Fundamental Equivalences for Square matrices

Fundamental Equivalences for Square matrices#

Yes, you’re absolutely right — these statements are all equivalent and form a core set of if-and-only-if conditions that characterize invertibility of a square matrix. Let’s list them more formally and then prove that they are all equivalent:

Theorem (Fundamental Equivalences)

Let \(\mathbf{A} \in \mathbb{R}^{n \times n}\). The following statements are equivalent — that is, they are all true or all false together:

  1. \(\mathbf{A}\) is invertible

  2. \(\det(\mathbf{A}) \neq 0\)

  3. \(\mathbf{A}\) is full-rank, i.e., \(\operatorname{rank}(\mathbf{A}) = n\)

  4. The columns of \(\mathbf{A}\) are linearly independent

  5. The rows of \(\mathbf{A}\) are linearly independent

  6. \(\mathbf{A}\) is row-equivalent to the identity matrix

  7. The system \(\mathbf{A}\mathbf{x} = \mathbf{b}\) has a unique solution for every \(\mathbf{b} \in \mathbb{R}^n\)

Proof. We’ll prove the chain of implications in a circular fashion, which implies all are equivalent.

(1) ⇒ (2): Invertible ⇒ Determinant nonzero

If \(\mathbf{A}^{-1}\) exists, then

\[ \det(\mathbf{A} \mathbf{A}^{-1}) = \det(\mathbf{I}) = 1 = \det(\mathbf{A}) \det(\mathbf{A}^{-1}) \Rightarrow \det(\mathbf{A}) \neq 0 \]

(2) ⇒ (3): \(\det(\mathbf{A}) \neq 0\) ⇒ Full-rank

A square matrix has full rank \(\iff\) its rows/columns span \(\mathbb{R}^n\), and this happens exactly when \(\det(\mathbf{A}) \neq 0\).

If \(\operatorname{rank}(\mathbf{A}) < n\), then one row or column is linearly dependent, making \(\det(\mathbf{A}) = 0\).

(3) ⇒ (4) and (5): Full-rank ⇒ Linear independence of rows and columns

A matrix with rank \(n\) must have linearly independent rows and columns by the definition of rank.

(4) ⇒ (6): Independent columns ⇒ Row-equivalent to identity

If the columns are linearly independent, Gaussian elimination can reduce \(\mathbf{A}\) to the identity matrix \(\mathbf{I}\) using row operations.

This means \(\mathbf{A}\) is row-equivalent to \(\mathbf{I}\).

(6) ⇒ (7): Row-equivalent to \(\mathbf{I}\) ⇒ Unique solution for all \(\mathbf{b}\)

If \(\mathbf{A} \sim \mathbf{I}\), then solving \(\mathbf{A} \mathbf{x} = \mathbf{b}\) is equivalent to solving \(\mathbf{I} \mathbf{x} = \mathbf{b}'\), which always has the unique solution \(\mathbf{x} = \mathbf{b}'\).

(7) ⇒ (1): Unique solution for all \(\mathbf{b}\) ⇒ \(\mathbf{A}\) is invertible

If \(\mathbf{A} \mathbf{x} = \mathbf{b}\) has a unique solution for all \(\mathbf{b}\), then the inverse mapping \(\mathbf{b} \mapsto \mathbf{x}\) is well-defined and linear, so \(\mathbf{A}^{-1}\) exists.

Conclusion

All these statements are equivalent:

\[ \boxed{ \mathbf{A} \text{ invertible } \iff \det(\mathbf{A}) \neq 0 \iff \operatorname{rank}(\mathbf{A}) = n \iff \text{cols/rows lin. independent} \iff \mathbf{A} \sim \mathbf{I} \iff \text{unique solution for all } \mathbf{b} } \]