Question

If \(V\) is any vector space, let \(V^{n}\) denote the space of all \(n\) -tuples \(\left(\mathbf{v}_{1}, \mathbf{v}_{2}, \ldots, \mathbf{v}_{n}\right),\) where each \(\mathbf{v}_{i}\) lies in \(V\). (This is a vector space with component-wise operations; see Exercise 6.1.17.) If \(C_{j}(A)\) denotes the \(j\) th column of the \(m \times n\) matrix \(A,\) show that \(T: \mathbf{M}_{m n} \rightarrow\left(\mathbb{R}^{m}\right)^{n}\) is an isomorphism if \(T(A)=\left[\begin{array}{llll}C_{1}(A) & C_{2}(A) & \cdots & C_{n}(A)\end{array}\right] .\) (Here \(\mathbb{R}^{m}\) con- sists of columns.)

Short answer

The transformation \( T \) is an isomorphism because it is bijective.

Step-by-step solution

Understand the Problem

We need to show that the linear transformation \( T: \mathbf{M}_{mn} \rightarrow \left(\mathbb{R}^{m}\right)^{n} \), which maps an \( m \times n \) matrix \( A \) to a tuple of its columns, is an isomorphism.

Define Isomorphism

An isomorphism in linear algebra is a linear transformation that is bijective. This means that the transformation must be both injective (one-to-one) and surjective (onto). We need to verify both these properties for \( T \).

Show Linearity of T

For \( T \) to be linear, it must satisfy \( T(A + B) = T(A) + T(B) \) and \( T(kA) = kT(A) \) for any matrices \( A, B \) in \( \mathbf{M}_{mn} \) and any scalar \( k \). Since \( T \) maps columns linearly, these properties hold.

Prove Injectivity

To prove that \( T \) is injective, assume \( T(A) = T(B) \). This implies that the columns of \( A \) and \( B \) are identical, i.e., \( C_j(A) = C_j(B) \) for all \( j \). Thus, \( A = B \), confirming that \( T \) is injective.

Prove Surjectivity

For surjectivity, we need to show that for any \( (x_1, x_2, \ldots, x_n) \in \left(\mathbb{R}^m\right)^n \), there exists a matrix \( A \) such that \( T(A) = (x_1, x_2, \ldots, x_n) \). Construct a matrix \( A \) whose columns are \( x_1, x_2, \ldots, x_n \), then \( T(A) = (x_1, x_2, \ldots, x_n) \). Thus, \( T \) is surjective.

Conclusion

Since \( T \) is both injective and surjective, it is bijective. Therefore, \( T \) is an isomorphism from \( \mathbf{M}_{mn} \) to \( \left(\mathbb{R}^{m}\right)^{n} \).

Key concepts

linear transformation

Linear transformations are a key concept in linear algebra. They are essentially functions that map one vector space to another while preserving vector addition and scalar multiplication. In simple terms, if you have two operations like adding vectors or multiplying a vector by a scalar, a linear transformation will apply these operations in the same linear way in another vector space.
For any vectors \( \mathbf{u} \) and \( \mathbf{v} \) in a vector space \( V \) and a scalar \( k \), a linear transformation \( T \) satisfies the following two properties:

• \( T(\mathbf{u} + \mathbf{v}) = T(\mathbf{u}) + T(\mathbf{v}) \)
• \( T(k \mathbf{u}) = k T(\mathbf{u}) \)

In our exercise, the transformation \( T \) maps an \( m \times n \) matrix to a sequence of \( n \) column vectors in \( (\mathbb{R}^m)^n \). It's essential to understand that this transformation preserves the linear structure of the original space. When you add two matrices and then apply \( T \), it's the same as applying \( T \) to each and then adding the results. This behavior is what defines \( T \) as linear.

bijective mapping

To fully understand vector space isomorphism, it's crucial to grasp the idea of a bijective mapping. In layman's terms, a bijection is a perfect pair-up between two sets, where every element in one set corresponds to exactly one element in the other set, and vice versa.
A linear transformation \( T \) is bijective if it is both:

• Injective (one-to-one): Each element of the domain maps to a unique element of the codomain. This means no two different elements in the domain end up in the same spot in the codomain. For our transformation \( T(A) = T(B) \), this results in \( A = B \).
• Surjective (onto): Every element of the codomain is hit by some element of the domain. This means, for every output we want, there's an input that produces it. You can think of this as "covering" the whole target space.

In the context of our exercise, demonstrating that our transformation \( T \) is both injective and surjective is what confirms it's an isomorphism. Having this bijective property ensures that two vector spaces are structurally the same in terms of dimension and operations.

matrix operations

Working with matrices involves various operations, such as addition, scalar multiplication, and more complex tricks like finding determinants or inverses. However, the focus here is on how we use matrices in linear transformations, particularly highlighting their column representation.
Imagine a matrix \( A \) as a collection of column vectors. When we perform matrix operations like addition or scalar multiplication, each operation is done component-wise, meaning it happens within each corresponding entry of the columns, exactly like how you'd add vectors or multiply them by a scalar.
In our problem, the transformation \( T \) maps a matrix \( A \) to a set of its columns, treating each column as a vector. This transformation respects matrix operations:

• Adding two matrices \( A \) and \( B \) simply adds each corresponding pair of columns.
• Multiplying a matrix \( A \) by a scalar scales each column by that scalar.

These operations might seem straightforward, but they play a pivotal role in maintaining the structure of vector spaces under transformations and proving properties like linearity in transformations.