Erwin Kreyszig's Introductory Functional Analysis With Applications, Section 2.6, Problem 11

Question

Let $X$ be the vector space of all complex $n \times n$ matrices and define $T \colon X \to X$ by $Tx \colon= bx$, where $b \in X$ is fixed and $bx$ denotes the usual product of matrices. I know that $T$ is linear.

Under what conditions does $T^{-1}$ exist?

If $b$ is an invertible matrix, then of course $T^{-1}$ exists.

But does the existence of $T^{-1}$ necessarily imply the invertibility of the matrix $b$?

What condition(s) (other than invertibility), if any, should $b$ satisfy in order for $T^{-1}$ to exist?

Answer 1

Suppose $T$ has inverse $T^{-1}$. Then $T$ is an injective map between finite dimensional vector spaces. The rank-nullity theorem then implies the map is surjective, so PhoemueX's argument may be applied.

Answer 2

If $T$ is invertible, there is some $x$ such that $bx = Tx = I$, where $I$ is the identity matrix (why?).

Hence?