Equivalent Characterizations of Invertible Matrix

Theorem

Let $\mathbf A$ be a square matrix of order $n$ over a field $K$.

The following statements are equivalent:

$(1):\quad$ $\mathbf A$ is invertible

$(2):\quad$ The transpose of $\mathbf A$ is invertible

$(3):\quad$ $\mathbf A$ row-reduces to the identity matrix $\mathbf I_n$

$(4):\quad$ The null space of $\mathbf A$ contains only the zero vector $\mathbf 0$

$(5):\quad$ The homogeneous linear system $\mathbf A \mathbf x = \mathbf 0$ has only the trivial solution $\mathbf 0$

$(6):\quad$ The linear system $\mathbf A \mathbf x = \mathbf b$ has a unique solution for every possible choice of $\mathbf b$

$(7):\quad$ The columns of $\mathbf A$ are linearly independent

$(8):\quad$ The column space of $\mathbf A$ is $K^n$

$(9):\quad$ The columns of $\mathbf A$ are a basis for $K^n$

$(10):\quad$ The rows of $\mathbf A$ are linearly independent

$(11):\quad$ The row space of $\mathbf A$ is $K^n$

$(12):\quad$ The rows of $\mathbf A$ are a basis for $K^n$

$(13):\quad$ The rank of $\mathbf A$ is $n$

$(14):\quad$ The nullity of $\mathbf A$ is zero

$(15):\quad$ The determinant of $\mathbf A$ is not zero

$(16):\quad$ $0_K$ is not an eigenvalue of $\mathbf A$

Proof

$(3)$ iff $(5)$

See Elementary Row Operation on Augmented Matrix leads to Equivalent System of Simultaneous Linear Equations/Corollary 5.

$\Box$

$(1)$ iff $(3)$

See Matrix Inverse Algorithm.

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$(4)$ iff $(5)$

Follows directly from the definition of a null space.

$\Box$

$(3)$ implies $(6)$

Suppose that $\mathbf A$ row-reduces to the identity matrix $\mathbf I_n$.

Let $\mathbf b$ be arbitrary.

Let $\begin {pmatrix} \mathbf I_n & \mathbf s \end {pmatrix}$ be a reduced echelon matrix derived from $\begin {pmatrix} \mathbf A & \mathbf b \end {pmatrix}$.

By Elementary Row Operation on Augmented Matrix leads to Equivalent System of Simultaneous Linear Equations, the linear system $\mathbf A \mathbf x = \mathbf b$ has the same solutions as $\mathbf I_n \mathbf x = \mathbf s$.

But $\mathbf I_n \mathbf x = \mathbf x$, so $\mathbf x = \mathbf s$ is the unique solution to $\mathbf I_n \mathbf x = \mathbf s$.

Thus $\mathbf x = \mathbf s$ is the also the unique solution to $\mathbf A \mathbf x = \mathbf b$.

As $\mathbf b$ is arbitrary, the result follows.

$\Box$

$(6)$ implies $(5)$

The result follows from setting $\mathbf b = \mathbf 0$.

$\Box$

$(4)$ iff $(7)$

See Null Space Contains Only Zero Vector iff Columns are Independent.

$\Box$

$(7)$ iff $(8)$ iff $(9)$

Follows directly from Sufficient Conditions for Basis of Finite Dimensional Vector Space.

$\Box$

$(3)$ iff $(10)$

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$(10)$ iff $(11)$ iff $(12)$

Follows directly from Sufficient Conditions for Basis of Finite Dimensional Vector Space.

$\Box$

$(8)$ iff $(13)$

Follows from the definition of rank of matrix and $\map \dim {K^n} = n$.

$\Box$

$(4)$ iff $(14)$

Follows from the definition of nullity and $\map \dim {\set {\mathbf 0} } = 0$.

$\Box$

$(1)$ iff $(15)$

See Matrix is Invertible iff Determinant has Multiplicative Inverse.

$\Box$

$(5)$ iff $(16)$

From the definition of an eigenvalue:

$0_K$ is an eigenvalue of $\mathbf A$ if and only if there exists a non-zero vector $\mathbf v$ such that $\mathbf {A v} = 0_K \mathbf v = \mathbf 0$

Therefore $\mathbf A \mathbf v = \mathbf 0$ has only the trivial solution $\mathbf 0$ if and only if $0_K$ is not an eigenvalue of $\mathbf A$.

$\Box$

$(1)$ iff $(2)$

By Matrix is Invertible iff Determinant has Multiplicative Inverse:

$\mathbf A$ is invertible if and only if the determinant of $\mathbf A$ is not zero

By Determinant of Transpose:

$\map \det {\mathbf A} = \map \det {\mathbf A^\intercal}$

where $\map \det {\mathbf A}$ is the determinant of $\mathbf A$ and $\mathbf A^\intercal$ is the transpose of $\mathbf A$.

Therefore the determinant of $\mathbf A$ is not zero if and only if the determinant of $\mathbf A^\intercal$ is not zero.

By Matrix is Invertible iff Determinant has Multiplicative Inverse again:

$\mathbf A^\intercal$ is invertible if and only if the determinant of $\mathbf A^\intercal$ is not zero

Hence $\mathbf A$ is invertible if and only if $\mathbf A^\intercal$ is invertible.

$\Box$

Hence all sixteen statements are logically equivalent.

$\blacksquare$