When an operator/matrix is invertible? Let $T$ be a linear operator on a finite-dimensional (n-dimensional) vector space $V$ over a field $K.$ Suppose $A$ is the matrix representation of $T$ with respect to a given basis for $V$. We can see that the following statements are equivalent:


*$T$ is invertible.


*$T$ is an injection ie. the kernel of $T$ is trivial.


*$T$ is a surjection.


*$T$ is a bijection.


*The rank of the matrix/ rank of the operator/dimension of the image of $T$ is $n$.


*The determinant of $T$ is non-zero i.e. $T$ is not singular.


*All the column vectors of $T$ are linearly independent.


*None of the eigenvectors of $T$ is zero.


*$T$ maps any set of independent vectors into another set of independent vectors.


*There exists a natural number m such that $det(T^m) \neq 0.$


*dim (range $T$) = rank($T$) = dim($V$).


*The reduced row echelon form of $A$ is $I_n.$


*The equation $T(x)=0$ has only one solution namely $0.$


*$Ax=b$ has a unique solution for any given vector $b$.


*The span of the columns of $A$= column space = $K^n.$


*The span of rows of $A$= row space of $A$= $K^n.$


*There exists an $(n\times n)$ matrix $B$ such that $AB = I_n = BA.$


*The transpose of $A$ namely $A^t$ is invertible.


*The matrix $A$ can be expressed as a finite product of elementary matrices.
This list can be extended. I can clearly see their equivalence. Using three basic concepts:
(a) the rank-nullity theorem/dimension theorem,
(b) $det(AB)= det(A) det(B)$, and
(c) $det(A)$= product of eigenvalues of $T$
one can easily prove the equivalence of these statements. Now, could you provide some "less obvious" statements which are also equivalent to the invertibility of $T$? Thank you for your time.
Thanks so much.
 A: Compiling statements into a community wiki answer that can be expanded.
Topology / Norms, $T$ is invertible iff:

*

*$T$ is bounded below, meaning for any norm on the space there is a $C$ so that $\|T v\|≥ C\|v\|$ for all vectors $v$.

*$T$ is an open map.

*For any linear map $A$ there is an $\epsilon >0$ so that $T+ rA$ is invertible (ie satisfies any of the other criteria) for all $|r|≤\epsilon$.

Consider a scalar product $(,)$ on the vector space. Call a decomposition $T=UA$ with $U$ unitary and $A≥0$ a polar decomposition. $T$ is invertible iff:

*

*The unitary in the polar decomposition is unique.

*$A$ in the polar decomposition is strictly positive, ie $(x,Ax)\neq0$ for all $x\neq0$.

Algebra, $T$ is invertible iff:

*

*The linear map $L_T: \mathrm{End}(V)\to\mathrm{End}(V)$, $A\mapsto TA$ is invertible. The same for the map $R_T(A)=AT$.

*The ideal $\{ TAT\mid A\in\mathrm{End}(V)\}$ is all of $\mathrm{End}(V)$.

*One has $\mathrm{Tr}(\rho T^*T)>0$ for any non-zero matrix $\rho≥0$.

