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Chapter 14: Problem 22

Show that the tensor product \(A \otimes B\) is bilinear in both \(A\) and \(B\).

Short Answer

Expert verified
In summary, to show that the tensor product \(A \otimes B\) is bilinear, we demonstrated linearity in both the first argument (A) and the second argument (B). We calculated and compared the left-hand side and right-hand side of equations for each case, and the results confirmed that the tensor product satisfies the properties of bilinearity.

Step by step solution

01

Linearity in the first argument

To check if the tensor product is linear in the first argument, we need to show that \(A \otimes (B_1 + B_2) = A \otimes B_1 + A \otimes B_2\). Let's take the components of A, B1, and B2, which are \(a_{ij}\), \(b^1_{kl}\), and \(b^2_{kl}\) respectively. The tensor product is defined as: \((A \otimes B)_{ijkl} = a_{ij}b_{kl}\). Now, let's calculate the left-hand side and right-hand side of the equation. Left-hand side (LHS) of the equation: \((A \otimes (B_1 + B_2))_{ijkl} = a_{ij}(b^1_{kl} + b^2_{kl})\). Right-hand side (RHS) of the equation: \((A \otimes B_1)_{ijkl} = a_{ij}b^1_{kl}\), \((A \otimes B_2)_{ijkl} = a_{ij}b^2_{kl}\), So, \((A \otimes B_1 + A \otimes B_2)_{ijkl} = a_{ij}b^1_{kl} + a_{ij}b^2_{kl}\). Comparing LHS and RHS, we see that \(a_{ij}(b^1_{kl} + b^2_{kl}) = a_{ij}b^1_{kl} + a_{ij}b^2_{kl}\), which implies that the tensor product is linear in the first argument.
02

Linearity in the second argument

To show that the tensor product is linear in the second argument, we need to demonstrate that \((A_1 + A_2) \otimes B = A_1 \otimes B + A_2 \otimes B\). Let's take the components of A1, A2, and B, which are \(a^1_{ij}\), \(a^2_{ij}\), and \(b_{kl}\) respectively. The tensor product is defined as: \((A \otimes B)_{ijkl} = a_{ij}b_{kl}\). Now, let's calculate the left-hand side and right-hand side of the equation. Left-hand side (LHS) of the equation: \(((A_1 + A_2) \otimes B)_{ijkl} = (a^1_{ij} + a^2_{ij})b_{kl}\). Right-hand side (RHS) of the equation: \((A_1 \otimes B)_{ijkl} = a^1_{ij}b_{kl}\), \((A_2 \otimes B)_{ijkl} = a^2_{ij}b_{kl}\), So, \((A_1 \otimes B + A_2 \otimes B)_{ijkl} = a^1_{ij}b_{kl} + a^2_{ij}b_{kl}\). Comparing LHS and RHS, we can deduce that \((a^1_{ij} + a^2_{ij})b_{kl} = a^1_{ij}b_{kl} + a^2_{ij}b_{kl}\), which means that the tensor product is linear in the second argument. In conclusion, since the tensor product \(A \otimes B\) is linear in both the first argument and the second argument, it is bilinear.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Linear Algebra
Linear algebra is a fundamental field within mathematics, dealing with vector spaces, linear mappings between these spaces, matrices, and the algebraic structures that emerge from these concepts. One of the strengths of linear algebra is its ability to facilitate the solution of systems of linear equations, provide ways to operate on mathematical objects, and its broad application across sciences including physics, engineering, computer science, and more. At its heart, linear algebra involves objects that behave 'linearly,' meaning they adhere to the properties of additivity and scalar multiplication.
Tensor Product Properties
The tensor product, denoted by \( \otimes \), is a way of combining two mathematical objects (such as vectors, matrices, or even more abstract entities) into a new, higher-dimensional object, which encapsulates information about both original objects.
One of the key attributes of tensor products is their ability to adhere to bilinearity. Bilinearity refers to the property where a product is linear in both arguments separately. This means that for two vectors \( u \) and \( v \), the operation \( u \otimes v \) will satisfy the bilinearity condition if it is linear in \( u \) when \( v \) is held constant and linear in \( v \) when \( u \) is held constant.
In the context of matrices, this attribute of the tensor product allows for the combination of different matrices in ways that are consistent with the principles of linear algebra. The tensor product can distribute across matrix addition, and is compatible with scalar multiplication, two fundamental aspects of vector space operations.
Multilinearity
Multilinearity is an extension of bilinearity. A function is multilinear if it is linear in each of its arguments independently. In the case of tensor products, this means that the product will maintain linearity in each argument through operations such as addition and scalar multiplication, examined separately for each input entity.
To put it simply, if we have an operation\( f(u,v) \) that's multilinear, then keeping \( v \) constant, \( f(u + w, v) \) would equal \( f(u,v) + f(w,v) \). Similarly, keeping \( u \) constant, if we multiply \(v\) by a scalar \(\alpha\), then \( f(u, \alpha v) \) would result in \( \alpha f(u, v) \).
This characteristic is crucial for ensuring that tensor products provide predictable and cohesive results, reflecting their components' behaviors while preserving linear algebra's structure and rules.

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Most popular questions from this chapter

Show that the tensor product is not, in general, commutative.

Let \(S \subseteq U\) and \(T \subseteq V\) be subspaces of vector spaces \(U\) and \(V\), respectively. Show that $$ (S \otimes V) \cap(U \otimes T) \approx S \otimes T $$

Let \(\mathcal{B}=\left\\{u_{i} \mid i \in I\right\\}\) be a basis for \(U\) and let \(\mathcal{C}=\left\\{v_{j} \mid j \in J\right\\}\) be a basis for \(V\). Show that the set $$ \mathcal{D}=\left\\{u_{i} \otimes v_{j} \mid i \in I, j \in J\right\\} $$ is a basis for \(U \otimes V\) by showing that it is linearly independent and spans.

Suppose that \(\tau_{1}: U \rightarrow V, \tau_{2}: V \rightarrow W\) and \(\sigma_{1}: U^{\prime} \rightarrow V_{K}, \sigma_{2}: V_{K} \rightarrow W^{\prime}\). Prove that $$ \left(\tau_{2} \circ \tau_{1}\right) \odot\left(\sigma_{2} \circ \sigma_{1}\right)=\left(\tau_{2} \odot \sigma_{2}\right) \circ\left(\tau_{1} \odot \sigma_{1}\right) $$

Let \(\mathcal{B}=\left\\{b_{i}\right\\}\) be a basis for \(U\) and \(\mathcal{C}=\left\\{c_{i}\right\\}\) be a basis for \(V\). Show that any function \(f: \mathcal{B} \times \mathcal{C} \rightarrow W\) can be extended to a linear function \(\bar{f}: U \otimes V \rightarrow W\). Deduce that the function \(f\) can be extended in a unique way to a bilinear map \(\widehat{f}: U \times V \rightarrow W\). Show that all bilinear maps are obtained in this way.
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