⊗ W⟨x, x⟩ = 0if and only ifx = 0
Where 0 represents the zero vector in V ⊗ W.
Part 1: Non-negativity
Let's start with the first part: showing that ⟨x, x⟩ ≥ 0 for all x ∈ V ⊗ W. Since any tensor x in V ⊗ W can be written as a linear combination of elementary tensors, let's express x as:
x = Σᵢ aᵢ(vᵢ ⊗ wᵢ)
Where aᵢ are scalars, vᵢ ∈ V, and wᵢ ∈ W. Now, let's compute ⟨x, x⟩ using the definition we established earlier:
⟨x, x⟩ = ⟨Σᵢ aᵢ(vᵢ ⊗ wᵢ), Σⱼ aⱼ(vⱼ ⊗ wⱼ)⟩ = Σᵢ Σⱼ aᵢ āⱼ ⟨vᵢ, vⱼ⟩₁ ⟨wᵢ, wⱼ⟩₂
This double sum might seem daunting, but we're going to massage it into a more manageable form. To do this, let's leverage the fact that V and W are inner product spaces. This means we can choose orthonormal bases for V and W. Let {e₁,..., eₘ} be an orthonormal basis for V and {f₁,..., fₙ} be an orthonormal basis for W. Then, we can express each vᵢ and wᵢ in terms of these bases:
vᵢ = Σₖ αᵢₖ eₖ
wᵢ = Σₗ βᵢₗ fₗ
Where αᵢₖ and βᵢₗ are scalars. Substituting these expressions into our formula for ⟨x, x⟩, we get:
⟨x, x⟩ = Σᵢ Σⱼ aᵢ āⱼ ⟨Σₖ αᵢₖ eₖ, Σₚ αⱼₚ eₚ⟩₁ ⟨Σₗ βᵢₗ fₗ, Σᵦ βⱼᵦ fᵦ⟩₂
Now, we can use the linearity of the inner product and the orthonormality of the bases to simplify this expression. Remember that for an orthonormal basis, ⟨eₖ, eₚ⟩₁ = δₖₚ (Kronecker delta, which is 1 if k = p and 0 otherwise) and ⟨fₗ, fᵦ⟩₂ = δₗᵦ. Therefore, our expression becomes:
⟨x, x⟩ = Σᵢ Σⱼ aᵢ āⱼ (Σₖ Σₚ αᵢₖ ᾱⱼₚ ⟨eₖ, eₚ⟩₁) (Σₗ Σᵦ βᵢₗ β̄ⱼᵦ ⟨fₗ, fᵦ⟩₂)
= Σᵢ Σⱼ aᵢ āⱼ (Σₖ αᵢₖ ᾱⱼₖ) (Σₗ βᵢₗ β̄ⱼₗ)
This is already looking much better! Now, let's make a clever observation. Define two matrices, A and B, with entries Aᵢₖ = aᵢ αᵢₖ and Bᵢₗ = aᵢ βᵢₗ. Then, we can rewrite the expression for ⟨x, x⟩ in terms of these matrices:
⟨x, x⟩ = Σₖ (Σᵢ Aᵢₖ) (Σⱼ Āⱼₖ) Σₗ (Σᵢ Bᵢₗ) (Σⱼ B̄ⱼₗ) = Σₖ |Σᵢ Aᵢₖ|² Σₗ |Σᵢ Bᵢₗ|²
Here, we've used the fact that z z̄ = |z|² for any complex number z. Now, we have a sum of squares, which is inherently non-negative:
⟨x, x⟩ = Σₖ |Σᵢ Aᵢₖ|² Σₗ |Σᵢ Bᵢₗ|² ≥ 0
Thus, we've successfully shown that ⟨x, x⟩ ≥ 0 for all x ∈ V ⊗ W. High five, guys!
Part 2: The Zero Vector Condition
Now, let's tackle the second part of the positive definiteness proof: showing that ⟨x, x⟩ = 0 if and only if x = 0. We've already shown that ⟨x, x⟩ ≥ 0, so we need to prove that ⟨x, x⟩ = 0 only when x is the zero vector and vice versa.
The “if” part is trivial: if x = 0, then ⟨x, x⟩ = ⟨0, 0⟩ = 0 by the properties of inner products. So, let's focus on the “only if” part: if ⟨x, x⟩ = 0, then x = 0.
From our previous derivation, we know that:
⟨x, x⟩ = Σₖ |Σᵢ Aᵢₖ|² Σₗ |Σᵢ Bᵢₗ|²
If ⟨x, x⟩ = 0, then this sum of squares must be zero. Since each term in the sum is non-negative, this implies that each term must be zero individually:
Σₖ |Σᵢ Aᵢₖ|² = 0 and Σₗ |Σᵢ Bᵢₗ|² = 0
This, in turn, implies that:
Σᵢ Aᵢₖ = 0 for all k and Σᵢ Bᵢₗ = 0 for all l
Remember that Aᵢₖ = aᵢ αᵢₖ and Bᵢₗ = aᵢ βᵢₗ. So, we have:
Σᵢ aᵢ αᵢₖ = 0 for all k and Σᵢ aᵢ βᵢₗ = 0 for all l
Recall that we expressed vᵢ and wᵢ in terms of the orthonormal bases {eₖ} and {fₗ}:
vᵢ = Σₖ αᵢₖ eₖ
wᵢ = Σₗ βᵢₗ fₗ
Now, consider the tensor x again:
x = Σᵢ aᵢ(vᵢ ⊗ wᵢ) = Σᵢ aᵢ(Σₖ αᵢₖ eₖ ⊗ Σₗ βᵢₗ fₗ)
We can rearrange this sum to get:
x = Σₖ Σₗ (Σᵢ aᵢ αᵢₖ βᵢₗ) (eₖ ⊗ fₗ)
But we know that Σᵢ aᵢ αᵢₖ = 0 and Σᵢ aᵢ βᵢₗ = 0, which means Σᵢ aᵢ αᵢₖ βᵢₗ = 0 for all k and l. Therefore:
x = Σₖ Σₗ (0) (eₖ ⊗ fₗ) = 0
Thus, we've shown that if ⟨x, x⟩ = 0, then x = 0. This completes the proof of positive definiteness! Give yourselves a pat on the back, guys.
Conclusion
In this deep dive, we've successfully proven that the inner product defined on the tensor product of two complex inner product spaces is indeed positive definite. We started by defining the inner product on elementary tensors and extending it linearly. Then, we used the properties of inner products and orthonormal bases to show that ⟨x, x⟩ ≥ 0 for all tensors x. Finally, we demonstrated that ⟨x, x⟩ = 0 if and only if x = 0, completing the positive definiteness proof.
This result is not just an abstract mathematical curiosity; it has significant implications in various areas of mathematics and physics, including quantum mechanics and functional analysis. Understanding the properties of tensor products and inner products is crucial for anyone working in these fields. Keep exploring, guys, and you'll uncover even more fascinating connections!
