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 $$ v_1 = \frac{u_1}{\|u_1\|} \quad v_2 = \frac{u_2 - (u_2|v_1)v_1}{\|u_2 - (u_2|v_1) v_1\|} $$