📂Matrix Algebra

Properties of Full Rank Matrices

Theorem¹

Let’s refer to $A$ as matrix $m \times n$. Then, the necessary and sufficient condition for $A$ to have a full rank is for $A^{T}A$ to be an invertible matrix.

Proof

• $(\Longrightarrow)$

  Assume that $A$ has a full rank. Since $A^{T}A$ is a square matrix $n \times n$, showing that the linear system $A^{T}A \mathbf{x} = \mathbf{0}$ only has the trivial solution, according to the equivocal condition of being an invertible matrix, suffices. Let $\mathbf{x}$ be any solution. Then, $A \mathbf{x}$ belongs to the null space of $A^{T}$. Moreover, $A \mathbf{x}$ belongs to the column space of $A$. However, these two are orthogonal complements to each other.

  Property of Orthogonal Complements

  $$ W \cap W^{\perp} = \left\{ \mathbf{0} \right\} $$

  Therefore, $A \mathbf{x} = \mathbf{0}$ holds. But since we assumed that $A$ has a full rank, the only $\mathbf{x}$ that satisfies this is the trivial solution. Hence, $A^{T}A$ is invertible.

• $(\Longleftarrow)$

  Assume that $A^{T}A$ is invertible. Then, the following linear system only has the trivial solution.

  $$ A^{T}A \mathbf{x} = \mathbf{0} $$

  Then, $A \mathbf{x} = \mathbf{0}$ holds, and this linear system also can only have the trivial solution. Therefore, by the equivalence condition, $A$ has a full rank.

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