Prove that if $n\times n$ matrix $A$ is nonsingular, then it is invertible. I know that nonsingularity means having unique soln. $x$ for every RHS b in the system $Ax= b$. But how do I use this definition to come up with a proof for the problem ? I really have tried but to no avail.
 A: Consider the linear transformation associated to $A$ where $L_{A}:\textbf{F}^{n}\to\textbf{F}^{n}$ is given by $L_{A}x = Ax$.
Thus $A$ is nonsingular iff $L_{A}$ is injective. Consequently, due to the rank-nullity theorem, $L_{A}$ is invertible, which is only possible if $[L_{A}]_{\mathcal{B}}$ is invertible. But $[L_{A}]_{\mathcal{B}} = A$, whence we conclude that $A$ is invertible.
Here $\mathcal{B} = \{e_{1},e_{2},\ldots,e_{n}\}$ is the standard basis of $\textbf{F}^{n}$.
A: Recall that
$$\mathbf{M}^{-1}=\frac{1}{\det(\mathbf{M})}\operatorname{adj}(\mathbf{M})$$
Simply show that $\operatorname{adj}(\mathbf{M})$ always exists (this is not difficult), and thus $\mathbf{M}^{-1}$ always exists if $\det(\mathbf{M}) \neq 0$.
A: An even more direct proof without presuming knowledge of as much as determinants:
You can solve $A\vec{x}_1=[1,0,\dotsc,0]^T$ for $\vec{x}_1$. Consider the matrix formed by $\vec{x}_1$ then $n-1$ columns of zeros, call it $B$.
Then it is clear that $AB=\begin{bmatrix}1&0&\dotsc\\0&0&\dotsc\\\vdots&\vdots&\ddots\end{bmatrix}$ by the properties of matrix multiplication.
Next, solve for $A\vec{x}_2=[0,1,\dotsc,0]^T$, $A\vec{x}_3=[0,0,1,\dotsc,0]^T$ and so on.
Constructively, the matrix $A^{-1}=\begin{bmatrix}\vec{x}_1&\vec{x}_2&\vec{x}_3&\dotsc\end{bmatrix}$
satifies $AA^{-1}=I$.
