Writing direct sum of a normed space as $X=\mathcal{N}(T_{\lambda}^{r})\bigoplus T_{\lambda}^{r} (X)$

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I'm reading about some spectral properties of compact operators and I'm doing some problems about it. There is one that needs next theorem:

Theorem. Let $T: X ~ X$ be a compact linear operator on a normed space $X$, and let $\lambda\neq 0.$ Let $n=r$ be the smallest integer (depending of $\lambda$) such that $$\mathcal{N}(T_{\lambda}^{r})=\mathcal{N}(T_{\lambda}^{r+1})=\mathcal{N}(T_{\lambda}^{r+2})=\ldots$$ and $$T_{\lambda}^{r} (X)=T_{\lambda}^{r+1} (X)=T_{\lambda}^{r+2} (X)=\ldots.$$

Then $X$ can be represented in the form $X=\mathcal{N}(T_{\lambda}^{r})\bigoplus T_{\lambda}^{r} (X).$

The problem says this: Derive the representation of the above theorem in the case of the linear operator $T: \mathbb{R}^2\rightarrow\mathbb{R}^2$ represented by the matrix \begin{pmatrix} 1 & -1 \\ -1 & 1 \end{pmatrix}

I'm stuck solving this. I know that $T$ is a linear bounded operator with finite rank, so $T$ is compact with linear form $T(x,y)=(x-y,-y+x).$ I'm tryng to compute $\mathcal{N}(T_{\lambda}^{r})$ and $T_{\lambda}^{r} (X)$ but I don't get any useful.

Any kind of help is thanked in advanced.

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First show that $T$ has the eigenvalues $0$ and $2$.

Then $T- \lambda I$ is inverible if $0 \ne \lambda \ne 2$, hence $r=0$ and we have

$\mathbb R^2=\mathcal{N}(T_{\lambda}^{0})\bigoplus T_{\lambda}^{0} (X)$ if $0 \ne \lambda \ne 2$.

Now let $ \lambda=0$. Then $T - \lambda I=T$. Show that $T^2=2T$. Therefore $r=1$ , hence $\mathbb R^2=\mathcal{N}(T)\bigoplus T(X)$.

Now let $ \lambda=2$. Then $T - \lambda I=T-2I$. Show that $(T-2I)^2=-2(T-2I)$. Therefore $r=1$ , hence $\mathbb R^2=\mathcal{N}(T-2I)\bigoplus (T-2I)(X)$.