Let $V$ be a finite dimensional vector space over a field $F$ and $T$ a linear operator on $V$ such that $T^2 = I_V$.

Question

Let $V$ be a finite dimensional vector space over a field $\mathbb{F}$ and $T$ a linear operator on $V$ such that $T^2 = I_V$. If $\mathbb{F} = \mathbb{R}$ or $\mathbb{C}$, show that $T$ is diagonalizable!

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I have no idea how to do this question, the best i did was to have take any basis $B$, we have $$[T^2]_B = [T]_B[T]_B = [I_V]_B$$ which is pretty meaningless.

My answer sheet used a way in which i do not understand, it claimed that any $V$ in this situation will be a direct sum of the eigenspace of $1$ and $-1$.

Anyone has a better proof or direct me to the right direction!

THanks

Answer 1

Let $v\in V$ and define$$v_1=T(v)+v=(T+\operatorname{Id})(v)\text{ and }v_{-1}=T(v)-v=(T-\operatorname{Id})(v).$$Then$$(T-\operatorname{Id})(v_1)=(T^2-\operatorname{Id})(v)=0$$and therefore, $v_1$ is an eigenvector of $T$ with eigenvalue $1$ (unless $v_1=0$). For the same reason, $v_{-1}$ is an eigenvector of $T$ with eigenvalue $-1$ (again, unless $v_{-1}=0$). On the other hand,$$v_1-v_{-1}=2v\text{ and so }v=\frac12v_1-\frac12v_{-1}.$$So, this says that $V$ is the sum of the eigenspace which corresponds to the eigenvalue $1$ with the eigenspace which corresponds to the eigenvalue $-1$. Since it is clearly a direct sum, the problem is solved.

Answer 2

Proof:

We have

$$ \langle T^2x,x\rangle =\langle TTx,x\rangle = \langle x,T^*T^*x\rangle = \langle x,(T^2)^*x\rangle =\langle x,{I_V}^*x\rangle = \langle x,{I_V}x\rangle = \langle x,T^2x\rangle$$

and hence $T$ is self-adjoint (which implies it is also normal).

For $\mathbb{F} =\mathbb{C}$:

by fundamental theorem of algebra $\det(T-tI_V)$ splits over $\mathbb{C}$. Then by Schur's theorem there exists an orthonormal basis $\beta$ such that $[T]_\beta$ is uppertriangular. Since $T$ is normal $TT^* = T^*T$, then $[T]_\beta [T]_{\beta}^* = [T]_{\beta}^* [T]_\beta$ and hence $[T]_\beta$ is diagonal.

For $\mathbb{F} = \mathbb{R}$:

Since $T$ is self-adjoint then $\det(T-tI_V) = \prod_{i=1}^{\text{dim} V} (\lambda_i-t)$ splits over $\mathbb{R}$ where $\lambda_i \in \mathbb{R}$. Then by Schur's theorem there exists an orthonormal basis $\beta$ such that $[T]_\beta$ is uppertriangular and hence $[T]_\beta$ is diagonal.

Answer 3

It suffices to show that every generalised eigenvector of $T$ is an eigenvector of $T.$ Let $v \neq 0$ be a generalised eigenvector of $T$ with corresponding eigenvalue $c.$ By induction, we only need consider the case where $(T-cI)^2(v)=0.$ (why?) Clearly $c \neq 0$ since if $c=0$ then we'd have $0 = T^2v =v,$ contradiction. Since $c$ is an eigenvalue and $T^2 =I,$ we have $c^2 =1.$

Let $w =(T-cI)v$ so that $Tw =cw$. Then we have:

$0=(T^2 -I)(v)= (T+cI)(T-cI)v =(T+cI)w = 2cw$ which implies $w=0,$ that is $Tv =cv$ and $v$ is an eigenvector of $T,$ from which it follows that $T$ is diagonalisable.

Answer 4

Note that $T$ is an involutive matrix, so the only eigen values of $T$ are $-1$ and $1$. Thus the characteristic polynomial is $\text{char}(x)=(x-1)^{\text{AM}(1)}(x+1)^{\text{AM}(-1)}$, where $\text{AM}(1)+\text{AM}(-1)=\text{dim} V$, while minimal polynomial $m(x)=(x-1)(x+1)$. We then see that the minimal polynomial splits into linear factors. It is an equivalent statement of saying that the matrix is diagonalizable.