Problem 17 - Differential Equations and Linear Algebra

Question

If \(T: M_{23}(\mathbb{R}) \rightarrow P_{6}(\mathbb{R})\) is one-to-one, what is \(\operatorname{dim}[\operatorname{Rng}(T)] ?\)

Step-by-Step Solution

Answer

The dimension of the range of the linear transformation \(T: M_{23}(\mathbb{R}) \rightarrow P_{6}(\mathbb{R})\) is 6.

Step 1: Identify the dimensions of domain and kernel

We are given that T is a linear transformation from \(M_{23}(\mathbb{R})\) (the space of 2x3 matrices) to \(P_{6}(\mathbb{R})\) (the space of polynomials of degree at most 6). The dimension of \(M_{23}(\mathbb{R})\) is 2 * 3 = 6, since each entry is an independent component. We also know that T is a one-to-one linear transformation, so the kernel of T (the set of vectors v in the domain such that T(v) = 0) is trivial and contains only the zero vector. Thus, the dimension of the kernel is 0.

Step 2: Apply the rank-nullity theorem

We will now apply the rank-nullity theorem to find the dimension of the range of T. The theorem states: \[\operatorname{dim}(\operatorname{Dom}(T)) = \operatorname{dim}(\operatorname{Rng}(T)) + \operatorname{dim}(\operatorname{Ker}(T))\] Substitute the dimensions of the domain and kernel that we found in Step 1: \[6 = \operatorname{dim}(\operatorname{Rng}(T)) + 0\]

Step 3: Solve for the dimension of the range

Since the equation becomes \(6 = \operatorname{dim}(\operatorname{Rng}(T))+0\), we can easily solve for the dimension of the range of T: \[\operatorname{dim}(\operatorname{Rng}(T)) = 6\] Therefore, the dimension of the range of the linear transformation T is 6.

Key Concepts

Matrix Space

Matrix space is a fascinating concept in linear algebra. It refers to the set of all matrices of a certain size over a given field. For example, in the previous problem, we discussed the space of 2x3 matrices, denoted as \(M_{23}(\mathbb{R})\). This space consists of all matrices that have 2 rows and 3 columns with real number entries.

Matrix spaces have dimensions similar to vectors - think of them as vectors with more structure due to their shape. For a matrix space \(M_{mn}(\mathbb{R})\), the dimension is calculated by the product of the number of rows and columns (\(m \times n\)). Thus, the dimension of \(M_{23}(\mathbb{R})\) is 6, since it has 6 individual entries that can vary independently.

• Each entry is an independent component.
• Matrix spaces are essential in defining transformations between different vector spaces.
• They help represent linear transformations in a compact form.

Understanding matrix spaces is crucial for exploring how linear transformations, like \(T\), act on different spaces and how they change dimensions between different settings.

Rank-Nullity Theorem

The rank-nullity theorem is a fundamental theorem in linear algebra that links dimensions of different pieces of a linear transformation. For a linear transformation \(T\) from one vector space to another, this theorem connects the dimension of the domain, the range, and the kernel.

The theorem is mathematically expressed as:\[\operatorname{dim}(\operatorname{Dom}(T)) = \operatorname{dim}(\operatorname{Rng}(T)) + \operatorname{dim}(\operatorname{Ker}(T))\]This equation essentially says the dimension of the domain is the sum of the dimensions of the range and the kernel. It is useful because:

• If you know the dimensions of any two spaces, you can find the third.
• It helps verify properties of \(T\), such as injectivity (one-to-one) or surjectivity (onto).
• It's pivotal in determining the possible outcomes of a transformation.

For the problem at hand, since \(T\) is one-to-one (injective), the kernel is trivial (dimension zero), leading directly to the conclusion that the range's dimension equals the domain's dimension. This highlights the elegance and power of the rank-nullity theorem.

Polynomial Space

Polynomial spaces are sets of polynomials with coefficients in a particular field, such as the real numbers \(\mathbb{R}\). In our problem, the polynomial space \(P_{6}(\mathbb{R})\) consists of polynomials with real coefficients of degree at most 6.

Such a polynomial space has a dimension equal to one more than the highest polynomial degree. So, \(P_{6}(\mathbb{R})\) has dimension 7, with basis elements

• \(1\)
• \(x\)
• \(x^2\)
• \(x^3\)
• \(x^4\)
• \(x^5\)
• \(x^6\)

Each polynomial in this space can be expressed as a linear combination of these basis elements. Polynomial spaces are central to the study of functions in algebra and calculus because they offer a structured way to approach problems involving unknowns and variation.

Grasping polynomial spaces allows us to see how abstract operations in calculus relate to hands-on computations in algebra, such as transformations and solving equations.