Problem 5 - Advanced Linear Algebra

Question

Prove that the following property of a pair \((W, g: U \times V \rightarrow W)\) with \(g\) bilinear characterizes the tensor product \((U \otimes V, t: U \times V \rightarrow U \otimes V)\) up to isomorphism, and thus could have been used as the definition of tensor product: For a pair \((W, g: U \times V \rightarrow W)\) with \(g\) bilinear if \(\left\\{u_{i}\right\\}\) is a basis for \(U\) and \(\left\\{v_{i}\right\\}\) is a basis for \(V\), then \(\left\\{g\left(u_{i}, v_{j}\right)\right\\}\) is a basis for \(W\).

Step-by-Step Solution

Answer

To prove that the given property characterizes the tensor product up to isomorphism, we defined a linear map \(f: U \otimes V \rightarrow W\) by \(f(u\otimes v)=g(u,v)\) for all \(u \in U\) and \(v \in V\). We then showed that \(f\) is onto and one-to-one, which implies that it is an isomorphism between \(U \otimes V\) and \(W\). Finally, we observed that the isomorphism preserves the bilinear property of \(g\) (and hence, \(t\)). Therefore, the given property could have been used as the definition of the tensor product.

Step 1: Define the linear map

We define a linear map \(f: U \otimes V \rightarrow W\) by \(f(u\otimes v)=g(u,v)\) for all \(u \in U\) and \(v \in V\). It is well-defined since \(g\) is bilinear.

Step 2: Prove that f is onto

Suppose \(\left\\{u_{i}\right\\}\) is a basis for \(U\) and \(\left\\{v_{j}\right\\}\) is a basis for \(V\). Then the given condition tells us that \(\left\\{g(u_{i},v_{j})\right\\}\) is a basis for \(W\). Any element in \(W\) can be expressed as a linear combination of the basis elements: \[w = \sum_{i,j} a_{i,j} g(u_i, v_j).\] But by the definition of \(f\), we have \(f(u_i \otimes v_j) = g(u_i, v_j)\). So, we can also express \(w\) as follows: \[w = \sum_{i,j} a_{i,j} f(u_i \otimes v_j).\] This shows that every element in \(W\) can be reached by applying \(f\) to some element in \(U \otimes V\), which means that f is onto.

Step 3: Prove that f is one-to-one

Suppose that \(u_1, u_2 \in U\) and \(v_1, v_2 \in V\) with \(f(u_1 \otimes v_1) = f(u_2 \otimes v_2)\). This implies that \(g(u_1, v_1) = g(u_2, v_2)\). However, \(\left\{ g(u_{i},v_{j}) \right\}\) is a basis for \(W\), which means that the set of all \(g(u,v)\) is linearly independent. Therefore, the only way for \(g(u_1, v_1) = g(u_2, v_2)\) to hold is if \(u_1=u_2\) and \(v_1=v_2\). So, if \(f(u_1 \otimes v_1) = f(u_2 \otimes v_2)\), then \(u_1 \otimes v_1 = u_2 \otimes v_2\). This implies that \(f\) is one-to-one.

Step 4: Conclude that f is an isomorphism

Since we proved that \(f: U \otimes V \rightarrow W\) is a linear map, is one-to-one, and is onto, we can conclude that it is an isomorphism between \(U \otimes V\) and \(W\).

Step 5: Observe that f preserves bilinearity

Since \(f(u\otimes v) = g(u,v)\) for all \(u \in U\) and \(v \in V\), it is clear that the isomorphism preserves the bilinear property of \(g\) \((\)and hence, \(t)\). Thus, we have shown that the property given characterizes the tensor product up to isomorphism.

Key Concepts

Bilinear Map

A bilinear map is a function that takes two arguments from different vector spaces and outputs a value in another vector space, adhering to linearity with respect to each input. Think of it like a function that maintains the balance of linearity through two dimensions. This means that if you hold one of the inputs fixed, and vary the other, the output behaves just like a linear map.

For example, for a bilinear map \( g: U \times V \rightarrow W \), the rule is if \( u_1, u_2 \in U \) and \( v_1, v_2 \in V \) and scalars \( c_1, c_2 \), then:

• \( g(c_1 u_1 + c_2 u_2, v) = c_1 g(u_1, v) + c_2 g(u_2, v) \)
• \( g(u, c_1 v_1 + c_2 v_2) = c_1 g(u, v_1) + c_2 g(u, v_2) \)

These conditions ensure that the function is linear in both arguments independently, which is why it's called 'bilinear'.

In the context of the tensor product, the bilinear map helps describe how components from two spaces can be combined in a structured and consistent way to form elements of a new product space.

Isomorphism

Isomorphism is a fundamental concept in mathematics that indicates a kind of equivalence between two structures. If two mathematical objects are isomorphic, they can be considered as essentially the same for practical purposes because their structures are identical in some sense.

Formally, an isomorphism between two vector spaces, say \( U\otimes V \) and \( W \), is a bijective (one-to-one and onto) linear map \( f: U\otimes V \to W \). This means that every element in \( U\otimes V \) corresponds to exactly one element in \( W \) and vice versa.

To prove isomorphism, we usually show three things:

• Linear: The function \( f \) respects addition and scalar multiplication.
• Onto: Every element in \( W \) is reached by applying \( f \) to some element in \( U\otimes V \).
• One-to-One: If \( f(x) = f(y) \), then \( x = y \).

When such a map exists between two vector spaces, they are said to be isomorphic, implying they have the same dimension and structure.

Linear Map

A linear map is a function between two vector spaces that respects the operations of vector addition and scalar multiplication. This means the map behaves in ways we expect with regard to combining and scaling vectors. For a linear map \( f: U \to W \), it must satisfy:

\( f(u_1 + u_2) = f(u_1) + f(u_2) \) and \( f(cu) = c f(u) \) for any vectors \( u_1, u_2 \in U \) and scalar \( c \).

The beauty of linear maps is their simplicity and predictability, making them a cornerstone in the study of linear algebra. When we define \( f: U \otimes V \rightarrow W \) as a linear map by the nature of the bilinear operation \( f(u \otimes v) = g(u,v) \), we leverage the principle that the tensor product behaves predictably as if it were a product of its constituent spaces.

This regularity allows transformations from one vector space to another, making manipulations and analyses within vector spaces consistent and categorical.

Basis

A basis is a set of vectors in a vector space from which every element of the space can be uniquely expressed as a combination. It provides a framework or 'coordinate system' for the space. For a vector space \( V \), a basis is a set of vectors \( \{v_1, v_2, \ldots, v_n\} \) that is linearly independent and spans the entire space.

Linear independence means that no vector in the basis can be expressed as a combination of the others, ensuring each provides a distinct 'direction' in the space. Spanning the space means that any vector \( v \in V \) can be represented as:

\[ v = a_1 v_1 + a_2 v_2 + \ldots + a_n v_n \]
for some scalars \( a_1, a_2, \ldots, a_n \).

Bases are essential for comparing and transforming vectors because they provide a common reference point. In this exercise, the basis \( \{u_i\} \) for \( U \) and \( \{v_j\} \) for \( V \) helps transform or translate the bilinear map into the linear framework necessary for establishing isomorphism, permitting us to express any element of a vector space in a uniquely determined way.