Doubts on obtaining orthonormal basis In finding the matrix $P$ that orthogonally diagonalizes $A$ and to determine $P^TAP$, where $$A = 
    \begin{pmatrix}
    1 & -1 & 1 & -1\\
    -1 & 1 & -1 & 1\\
    1 & -1 & 3 & 1\\
    -1 & 1 & 1 & 3\\
    \end{pmatrix}
$$
I've found the eigenvalues to be $0$ and $4$. 
In order to find the orthonormal basis, i've considered two scenarios where $\lambda = 0$ and $\lambda = 4$:
For $\lambda = 0$, on solving the system, we have $    \begin{pmatrix}
w\\
x\\
y\\
z\\
    \end{pmatrix} $ = $s    \begin{pmatrix}
1\\
1\\
0\\
0\\
    \end{pmatrix} $ + $t    \begin{pmatrix}
2\\
0\\
-1\\
1\
    \end{pmatrix} $
For $\lambda = 0$, on solving the system, we have $    \begin{pmatrix}
w\\
x\\
y\\
z\\
    \end{pmatrix} $ = $s    \begin{pmatrix}
\frac{1}{2}\\
-\frac{1}{2}\\
1\\
0\\
    \end{pmatrix} $ + $t    \begin{pmatrix}
-\frac{1}{2}\\
\frac{1}{2}\\
0\\
1\\
    \end{pmatrix} $
My Doubts:


*

*As for $E_{4}$ as seen above, the answer considers the set of $    \begin{pmatrix}
\frac{1}{2}\\
-\frac{1}{2}\\
1\\
0\\
    \end{pmatrix} $, $    \begin{pmatrix}
-1\\
1\\
0\\
2\\
    \end{pmatrix} $ to be the basis for $E_{4}$. However, why is there a need to scale the basis by two into its integer basis of $    \begin{pmatrix}
-1\\
1\\
0\\
2\\
    \end{pmatrix} $ from $    \begin{pmatrix}
-\frac{1}{2}\\
\frac{1}{2}\\
0\\
1\\
    \end{pmatrix} $. Isn't it the same? 

*The answers present the orthonormal basis set of $E_{0}$ as $    \begin{pmatrix}
\frac{1}{\sqrt2}\\
\frac{1}{\sqrt2}\\
0\\
0\\
    \end{pmatrix} $, $    \begin{pmatrix}
\frac{1}{2}\\
-\frac{1}{2}\\
-\frac{1}{2}\\
\frac{1}{2}\\
    \end{pmatrix} $ and the orthonormal basis set for $E_{4}$ as $    \begin{pmatrix}
\frac{1}{\sqrt6}\\
-\frac{1}{\sqrt6}\\
\frac{2}{\sqrt6}\\
0\\
    \end{pmatrix} $, $    \begin{pmatrix}
-\frac{1}{\sqrt12}\\
\frac{1}{\sqrt12}\\
\frac{1}{\sqrt12}\\
\frac{3}{\sqrt12}\\
    \end{pmatrix} $
They stated that the orthonormal basis can be found by using the Gram-Schmidt Process. However, i'm not really sure how it can be computed. As for $E_{0}$, isn't it true that we divide each integer by $\sqrt (1^2 + 1^2)$ to get $    \begin{pmatrix}
\frac{1}{\sqrt2}\\
\frac{1}{\sqrt2}\\
0\\
0\\
    \end{pmatrix} $ and $\sqrt (2^2+-1^2+1^2)$ to get  $    \begin{pmatrix}
\frac{2}{\sqrt6}\\
0\\
-\frac{1}{\sqrt6}\\
\frac{1}{\sqrt6}\\
    \end{pmatrix} $ but it was apparently not the case as seen above. Why is that so? 
Thanks. 

If you have two vectors $u_1$ and $u_2$, the Gram-Schmidt algorithm produces two vectors $v_1$ and $v_2$ defined by
$$
v_1 = \frac{u_1}{\|u_1\|}
\quad
v_2 = \frac{u_2 - (u_2|v_1)v_1}{\|u_2 - (u_2|v_1) v_1\|}
$$
