If $T$ is a linear operator. Prove that this 3 conditions are equivalent

Question

I have to prove this problem. I would like you to check if my proof is correct (or not) please.

Let $V$ be a vector space over $\mathbb{R}$ and $T$ a linear operator on $V$. Prove that the following conditions are equivalent:

1. $T^{2}= \text{id}_V$,
2. $V$ is the direct sum of the null space (kernel) of $T- \text{id}_V$ and the null space of $T+ \text{id}_V$,
3. there exist two subspaces, $W$ and $X$, of $V$ such than $V=W \oplus X$ and $T(w+x)=w-x$ for all $w \in W$ and all $x \in X$.

Here, my attempt:

• 3 $\Longrightarrow$ 1

$\forall \alpha \in V$, we have that $\alpha=w+x $ with $ w\in W$, and $x\in X$

\begin{align} \text{If} \ \ \ \ T(\alpha)=T(w+x)&=w-x\\ \Rightarrow \ \ \ T^{2}(\alpha)=T^{2}(w+x)&=T(T(w+x))\\&=T(w-x)\\&=T(w+(-x))\\&=w-(-x)\\&=w+x\\&=\alpha \\ \therefore T^{2}=id_{V} \end{align}

• 1 $\Longrightarrow$ 2

Firt, we have to see this \begin{align} T^{2}=id_{V} &\iff (T^{2}-id_{V})=0\\&\iff(T-id_{V})(T+id_{V})=0\\&\iff (T-id_{V})=0 \ \ \ \ \text{or} \ \ \ \ (T+id_{V})=0\\&\iff \alpha \in N(T-id_{V}) \ \ \ \ \text{or} \ \ \ \ \alpha \in N(T+id_{V}) \end{align} Now, we suppose that any $\alpha \in V$ is of the form $\alpha=w+x$ with $w\in N(T-id_{V})$, and $x \in N(T+id_{V})$. \begin{align} T^{2}(\alpha)=T^{2}(w+x)=T^{2}(w)+T^{2}(x)=0 & \iff \alpha =0 \\& \iff w=x=0\\ \end{align}

\begin{align} \therefore V=N(T-id_{V}) \oplus N(T+id_{V}) \end{align}

• 2 $\Longrightarrow$ 3

First, we have that $V=N(T-id_{V}) \oplus N(T+id_{V})$

$\Longrightarrow$ There exists two subspaces, $W,X$ such that $V=W \oplus X$

Now, if $w \in N(T-id_{V})$

\begin{align} T(w)-id_{V}(w)= 0 \iff T(w)=w \end{align}

And, if $x \in N(T+id_{V})$

\begin{align} T(x)+id_{V}(x)=0 \iff T(x)=-x \end{align}

By last, any vector $\alpha \in V$ is of the form $\alpha=w+x$ (because of the direct sum)

\begin{align} \therefore T(\alpha)=T(w+x)=w-x \end{align}

I hope you can help me to check this, and I really really appreciate your help.

Answer 1

$1\implies2$ doesn't look correct. $(T-I)(T+I)=0\not\Rightarrow T-I=0\vee T+I=0$. Remember that the product of two non-zero matrices may be zero matrix.

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First you need to show that $\ker(T-I)\cap\ker(T+I)=\{0\}$. Suppose $v\in\ker(T-I)\cap\ker(T+I)$, then $(T-I)v=0=(T+I)v$. This gives $Tv=v=-v\iff v=0$.

Next you need to show that every $\alpha\in V$ can be written as the sum of some $v^-\in\ker(T-I)$ and $v^+\in\ker(T+I)$. From $T^2\alpha=I\alpha$ we get $(T-I)(T+I)\alpha=0$. Thus $(T+I)\alpha=u_1\in\ker(T-I)$ which gives$$\alpha=u_1-T\alpha$$Similar we also have $(T+I)(T-I)\alpha=0$. Thus $(T-I)\alpha=u_2\in\ker(T+I)$ which gives$$\alpha=T\alpha-u_2$$Adding the two equations we get$$\alpha=\frac{u_1-u_2}2$$where $u_1/2=v^-\in\ker(T-I)$ and $-u_2/2=v^+\in\ker(T+I)$.