I haven’t done this in quite some time, so this solution is probably unnecessary complicated:
We identify $\mathbb{R}^{2 \times 2}$ with $\mathbb{R}^4$ via
$$
\mathbb{R}^{2 \times 2} \to \mathbb{R}^4, \,
\begin{pmatrix}
x & y \\
z & t
\end{pmatrix}
\mapsto
(x,y,z,t)^T.
$$
(So the “default basis” you used corresponds to the standard basis $(e_1, e_2, e_3, e_4)$ of $\mathbb{R}^4$.) If we understand $L$ as a linear map $\hat{L} \colon \mathbb{R}^4 \to \mathbb{R}^4$ then $\hat{L}$ is (with respect to the standard basis on both sides) given by the matrix
$$
A =
\begin{pmatrix}
1 & 1 & 0 & 1 \\
1 & 1 & 1 & 0 \\
0 & 1 & 1 & 1 \\
1 & 0 & 1 & 1
\end{pmatrix}.
$$
Also notice that the inner product on $\mathbb{R}^{2 \times 2}$ corresponds to the standard scalar product on $\mathbb{R}^4$ because
$$
\left\langle
\begin{pmatrix}
a_{11} & a_{12} \\
a_{21} & a_{22}
\end{pmatrix},
\begin{pmatrix}
b_{11} & b_{12} \\
b_{21} & b_{22}
\end{pmatrix}
\right\rangle
= a_{11} b_{11} + a_{12} b_{12} + a_{21} b_{21} + a_{22} b_{22}.
$$
(This also justifies called is the default inner product.) So to find an orthonormal basis of $\mathbb{R}^{2 \times 2}$ with respect to which $L$ is diagonal is the same as finding an orthogonal basis of $\mathbb{R}^4$ with respect to which $\hat{L}$ is represented a diagonal matrix.
There are now different ways to solve this problem. We will first calculate the eigenspaces of $\hat{L}$; because $A$ is symmetric we know that $\hat{L}$ is diagonalizable. Then we will use the following fact:
Proposition: Let $S \in \mathbb{R}^{n \times n}$ be symmetric and $x,y \in \mathbb{R}^n$ eigenvalues of $S$ to eigenvalues $\lambda \neq \mu$. Then $x$ and $y$ are orthogonal.
Proof: Notice that
\begin{align*}
\lambda \langle x,y \rangle
&= \langle \lambda x, y \rangle
= \langle Ax, y \rangle
= (Ax)^T y
= x^T A^T y
= x^T A y \\
&= \langle x, A y \rangle
= \langle x, \mu y \rangle
= \mu \langle x, y \rangle.
\end{align*}
Because $\lambda \neq \mu$ it follows that $\langle x,y \rangle = 0$.
So the eigenspaces of different eigenvalues are orthogonal to each other. Therefore we can compute for each eigenspace an orthonormal basis and them put them together to get one of $\mathbb{R}^4$; then each basis vectors will in particular be an eigenvectors $\hat{L}$.
By some lengthy calculation it can be shown that the characteristic polynomial of $A$ is given by
$$
\chi_A(t) = t^4 - 4 t^3 + 2 t^2 + 4t - 3.
$$
It is easy to guess the roots $1$ and $-1$, so we can factor $\chi_A$ and get
$$
\chi_A(t) = (t-1)^2 (t+1) (t-3).
$$
The eigenspaces can now be calculated as usual, and we find that
$$
E_1 = \langle (0,-1,0,1)^T, (-1,0,1,0)^T \rangle, \;
E_{-1} = \langle (-1,1,-1,1)^T \rangle, \;
E_3 = \langle (1,1,1,1)^T \rangle,
$$
where $E_\lambda$ denotes the eigenspace with respect to the eigenspace $\lambda$.
Next we need to find orthonormal basis for each eigenspace. We can always do this by picking some basis and then using Gram–Schmidt. But here we are pretty lucky:
We know the basis $((0,-1,0,1)^T, (-1,0,1,0)^T)$ of $E_1$. Because both basis vectors are already orthogonal to each other we only need to normalize them. So we get $b_1 = \frac{1}{\sqrt{2}}(0,-1,0,1)^T$ and $b_2 = \frac{1}{\sqrt{2}}(-1,0,1,0)^T$.
In the case of $E_{-1}$ and $E_3$ we are even luckier, as they are both one-dimensional. So here too we only need to normalize and thus get $b_3 = \frac{1}{2} (-1,1,-1,1)^T$ and $b_4 = \frac{1}{2}(1,1,1,1)^T$.
Putting these together we have now found a basis $(b_1, b_2, b_3, b_4)$ of $\mathbb{R}^4$ given by
$$
b_1 = \frac{1}{\sqrt{2}} \begin{pmatrix} 0 \\ -1 \\ 0 \\ 1 \end{pmatrix}, \;
b_2 = \frac{1}{\sqrt{2}} \begin{pmatrix} -1 \\ 0 \\ 1 \\ 0 \end{pmatrix}, \;
b_3 = \frac{1}{2} \begin{pmatrix} -1 \\ 1 \\ -1 \\ 1 \end{pmatrix}, \;
b_4 = \frac{1}{2} \begin{pmatrix} 1 \\ 1 \\ 1 \\ 1 \end{pmatrix},
$$
which is orthonormal and cosists of eigenvectors of $\hat{L}$. The corresponding $2 \times 2$ matrices are
\begin{align*}
B_1 &= \frac{1}{\sqrt{2}} \begin{pmatrix} 0 & -1 \\ 0 & 1 \end{pmatrix}, &
B_2 &= \frac{1}{\sqrt{2}} \begin{pmatrix} -1 & 0 \\ 1 & 0 \end{pmatrix}, \\
B_3 &= \frac{1}{2} \begin{pmatrix} -1 & 1 \\ -1 & 1 \end{pmatrix}, &
B_4 &= \frac{1}{2} \begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix}.
\end{align*}
Best Answer
I think you must have written something wrong.
When you are permuting rows, that tells you what $P$ is.
You have, the permutation matrix associated with the row interchange $(E_1) \rightarrow (E_4)$ and $(E_2) \rightarrow (E_1)$ and $(E_4) \rightarrow (E_2)$. Row $E_3$ stayed put.
The way you would have come up with this permutation is by creating the augmented system using $A$, doing row eliminations, and swapping rows as the $P$ matrix says to get an upper triangular matrix.
Does that make sense? Take the matrix $A$, do Gaussian elimination on it and see how you would swap rows and you should get that P.
What you are actually doing is trying to write:
$$PAx = Pb.$$
Since $P^{-1} = P^{T}$, you end up with:
$$A = (P^T L) U$$
You show a $U$ above. If they gave you $A, P \text{and} U$, it is simple to find $L$.
However, you should be able to do this entire process just from $A$. If no $b$ is give, just skip that part of the problem and simply find $A = P^TLU$.