# Proving that a square matrix whose kernel is $\{0\}$ is invertible?

This is not an assignment or anything from my book, just a pure interest. If I wanted to show this, wouldn't I have to say that the square matrix can be reduced into the identity square matrix and then say that the determinant is not zero. After I say that and use the fact that the only element in the kernel is the zero vector, could I conclude that the matrix is invertible?

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I would do it using the rank-nullity theorem and properties of linear maps. If $Ker(A) = {0}$ Then the linear map which the matrix represents is injective and has nullity zero. Hence the rank of the matrix is $n$ (if the matrix is $n\times n$,) which means that it is a surjective map. Since the map is linear and a bijection, it is an isomorphism which means it has an inverse. Hence the matrix must be invertible.
Imagine doing Gauss-Jordan elimination on the matrix $A$. The result is a matrix $GA$ in reduced row echelon form, where $G$ is invertible. If $GA=I$, then $A$ is invertible (and $G$ is its inverse). Otherwise there is a non-pivot column, and you can use that to construct a nonzero column vector $X$ such that $(GA)X=0$. But since $G$ is invertible, $G(AX)=0$ implies $AX=0$ so $X$ is in the kernel of $A$, and the kernel is therefore nontrivial.