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What is the dimension of the vector space \(\mathcal{M}_{m \times n}\) ? Give a basis.

Short Answer

Expert verified
The dimension of the vector space \(\mathcal{M}_{m \times n}\) is \(m \times n\). A basis for this vector space is given by the set of matrices \(E_{ij}\) with a single entry equal to 1 at the i-th row and j-th column, and all other entries being 0: \[ \mathcal{B} = \{E_{11}, E_{12}, \cdots, E_{1n}, E_{21}, E_{22}, \cdots, E_{2n}, \cdots, E_{m1}, E_{m2}, \cdots, E_{mn}\} \]

Step by step solution

01

Understanding the definition of dimension and basis

The dimension of a vector space is the number of elements in any basis of that vector space. A basis is a set of linearly independent vectors that can be combined through linear combinations to generate all the vectors belonging to the vector space.
02

Representation of elements in \(\mathcal{M}_{m \times n}\)

Each element in the \(\mathcal{M}_{m \times n}\) can be represented as an m × n matrix, with each entry being an element of the underlying field, usually denoted as \(\mathbb{F}\) (e.g., \(\mathbb{R}\) for real numbers and \(\mathbb{C}\) for complex numbers). So, a matrix \(A \in \mathcal{M}_{m \times n}\) can be written as: $$ A = \begin{bmatrix} a_{11} & a_{12} & \cdots & a_{1n} \\ a_{21} & a_{22} & \cdots & a_{2n} \\ \vdots & \vdots & \ddots & \vdots \\ a_{m1} & a_{m2} & \cdots & a_{mn} \end{bmatrix} $$
03

Finding the basis for \(\mathcal{M}_{m \times n}\)

We can find a basis for \(\mathcal{M}_{m \times n}\) by constructing m × n matrices that have a single entry equal to 1 and all other entries are 0. If we denote these matrices by \(E_{ij}\), with i representing the row number and j representing the column number, the basis is given by: $$ \mathcal{B} = \{E_{11}, E_{12}, \cdots, E_{1n}, E_{21}, E_{22}, \cdots, E_{2n}, \cdots, E_{m1}, E_{m2}, \cdots, E_{mn}\} $$ For example, \(E_{11}\) will be a matrix with a 1 in the first row, first column, and all other entries are 0: $$ E_{11} = \begin{bmatrix} 1 & 0 & \cdots & 0 \\ 0 & 0 & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & 0 \end{bmatrix} $$ Note that these matrices form a basis because they are linearly independent and can be combined through linear combinations to generate any matrix in \(\mathcal{M}_{m \times n}\).
04

Calculating the dimension of \(\mathcal{M}_{m \times n}\)

To find the dimension of \(\mathcal{M}_{m \times n}\), we just need to count the number of elements in the basis \(\mathcal{B}\). Since there is one basis matrix for each row and column combination, the total number of basis matrices is m × n. Therefore, the dimension of \(\mathcal{M}_{m \times n}\) is: $$ \text{dim} \, (\mathcal{M}_{m \times n}) = m \times n $$

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Key Concepts

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

Basis of Vector Spaces
When we talk about the **basis of a vector space**, we refer to a special set of vectors that play a crucial role in defining the space. A basis consists of:

  • Linearly independent vectors: No vector in the set can be expressed as a combination of the others.
  • A set that spans the vector space: Every vector in the space can be formed using a linear combination of the basis vectors.

For example, in the context of the vector space of matrices denoted as (\(\mathcal{M}_{m \times n}\)), the basis is typically made up of matrices with one element equal to 1 and all other elements zero. These basis matrices are called unit matrices and might be denoted as \(E_{ij}\) indicating a 1 at position (i, j). Each choice of i and j gives us a different basis matrix.

This concept of a basis is incredibly useful because once you have a basis for a vector space, it helps you easily work with, transform, or understand all possible elements (vectors) in that space.
m × n Matrices
Matrices are a powerful way to represent data and transformations. An **m × n matrix** consists of m rows and n columns:

``` \[ A = \begin{bmatrix} a_{11} & a_{12} & \cdots & a_{1n} \ a_{21} & a_{22} & \cdots & a_{2n} \ \vdots & \vdots & \ddots & \vdots \ a_{m1} & a_{m2} & \cdots & a_{mn} \end{bmatrix}\] ```
Each element, \(a_{ij}\), represents the data at row \(i\) and column \(j\). These matrices are crucial in various areas such as computer graphics, statistics, and physics.

In this context, every matrix belongs to the vector space \(\mathcal{M}_{m \times n}\) where each element is a member of a field (e.g., real numbers \(\mathbb{R}\) or complex numbers \(\mathbb{C}\)). This allows us to perform operations like addition and scalar multiplication with matrices, essential for solving systems of equations and linear transformations in higher algebra.
Linear Independence
**Linear independence** is a fundamental concept that ensures vectors (or matrices, in our case) do not overlap in their contributions to vector space.

For a set of vectors (or matrices) to be linearly independent:
  • No vector (or matrix) in the set can be described as a linear combination of the others.
  • The only solution to a combination of them equating to zero is when all coefficients of the combination are zero.

This is important because it ensures the set of vectors provides unique directions or dimensions within the space. It essentially means each vector (or matrix) contributes something new.

In the vector space of matrices \(\mathcal{M}_{m \times n}\), the basis matrices \(E_{ij}\) are linearly independent. No matrix in this set can be recreated using a combination of others. Hence, they represent unique structural elements of the m × n matrices, allowing us to explore the entire space efficiently.

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Most popular questions from this chapter

The subspace \(\ell^{2} \subset \mathbb{R}^{\omega}\) defined by \(\ell^{2}=\left\\{\mathbf{x} \in \mathbb{R}^{\omega}: \sum_{k=1}^{\infty} x_{k}^{2}\right.\) exists \(\\}\) is an inner product space with inner product defined by \(\langle\mathbf{x}, \mathbf{y}\rangle=\sum_{k=1}^{\infty} x_{k} y_{k}\). (That this sum makes sense follows by "taking the limit" of the Cauchy-Schwarz Inequality.) Let $$ V=\left\\{\mathbf{x} \in \ell^{2}: \text { there is an integer } n \text { such that } x_{k}=0 \text { for all } k>n\right\\} . $$ Show that \(V\) is a (proper) subspace of \(\ell^{2}\) and that \(V^{\perp}=\\{0\\}\). It follows that \(\left(V^{\perp}\right)^{\perp}=\ell^{2}\), so Proposition \(3.6\) need not hold in infinite-dimensional spaces.

Let \(A\) be an \(m \times n\) matrix with rank \(r\). Suppose \(A=B U\), where \(U\) is in echelon form. Show that the first \(r\) columns of \(B\) give a basis for \(\mathbf{C}(A)\). (In particular, if \(E A=U\), where \(U\) is the echelon form of \(A\) and \(E\) is the product of elementary matrices by which we reduce \(A\) to \(U\), then the first \(r\) columns of \(E^{-1}\) give a basis for \(\mathbf{C}(A)\).)

Suppose \(\mathbf{v}_{1}, \ldots, \mathbf{v}_{k} \in \mathbb{R}^{n}\) form a linearly dependent set. Prove that either \(\mathbf{v}_{1}=\mathbf{0}\) or \(\mathbf{v}_{i} \in \operatorname{Span}\left(\mathbf{v}_{1}, \ldots, \mathbf{v}_{i-1}\right)\) for some \(i=2,3, \ldots, k\). (Hint: There is a relation \(c_{1} \mathbf{v}_{1}+\) \(c_{2} \mathbf{v}_{2}+\cdots+c_{k} \mathbf{v}_{k}=\mathbf{0}\) with at least one \(c_{j} \neq 0\). Consider the largest such \(j\).)

Continuing Exercise 3.2.10: Let \(A\) be an \(m \times n\) matrix. a. Use Theorem \(2.5\) to prove that \(\mathbf{N}\left(A^{\top} A\right)=\mathbf{N}(A)\). (Hint: If \(\mathbf{x} \in \mathbf{N}\left(A^{\top} A\right)\), then \(\left.A \mathbf{x} \in \mathbf{C}(A) \cap \mathbf{N}\left(A^{\top}\right) .\right)\) b. Prove that \(\operatorname{rank}(A)=\operatorname{rank}\left(A^{\top} A\right)\). c. Prove that \(\mathbf{C}\left(A^{\mathrm{T}} A\right)=\mathbf{C}\left(A^{\mathrm{T}}\right)\).

Let \(g_{1}(t)=t-1\) and \(g_{2}(t)=t^{2}+t\). Using the inner product on \(\mathcal{P}_{2} \subset \bigodot^{0}([0,1])\) defined in Example \(10(\mathrm{c})\), find the orthogonal complement of Span \(\left(g_{1}, g_{2}\right)\).

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