Trace of involution is the dimension of the space implies it is the identity.

Question

I was trying to write explicitly a linear automorphism $\phi$ on a 2-dimensional vector space $V$. I knew that $\phi^2=Id$ and $\text{Tr } \phi=2$. Using the matrix representation $\phi=\begin{pmatrix} a& b\\ c& d \end{pmatrix}$, I got the system of equations \begin{align} &a^2+bc=1\\ &ab+bd=0\\ &ac+dc=0\\ &bc+d^2=1\\ &a+b=2. \end{align} From this it is easy to see that only solution is $\phi=Id$. But now I ask wether we could prove this fact without working out a matrix, I mean, in a more elegant way? Also, does this extend to other dimensions; i.e. if $V$ is finite dimensional and $\phi\in\text{Aut}(V)$ is an involution whose trace equals $\text{dim }V$, then $\phi=Id$?

Answer 1

I don’t know what is your background in linear algebra.

But you could say:

• As $\phi^2 =Id$, the minimal polynomial of $\phi$ is either $x-1$, in which case $\phi$ is the identity or $x^2-1$.
• In later case, the minimal polynomial has simple roots, so $\phi$ is diagonalizable.
• $\phi$ being diagonalizable is similar to a diagonal matrix having $1$ and $-1$ for only possible values on the diagonal.
• As its trace is equal to $2$, the only option is that all values on the diagonal are equal to $1$.
• So $\phi$ is similar to the identity.
• But the only endomorphism similar to the identity is the identity itself.

Finally, the identity is the only solution to the problem in dimension two.

For other dimensions, a similar approach proves that the identity is also the only solution, supposing that $Tr \phi = \dim V$.