= Invertible matrix
{title2=$\operatorname{rank}A=n\iff\det A\ne0$}
{wiki}
An $n$-by-$n$ <square matrix> $A$ over a <field> is invertible when it has a two-sided <matrix inverse>. Equivalently, its <linear map> $x\mapsto Ax$ is a <bijection>: every output comes from exactly one input. For a <square matrix>, the following conditions are equivalent:
$$
A\text{ is invertible}
\iff\operatorname{rank}A=n
\iff\ker A=\{0\}
\iff\det A\ne0
\iff Ax=b\text{ has exactly one solution for every }b.
$$
Here <matrix rank> $n$ means both <full column rank> and <full row rank>. The square hypothesis matters: a rectangular <matrix> can have <full column rank> without reaching every output, or <full row rank> without determining every input.https://ocw.mit.edu/courses/18-06-linear-algebra-spring-2010/resources/lecture-3-multiplication-and-inverse-matrices/[Strang, MIT 18.06 Lecture 3: multiplication and inverse matrices]{ref}
The equivalences explain why invertibility is a question about retaining information. A <trivial kernel> makes the <linear map> <injective>; the <rank-nullity theorem> then gives <matrix rank> $n$, so its columns form a <basis> and the <linear map> is also <surjective>. Assigning to each output its unique input gives a <linear map>: the preimage of $\alpha b+\beta c$ is $\alpha A^{-1}b+\beta A^{-1}c$. The <determinant> is nonzero precisely when the columns are <linearly independent>. Thus algebraic solvability and independent directions describe the same phenomenon; over the <real numbers>, nonzero <oriented volume> supplies its geometric interpretation.
Back to article page
