Can I find the basis for the eigenspaces of an operator just having the diagonal matrix??

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I have been thinking this exercise for a while and I am really clueless about how to construct the basis of the eigenspaces . ( I point out as always, that english is not my language, though I manage to translate the exercise from spanish and understand the answers you give)

Exercise: Considering the canonical inner product, let T be an linear operator, $T:{R}^{4} \mapsto {R}^{4}$ and A and B non trivial subspaces such as:

$T_{ | A}=Id $ , $T_{ | B}= -Id , $ , $ N\left ( T-2Id \right ) = \left ( A+B \right )^{\perp }$ and $rank\left ( T-2Id \right )=2$

  1. Find a basis for ${R}^{4}$ consisting of eigenvectors of T
  2. Prove T is invertible
  3. Let $S:{R}^{4}\mapsto {R}^{4}$ such as $S=T^{3}+4T^{-1}+2Id$ . Prove S is diagonalizable and find the eigenvalues.
  4. If $\left \langle x,z \right \rangle=0, \forall x \in X, \forall x \in Z$, could it be possible to find an orthonormal basis of $T:{R}^{4}$ consisting of eigenvectors of T?

Here my thoughts:

  • A has at least dimension 1, and is associated to T eigenvalue 1
  • B has at least dimension 1, and is associated to T eigenvalue -1
  • T also has eigenvalue 2 with geometric multiplicity 2
  • Based on the assumptions above I say that T is associated to this diagonal matrix D= $\begin{bmatrix} 2 & 0 & 0 & 0 \\ 0 & 2 & 0 & 0\\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & -1 \end{bmatrix}$
    • As the matrix above is non singular, T is invertible.
    • It is easy to find S and it s diagonal matrix, with eigenvalues 12, 12, 7 and -3.

I still dont have the slightless idea of how to complete items 1 and 4 or how to relate this to any inner product.

Could somebody be so kind to give me a clue ? I am studying on my own and sitting next saturday for a test