AcademyLinear Systems
Academy
Invertible Matrix Theorem
Level 1 - Linear Algebra topic page in Linear Systems.
Principle
The invertible matrix theorem collects many equivalent ways to say that a square matrix is invertible. These tests connect row reduction, linear systems, determinants, pivots, kernels, and columns.
For a square matrix, proving any one condition proves all the others.
Notation
\(A\)
an n by n matrix
\(A^{-1}\)
the inverse of A
\(I_n\)
the n by n identity matrix
\(\det A\)
the determinant of A
\(A\mathbf x=\mathbf b\)
a square linear system
\(\mathbf 0\)
the zero vector
The Core Test
For an \(n\) by \(n\) matrix \(A\), the following conditions are equivalent:
- \(A\) is invertible.
- The RREF of \(A\) is \(I_n\).
- \(A\) has a pivot in every column.
- \(A\mathbf x=\mathbf 0\) has only the zero solution.
- \(A\mathbf x=\mathbf b\) has a unique solution for every \(\mathbf b\in\mathbb R^n\).
- The columns of \(A\) span \(\mathbb R^n\).
- \(\det A\ne0\).
In compact form:
Invertibility and determinant
\[A\text{ invertible}\Longleftrightarrow \det A\ne0\]
Worked Cases
Question
Suppose row reduction gives \(A\sim I_3\). What can you conclude about \(A\)?
Answer
If a square matrix row-reduces to the identity matrix, then it has a pivot in every column. By the invertible matrix theorem, \(A\) is invertible, \(A\mathbf x=\mathbf 0\) has only the zero solution, and \(A\mathbf x=\mathbf b\) has a unique solution for every \(\mathbf b\in\mathbb R^3\).
Examples
Question
A square matrix \(A\) has a non-zero vector \(\mathbf v\) such that \(A\mathbf v=\mathbf 0\). Can \(A\) be invertible?
Answer
No. If \(A\) were invertible, multiply \(A\mathbf v=\mathbf 0\) by \(A^{-1}\) to get \(A^{-1}A\mathbf v=A^{-1}\mathbf 0\), so \(I\mathbf v=\mathbf 0\), hence \(\mathbf v=\mathbf 0\). This contradicts \(\mathbf v\ne\mathbf 0\). Therefore \(A\) is not invertible.