$T:V→V$ is a linear transformation such that $T\circ T(x)$ is invertible. Prove that $T$ is also invertible.

Question

Let $V$ be a vector space with dimension $n\in\mathbb{N}$ and $T:V→V$ a linear transformation such that $T\circ T(x)$ is invertible. Prove that $T$ is also invertible.

I'm thinking to use the Theorem that states: If $T:V→W$ is an invertible linear transformation with inverse $T^{-1}:W→V$, then $T^{-1}$ is a linear transformation.

Any tips on how I should go about this problem?

Answer 1

Hint:

For any maps $f:X\to Y$, $g:Y\to Z$,

• $g\circ f$ injective implies $f$ injective,
• $g\circ f$ surjective implies $g$ surjective.

Answer 2

Hint: $\det(T^2)=(\det T)^2$.

Answer 3

If $T\circ T$ has inverse then exists $S$ which gives $(T\circ T)\circ S=1\!\!1$, but if we use composition's associativity then $T\circ (T\circ S)=1\!\!1$ too, therefore $T$ has inverse, which is $T\circ S$.

Answer 4

A proof by contraposition is swift and easy:

Suppose $T$ were not invertible; then

$\exists V \ni x \ne 0, \; Tx = 0; \tag 1$

then

$T^2(x) = T(Tx) = T(0) = 0, \tag 2$

and thus $T^2$ is not invertible either.

Now taking the contrapositive yields the desired result.