Inverse of Linear Transformation is Linear Transformation

Theorem

Let $K$ be a field.

Let $V$ and $U$ be vector spaces over $K$.

Let $A : V \to U$ be an invertible (in the sense of a mapping) linear transformation with inverse mapping $A^{-1} : U \to V$.

We aim to show that:

$\map {A^{-1} } {\alpha x + \beta y} = \alpha A^{-1} x + \beta A^{-1} y$

for all $x, y \in U$ and $\alpha, \beta \in K$.

Since $A$ is a linear transformation, we have:

$\map A {\alpha u + \beta v} = \alpha A u + \beta A v$

for each $u, v \in V$.

Note that we have $A^{-1} x \in V$ and $A^{-1} y \in V$, so that:

$\map A {\alpha A^{-1} x + \beta A^{-1} y} = \alpha x + \beta y$

for each $x, y \in U$ and $\alpha, \beta \in K$.

Composing with $A^{-1}$, we then have:

$\map {\paren {A^{-1} \circ A} } {\alpha A^{-1} x + \beta A^{-1} y} = \map {A^{-1} } {\alpha x + \beta y}$

From the definition of an inverse mapping, we therefore have:

$\ds \map {A^{-1} } {\alpha x + \beta y} = \alpha A^{-1} x + \beta A^{-1} y$

for all $x, y \in U$ and $\alpha, \beta \in K$.

So:

$\blacksquare$