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:
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$T$ is invertible.
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$T$ is an injection ie. the kernel of $T$ is trivial.
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$T$ is a surjection.
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$T$ is a bijection.
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The rank of the matrix/ rank of the operator/dimension of the image of $T$ is $n$.
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The determinant of $T$ is non-zero i.e. $T$ is not singular.
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All the column vectors of $T$ are linearly independent.
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None of the eigenvectors of $T$ is zero.
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$T$ maps any set of independent vectors into another set of independent vectors.
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There exists a natural number m such that $det(T^m) \neq 0.$
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dim (range $T$) = rank($T$) = dim($V$).
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The reduced row echelon form of $A$ is $I_n.$
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The equation $T(x)=0$ has only one solution namely $0.$
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$Ax=b$ has a unique solution for any given vector $b$.
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The span of the columns of $A$= column space = $K^n.$
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The span of rows of $A$= row space of $A$= $K^n.$
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There exists an $(n\times n)$ matrix $B$ such that $AB = I_n = BA.$
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The transpose of $A$ namely $A^t$ is invertible.
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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.
Best Answer
Compiling statements into a community wiki answer that can be expanded.
Topology / Norms, $T$ is invertible iff:
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:
Algebra, $T$ is invertible iff: