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*}
You could find the eigenvalues and eigenvectors algebraically, i.e. calculate the eigenvalues as the roots of the characteristic polynomial and solve a linear, homogeneous system per eigenvalue to find the corresponding eigenvector(s). Here however, they want you to use the geometrical interpretation of the reflection to find them by reasoning.
The basic question you should ask yourself is:
When reflecting about a plane, which vectors are mapped to a vector parallel with the vector you reflected; i.e. to a scalar multiple of that vector?
You should try to picture this geometrically. The different sub-questions guide you through this.
b) If $\textbf{n}$ is any normal to $\textit{P}$, find $T(n)$ use the result to find an eigenvalue of $A$
Take any normal to $P$ and find its reflection: you expect to find the opposite vector as a result, right? This means any normal $\bf n$ is reflected to the vector $\bf -n$. So any normal $\bf n$ is an eigenvector with eigenvalue...
You can try it out with a few normals such as $(1,2,-1)$ or $(-2,-4,2)$.
c) If $\textbf{v}$ is any vector in the plane $P$, find $T(v)$ using the fact that $T$ is a reflection about $P$. Use the result to find another eigenvalue of $A$.
If you take a vector lying in the plane, what do you expect will happen when you calculate its reflection with respect to that plane? It remains unchanged! This means any vector $\bf p$ in the plane is reflected to the vector $\bf p$ itself, so any vector in the plane is an eigenvector with eigenvalue...
d) Find eigenspaces corresponding to eigenvalues obtained in $(b)$ and $(c)$ using the fact that $T$ is refection about $P$.
With the reasoning in parts (b) and (c), you should ask yourself how many eigenvectors you can find in that situation:
- a normal to the plane is not unique, but all normals are... so you only have 1 linearly indepedent one;
- how many linearly independent vectors lying in the plane can you find?
Best Answer
Hint:
I suppose that ${e_i}$ is the standard basis. In this case you can find the matrix from a geometrical construction.
If $\vec u$ is the vector orthogonal to the plane, the projection of a vector $x=[x_1,x_2,x_3]^T$ on $\vec u$ is: $$ \vec v= \frac {\langle \vec x, \vec u \rangle}{|u|^2}\vec u $$ and the reflection of $\vec x$ in the plane gives a vector
$$\vec x'=\vec x - 2 \vec v$$
In your case $\vec u=[1,2,-2]^T$. can you do from this?