Prove that there is a base of $\mathbb R^4$ made of eigenvectors of matrix $A$ Matrix of linear operator $\mathcal A$:$\mathbb R^4$ $\rightarrow$ $\mathbb R^4$ is $$A=
        \begin{bmatrix}
        1 & 1 & 1 & 1 \\
        1 & 1 & -1 & -1\\
        1 & -1 & 1 & -1\\
        1 & -1 & -1 & -1\\
        \end{bmatrix}
$$
Prove that there is a base of $\mathbb R^4$ made of eigenvectors of matrix $A$. Using the new base, find matrix of that operator.
I hope I translated all correctly.
This is what I have done so far.


*

*I found characteristic polynomial of matrix $A$, so I can get eigenvalues and thus find eigenvectors. My characteristic polynomial is $$p_A(\lambda)=\lambda^4-2\lambda^3-6\lambda^2+16\lambda-8$$

*My eigenvalues are
$$\lambda_1=\lambda_2=2  $$ $$\lambda_3=-1-\sqrt3$$ $$
\lambda_4=-1+\sqrt3 $$

*After that I calculated my eigenvectors. This is where I need help understanding. Eigenvectors that belong to different eigenvalues are linearly independent so then they can make a base. In this case, I have two equal eigenvalues. But, when I calculate: $$A\overrightarrow v=\lambda_1 \overrightarrow v$$ where $\overrightarrow v=(x_1,x_2,x_3,x_4)$ is eigenvector for eigenvalue 2 I get this form (final):$$[A-\lambda_1I]=
    \begin{bmatrix}
    -1 & 1 & 1 & 1 \\
    0 & 0 & 0 & -2\\
    0 & 0 & 0 & 0\\
    0 & 0 & 0 & 0\\
    \end{bmatrix}$$


So, my vector $$\overrightarrow v=
    \begin{bmatrix}
    x_2+x_3 \\
    x_2\\
    x_3\\
    0\\
    \end{bmatrix}=x_2\begin{bmatrix}
    1 \\
    1\\
    0\\
    0\\
    \end{bmatrix}+x_3\begin{bmatrix}
    1 \\
    0\\
    1\\
    0\\
    \end{bmatrix}$$
So, I am not even sure how to ask this question. Even if eigenvalues where the same, I did get one vector that is actually a linear combination of two linearly independent vectors? Is this observation correct? 
After that I calculated eigenvectors for remaining eigenvalues. These were results:
$\overrightarrow v_3 = x'_4\begin{bmatrix}
    -\sqrt3 \\
    \sqrt3\\
    \sqrt3\\
    1\\
    \end{bmatrix}$ where  $\overrightarrow v_3=(x'_1,x'_2,x'_3,x'_4)$ for $\lambda_3=-1-\sqrt3$
$\overrightarrow v_4 = x''_4\begin{bmatrix}
    \sqrt3 \\
    -\sqrt3\\
    -\sqrt3\\
    1\\
    \end{bmatrix}$ where $\overrightarrow v_4=(x''_1,x''_2,x''_3,x''_4) $ for $\lambda_4=-1+\sqrt3$
So, in this case, is my base:
$$B=
    \begin{bmatrix}
    1 & 1 & -\sqrt3 & \sqrt3 \\
    1 & 0 & \sqrt3 & -\sqrt3\\
    0 & 1 & \sqrt3 & -\sqrt3\\
    0 & 0 & 1 & 1\\
    \end{bmatrix}$$?
and would new matrix of operator $\mathcal A$ be $B^{-1}AB$?
I also have one more question: Is there some shorter way in finding these results? I am not lazy to do these calculations, but it is easy to make a mistake when time is short. Could I conclude something by looking at matrix $A$ to help me find eigenvalues and eigenvectors faster?
Thank you all in advance.
 A: Note that your matrix $A$ is symmetric and hence diagonalizable. You don't even need to find the eigenvalues of $A$ to conclude that there exists a basis of eigenvectors for $A$. I don't see any calculation-free way to find the eigenvalues of $A$ but once you find them, you don't need to know the eigenvectors in order to know how the operator will look with respect to a basis of eigenvectors. If the eigenvectors are $v_1, \dots, v_4$ with $Av_i = \lambda_i v_i$ then with respect to $(v_1, \dots, v_4)$ the operator will be $\operatorname{diag}(\lambda_1, \dots, \lambda_4)$. 
If you are not asked explicitly to find a basis of eigenvectors for $A$, you can skip 3 entirely and say that $A$ can be represented as $\operatorname{diag}(2,2,-1-\sqrt{3},-1+\sqrt{3})$ (or by any matrix that is obtained by permuting the rows).
Last comment - the trace of your matrix is 2 and this should be the sum of the eigenvalues $\lambda_1 + \dots + \lambda_4$. This can be used for "sanity check" after calculating the eigenvalues to make sure you haven't done a computation error (this doesn't guarantee that you haven't made a mistake but provides some evidence for it).
A: Carl Meyer
Matrix Analysis and Applied Linear Algebra (2000)
$\S$ 7.2, eqn 7.2.5, p 512
Diagonalizability and Multiplicities
The matrix $\mathbf{A}\in\mathcal{C}^{n\times n}$ is diagonalizable iff 
$$
  geometric\ multiplicity _{\mathbf{A}} \left( \lambda \right) = 
  algebraic\ multiplicity _{\mathbf{A}} \left( \lambda \right)
$$
for each $\lambda\in\sigma \left( \mathbf{A} \right)$. That is, iff every eigenvalue is semisimple.
Application
You have identified the eigenvalues that their algebraic multiplicities. The issue is to quantify the geometric multiplicity of the eigenvalue $\lambda = 2$.
The geometric multiplicity 
$$
geometric\ multiplicity _{\mathbf{A}} \left( 2 \right) = 
\dim N \left( \mathbf{A} - 2 \mathbf{I}_{\,4} \right)
$$
$$
\mathbf{A} - 2 \mathbf{I}_{\,4} = 
\left[
\begin{array}{rrrr}
 -1 & 1 & 1 & 1 \\
 1 & -1 & -1 & -1 \\
 1 & -1 & -1 & -1 \\
 1 & -1 & -1 & -3 \\
\end{array}
\right]
$$
The row reduction process is immediate and leaves
$$
\left[
\begin{array}{rrrr}
 1 & -1 & -1 & 0 \\
 0 & 0 & 0 & 1 \\
 0 & 0 & 0 & 0 \\
 0 & 0 & 0 & 0 \\
\end{array}
\right].
$$
The rank of this matrix is 2; therefore the geometric multiplicity is 2. Therefore
$$
  geometric\ multiplicity _{\mathbf{A}} \left( 2 \right) = 
  algebraic\ multiplicity _{\mathbf{A}} \left( 2 \right)
$$
and $\mathbf{A}$ is diagonalizable.
