Understanding Matrix Rank and the Given Matrix A

The question asks us to determine the rank of matrix A, where the elements of the matrix, denoted as \(a_{ij}\), are defined by the product of their row index \(i\) and column index \(j\). Specifically, \(a_{ij} = i \cdot j\). The matrix A is an \(n \times n\) square matrix, meaning it has \(n\) rows and \(n\) columns. A crucial condition given is that \(n \ge 3\).

The rank of a matrix is a fundamental concept in linear algebra. It represents the maximum number of linearly independent row vectors or column vectors in the matrix. An important property is that the row rank is always equal to the column rank, and this common value is what we call the rank of the matrix. A set of vectors is linearly independent if no vector in the set can be expressed as a linear combination of the others.

Constructing the Matrix A

Let's visualize the structure of the matrix A based on the definition \(a_{ij} = i \cdot j\):

From this definition, the matrix A can be explicitly written as:

\[ A = \begin{pmatrix} 1 & 2 & 3 & \cdots & n \\ 2 & 4 & 6 & \cdots & 2n \\ 3 & 6 & 9 & \cdots & 3n \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ n & 2n & 3n & \cdots & n^2 \end{pmatrix} \]

Analyzing Matrix Rows for Linear Dependence

To determine the rank of A, we can examine the linear dependence among its row vectors. Let's denote the rows of the matrix A as \(\mathbf{R_1}, \mathbf{R_2}, \ldots, \mathbf{R_n}\):

• Row 1: \(\mathbf{R_1} = \begin{pmatrix} 1 & 2 & 3 & \cdots & n \end{pmatrix}\)
• Row 2: \(\mathbf{R_2} = \begin{pmatrix} 2 & 4 & 6 & \cdots & 2n \end{pmatrix}\)
• Row 3: \(\mathbf{R_3} = \begin{pmatrix} 3 & 6 & 9 & \cdots & 3n \end{pmatrix}\)
• ...
• Row \(i\): \(\mathbf{R_i} = \begin{pmatrix} i & 2i & 3i & \cdots & ni \end{pmatrix}\)

Let's observe the relationship between these rows by factoring out common terms:

• Notice that \(\mathbf{R_2} = \begin{pmatrix} 2 \cdot 1 & 2 \cdot 2 & 2 \cdot 3 & \cdots & 2 \cdot n \end{pmatrix} = 2 \cdot \begin{pmatrix} 1 & 2 & 3 & \cdots & n \end{pmatrix} = 2 \cdot \mathbf{R_1}\).
• Similarly, \(\mathbf{R_3} = \begin{pmatrix} 3 \cdot 1 & 3 \cdot 2 & 3 \cdot 3 & \cdots & 3 \cdot n \end{pmatrix} = 3 \cdot \begin{pmatrix} 1 & 2 & 3 & \cdots & n \end{pmatrix} = 3 \cdot \mathbf{R_1}\).
• In general, for any row \(i\), its elements are \(i \cdot j\). So, the row vector \(\mathbf{R_i}\) can be written as \(\mathbf{R_i} = i \cdot \begin{pmatrix} 1 & 2 & 3 & \cdots & n \end{pmatrix} = i \cdot \mathbf{R_1}\).

This shows that every row in the matrix A is a scalar multiple of the first row (\(\mathbf{R_1}\)). This implies that all rows are linearly dependent on the first row. The only linearly independent row is the first row itself, since \(n \ge 3\) guarantees that \(\mathbf{R_1}\) is not a zero vector (it contains non-zero elements like 1, 2, 3, etc.).

Determining the Rank of Matrix A

Since all rows are scalar multiples of a single non-zero row, the row space of matrix A is spanned by just one vector, which is the first row vector \(\begin{pmatrix} 1 & 2 & \cdots & n \end{pmatrix}\). The dimension of the row space is defined as the number of linearly independent rows.

In this particular case, there is only one linearly independent row. Therefore, the rank of A is 1.

We can also verify this by considering the columns. Let \(\mathbf{C_j}\) denote the j-th column of matrix A:

• \(\mathbf{C_1} = \begin{pmatrix} 1 \\ 2 \\ \vdots \\ n \end{pmatrix}\)
• \(\mathbf{C_2} = \begin{pmatrix} 2 \\ 4 \\ \vdots \\ 2n \end{pmatrix} = 2 \cdot \begin{pmatrix} 1 \\ 2 \\ \vdots \\ n \end{pmatrix} = 2 \cdot \mathbf{C_1}\)
• ...
• \(\mathbf{C_j} = \begin{pmatrix} j \\ 2j \\ \vdots \\ nj \end{pmatrix} = j \cdot \begin{pmatrix} 1 \\ 2 \\ \vdots \\ n \end{pmatrix} = j \cdot \mathbf{C_1}\)

Similarly, every column is a scalar multiple of the first column. This means the column space is also spanned by a single vector, \(\begin{pmatrix} 1 \\ 2 \\ \vdots \\ n \end{pmatrix}\). The number of linearly independent columns is 1.

As the row rank equals the column rank, and both are determined to be 1, the rank of A is 1. Since \(n \ge 3\), the matrix A is not a zero matrix (for example, \(a_{11}=1 \neq 0\)), so its rank cannot be 0.

Conclusion on Matrix Rank

Based on the thorough analysis of linear dependence among the rows and columns of matrix A, where each element \(a_{ij}\) is defined as \(i \cdot j\), we consistently find that there is only one linearly independent row (and one linearly independent column). This specific structure is characteristic of a rank-1 matrix, which can also be understood as the outer product of two non-zero vectors.

Thus, the rank of A is 1.