Linear transformation: Two dimensional rotation matrix

Question:
There is an ellipse:
\[
x^2 + \frac{y^2}{4} = 1
\]
If it is on a new coordinate that is rotated by 45$^o$, find the equation of the ellipse with the new coordinate, ($X, Y$).

Answer:
When the coordinate axes are rotated, the old coordinate ($x, y$) will be expressed by the new coordinate ($X, Y$).
\[
\left( \begin{array}{c} x \\ y \end{array} \right)
 =
\left( \begin{array}{cc} \cos\theta & -\sin\theta \\
                                      \sin\theta & \cos\theta
\end{array} \right)
\left( \begin{array}{c} X \\ Y  \end{array} \right)
\]
Therefore, we have
\[
\left\{
\begin{array}{l}
x = X\cos\theta -Y\sin\theta = X\cos 45^o -Y\sin 45^o =\frac{X-Y}{\sqrt{2}} \\
y = X\sin\theta + Y\cos\theta = X\sin 45^o + Y\cos 45^o = \frac{X+Y}{\sqrt{2}}
\end{array}
\right.
\]
Then, they are plugged into the original equation of ellipse.
\[
\frac{1}{2}(X-Y)^2 + \frac{1}{8}(X+Y)^2 = 1
\]
Or you can also express it as
\[
\frac{1}{8}(5X^2 -6XY +5Y^2) = 1
\]

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Matrix, eigenvalue, and eigenvector

Question:
(1) Find the eigenvalues and eigenvectors of a matrix,
\[
A=\left(
\begin{array}{c}
-5 & 6  \\
-4 & 5
\end{array}
\right)
\]
(2) If $C$ is constructed by the eigen vectors obtained above, find whether $C$ is a regular function and whether $C^{-1}AC$ can be a diagonal matrix.

Answer:
(1) The eigenvalues can be obtained from $|\lambda E - A|=0$ where $E$ is the unit matrix.
\begin{eqnarray}
|\lambda E - A| &=&
\left|
\begin{array}{c}
\lambda+5 & -6  \\
4 & \lambda-5
\end{array}
\right|
&=& (\lambda+5)(\lambda-5)+24  \\
&=& \lambda^2 -25 +24  \\
&=& \lambda^2 - 1
\end{eqnarray}
Therefore, the eigenvalues are $\lambda = \pm 1$. The eigen equation can be expressed as
\[
A{\bf{x}} = \lambda{\bf{x}} \quad \mathrm{where} \
{\bf{x}}=\left(
\begin{array}{c}
x  \\
y
\end{array}
\right) \neq 0
\]
Thus,
\begin{eqnarray*}
\left(
\begin{array}{c}
-5-\lambda & 6 \\
-4 & 5-\lambda
\end{array}
\right)
\left( \begin{array}{c}
x  \\
y
\end{array} \right) &=& 0   \\
\left(
\begin{array}{c}
-5-1 & 6 \\
-4 & 5-1
\end{array}
\right)
\left( \begin{array}{c}
x  \\
y
\end{array} \right) &=&
\left(
\begin{array}{c}
-6 & 6 \\
-4 & 4
\end{array}
\right)
\left( \begin{array}{c}
x  \\
y
\end{array} \right) &=& 0
\end{eqnarray*}
Since the vector is not zero, in order to hold the equation, the ratio must be $x:y=1:1$. Likewise, plug in $\lambda=-1$; then, we have $x:y=3:2$.  Therefore, the eigen vectors are
\[
{\bf p}=\left( \begin{array}{c}
x  \\
y
\end{array} \right) =
\left(
\begin{array}{c}
1\\
1
\end{array}
\right), \quad
{\bf q}=\left( \begin{array}{c}
x  \\
y
\end{array} \right) =
\left(
\begin{array}{c}
3 \\
2
\end{array}
\right)
\]

(2) A matrix $C$ is combined by two eigen vectors obtained in question (1). Namely, $C=({\bf p},{\bf q})$.
\[
C = \left(\begin{array}{c}
1 & 3 \\
1 & 2
\end{array}\right)
\]
Calculate the determinant.
\[
\det C = 1 \cdot 2 - 1 \cdot 3 = -1
\]
Since $\det \neq 0$, $C$ is regular. Now, calculate the following:
\[
C^{-1}AC =
\left(\begin{array}{c}
-2 & 3 \\
1 & -1
\end{array}\right)
\left(
\begin{array}{c}
-5 & 6  \\
-4 & 5
\end{array}
\right)
\left(\begin{array}{c}
1 & 3 \\
1 & 2
\end{array}\right)
=
\left(
\begin{array}{c}
1 & 0  \\
0 & -1
\end{array}
\right)
\]
This is a diagonal matrix and notice that the values correspond to the eigenvalues of matrix $A$.

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Reference,
The inverse matrix for a 2$\times$2 matix is
\[
M = \left(
\begin{array}{c}
a & b  \\
c & d
\end{array}
\right), \quad
M^{-1} = \frac{1}{\det M}
\left(
\begin{array}{c}
d & -b  \\
-c & a
\end{array}
\right)
\]