Prove that $f$ is bijective?

Question

$(e_1, e_2, e_3)$ is a the canonical basis of $\mathbb{R}^3$.

$f: \mathbb{R}^3 \rightarrow \mathbb{R}^3$ is a linear map such that:

$ \begin{cases} f(e_1) = \frac{1}{3}(-e_1 + 2e_2 + 2e_3) \\ f(e_2) = \frac{1}{3}(2e_1 - e_2 + 2e_3) \\ f(e_3) = \frac{1}{3}(2e_1 + 2e_2 - e_3) \end{cases} $

We have: $E_{-1} = \{ u \in \mathbb{R}^3 | f(u) = -u\}$ and $E_{1} = \{ u \in \mathbb{R}^3 | f(u) = u\}$

1. Prove that $E_{-1}$ and $E_{1}$ are sub-vector spaces of $\mathbb{R}^3$.
2. Prove that $e_1 - e_2 \in E_{-1}$ and $e_1 - e_3 \in E_{-1} $ and $e_1 + e_2 + e_3 \in E_{1}$.
3. What can we deduce about the dimensions of $E_{-1}$ and $E_{1}$?
4. Determine $E_{-1} \cap E_{1}$.
5. Do we have $E_{-1} \oplus E_{1} = \mathbb{R}^3 $ ?
6. Compute $f^2 = f \circ f$, deduce that $f$ is bijective and determine $f^{-1}$.

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I am stuck in question 3. I don't know how to deduce the dimensions of $E_{-1}$ and $E_{1}$ from $e_1 - e_2 \in E_{-1}$ and $e_1 - e_3 \in E_{-1} $ and $e_1 + e_2 + e_3 \in E_{1}$.

I have problem with question 6 too, I have written the matrix $A$ of the linear map which is:

$A = \frac{1}{3} \begin{pmatrix} -1 & 2 & 2 \\ 2 & -1 & 2 \\ 2 & 2 & -1 \\ \end{pmatrix} $

and $A^2 = I_3$

Which means $A^{-1} = A$.

and $f^{-1}(x,y,z) = \frac{1}{9}(-x + 2y + 2z , 2x - y + 2z , 2x + 2y - y)$

How can I deduce $f$ is bijective from $f^2$ ?

Although, is $f^{-1}$ correct?

Answer 1

As $E_{-1}$ contains two linearly independent vectors, certainly $\dim E_{-1}\ge 2$. Likewise, $\dim E_1\ge 1$. With the answer to 4., yoiu dhould be able to deduce that $\dim E_{-1}+\dim E_1\le 3$. This then implies $\dim E_{-1}=2$ and $\dim E_1=1$.

For question 6, you better do not express $A$ in terms of the standard basis - that leads to clumsy computations with awfully many numbers. Instead, what is $f(f(u))$ for $u\in E_{-1}$? What is $f(f(u))$ for $u\in E_1$ (both by the very definition of $E_{\pm1}$ as set)= What does this tell you about $f\circ f$?