Let $T$ be a linear transformation on a vector space $V$ ($\textrm{dim}\ V = n$). If $\textrm{rank}\ (T^2) = n$, is $T$ invertible?

Question

For a linear transformation $T$ on a finite dimensional vector space $V$ ($\textrm{dim}\ V = n$). If $\textrm{rank}\ (T^2) = n$, is $T$ invertible? Also, is it guaranteed to have an eigenvalue?

Answer 1

Since $\mathrm{rank}(T^2)=n$, we have $\det(T^2)\ne 0$. Hence $$ \det(T^2) = \det T\det T\ne 0,$$ which implies that $\det T\ne 0$, and thus that $T$ is invertible.

$T$ need not have a real eigenvalue though. Consider a rotation matrix, e.g. $$\begin{bmatrix}0&1\\-1&0\end{bmatrix}.$$

$T^2=I$ implies that $$0 = T^2 - I = (T-I)(T+I).$$ So $T-I$ is not invertible and $1$ is an eigenvalue or $T+I$ is not invertible and $-1$ is an eigenvalue.

Answer 2

Without using determinants: supose

$$x\in\ker T\implies Tx=0\implies T(Tx)=T^2x=0$$

But $\;T^2\;$ is regular (=invertible in this context) $\;\iff\ker T^2=0\;$ , so we get

$$x=0\implies \ker T=\{0\}\iff T\;\;\text{is regular}$$

As for eigenvalues: take for example

$$T=\begin{pmatrix}1&1\\\!\!-1&0\end{pmatrix}\implies T^2=\begin{pmatrix}0&1\\\!\!-1&\!\!-1\end{pmatrix}$$

and this last mapping has no (real) eigenvalues.

Answer 3

$\textrm{rank}\ (T^2) = n$ implies that $T^2$ is invertible, by the rank-nullity theorem.

If $S$ is the inverse of $T^2$, then $I=ST^2=(ST)T$ and $I=T^2S=T(TS)$. This proves that $T$ is both injective and surjective, and so invertible. (We only need injective or surjective, the other comes from the rank-nullity theorem.)

Answer 4

This is a way to solve this problem without using determinants.

Note that $R(T^2)\subset R(T)$ so that if $dim(R(T^2))\leq dim(R(T))\leq n=dim(V)$. Since $dim(R(T^2))=n$ we have $n\leq dim(R(T))\leq n$, so that $dim(R(T))=n$. This means that the operator $T$ has full rank.