Question

Let \(A\)be an\(m \times n\)matrix whose columns are linearly independent.

1. Use Exercise 19 to show that \({A^T}A\) is an invertible matrix.
2. Explain why\(A\)must have at least as many rows as columns.
3. Determine the rank of \(A\).

Short answer

1. It is proved that \({A^T}A\) is an invertible matrix.
2. Because \(n\) linearly independent columns of \(A\) belongs to \({\mathbb{R}^m}\).
3. The rank of \(A\) is \(n\).

Step-by-step solution

Rank of a matrix

(a).

Given that \(A\) is an \(m \times n\) matrix whose columns are linearly independent.

So, the above statement implies that \(\dim \left( {{\rm{Col}}\,A} \right) = n\) and \({\rm{rank}}\left( A \right) = n\). Now, it is known that if \(A\) is an \(m \times n\) matrix then there exist a vector \(x\) in \({\mathbb{R}^n}\) satisfies \(Ax = 0\) if and only if \({A^T}Ax = 0\). So, \(\dim \left( {{\rm{Nul}}\left( {{A^T}A} \right)} \right) = 0\).

This implies that the rank of matrix \({A^T}A\) is \(n\). Since, \({A^T}A\) is \(n \times n\) matrix and \({\rm{rank}}\left( {{A^T}A} \right) = n\). So, \({A^T}A\) is invertible.

Proof of part (b)

The reason is that the columns of \(A\) form a linearly independent set of vectors in \({\mathbb{R}^m}\). Therefore, the number of rows cannot be less than the number of columns.

Hence, the number of rows is at least the number of columns,

Rank of matrix A

(c).

It is known that the dimensions of column space of \(A\)and the rank of \(A\) are equal. Since, the number of columns that are linearly independent is \(n\).

Therefore, the rank of \(A\) is equal to \(n\).