# Show that a square matrix with linear transformation T(M) = A·M is bijective when A is inversible

Suppose $K$ is a body (a field), $n ≥ 1$ and $A ∈ M_n(K)$ a fixed matrix. Consider the linear transformation $T : M_n(K) → M_n(K)$ defined by $T(M) = A · M$ for $M ∈ M_n(K)$

The mark scheme says that if A is invertible then $\exists A^{−1} ∈ M_n(K)$ such that $A^{−1}A = I_n$. If $T(M) = AM = 0$, then we multiply by $A^{-1}$ on the left to get $M = A^{−1}AM = 0$.

How does this multiply by $A^{-1}$? Wouldn't it be $A^{-1}M = A^{-1}AM = 0$?

And how does what they obtain show that it is injective and therefore bijective?

We need two Lemmas:

1) A linear transformation $T:V \rightarrow W$ is injective iff $\ \forall v \in V \ \ \ \ \ Tv = 0 \implies v = 0$

2) If $T:V \rightarrow V$ is an injective linear transformation from a finite dimensional vector space $V$ to itself, then $T$ is a bijection.

To use (1) suppose $A \cdot M = 0$ for some matrix $M \ \$ (this is our $Tv = 0$) . As $A$ is invertible we can multiply on the left by $A^{-1}$ to get:

$$M = IM = A^{-1}A M = A^{-1} 0 = 0$$

To use (2) note our map is from the vector space $M_n(K)$ to itself, and we have just shown that it is injective.