Find all the linear involutions $f: E \to E$, where $E$ is a finite-dimensional real vector space

Question

Can someone help me?

I've been thinking about this question for a while and got stuck. At first I only found the Identity transformation ($I$) and the anti-Identity transformation ($-I$). But then I realized that every reflection is also an involution. The only relevant information I got about the transformation's matrix is that $A^{-1} = A$ and, of course, $(A - I)\cdot(A + I) = 0$.

Answer 1

You are looking for an $n\times n$ matrix $A$ whose square is equal to the identity. This means that the matrix satisfies the equation $x^2-1$, so its minimal polynomial must be $x-1$, $x+1$ or $x^2-1$. In the case of the first two, we have the two matrices you've already found, that is $I$ and $-I$.

Now, let's suppose the minimal polynomial is $x^2-1$. Notice this factors completely over $\mathbb{R}$ into distinct linear factors, which shows that $A$ is diagonalizable. This means there is an invertible matrix $P$ such that:

$$PAP^{-1}=D$$

where $D$ is a diagonal matrices, whose diagonal elements are the eigenvalues of $A$. In our case, the eigenvalues are $1$ and $-1$ (the roots of the minimal polynomial). Since the Jordan form representation is unique up to permutation of the Jordan blocks, we only need to know the algebraic multiplicity of the eigenvalues, in other words, how many $1$'s and how many $-1$'s are on the diagonal of $D$. Suppose $D_i$ is the $n\times n$ matrix with $i$ $1's$ on the diagonal, where $0\le i\le n$. Notice that $D_0=-I$ and $D_n=I$. Then all such involutions $A$ have the form:

$$A=P^{-1}D_iP$$

Answer 2

Hint: If $T=T^{-1}$ then $T^2-1=0$. So, the minimal polynomial of $T$ divides $x^2-1$, and thus is separable and splits over $\mathbb{R}$. So, $T$ is diagonalizable.

Answer 3

Geometrically, an involution $f:E \to E$ corresponds to a direct sum decomposition $E = E_1 \oplus E_{-1}$ into the eigenspaces of $f$.

In a bit more detail, if $E = V \oplus W$ is an arbitrary direct sum decomposition of $E$, there exists a unique involution $f:E \to E$ whose restriction to $V$ is the identity and whose restriction to $W$ is minus the identity.

Conversely, if $f:E \to E$ is an involution, then every vector $v$ in $E$ may be written $$v = \tfrac{1}{2}\bigl(v + f(v)\bigr) + \tfrac{1}{2}\bigl(v - f(v)\bigr),$$ and these summands are easily checked to be eigenvectors of $V$ with respective eigenvalues $1$ and $-1$. (This identity gives an explicit proof that $f$ is diagonalizable on $E$: Every vector in $E$ is a sum of eigenvectors of $f$.)