USAAIO
March 28, 2025, 5:07am
1
In the remaining parts of this problem, for all non-coding tasks, when you need to use those eigenvalues,
\color{red}{\textbf{keep them as $\lambda_0$ and $\lambda_1$. No need to apply their formulae.}}
Part 6 (10 points, non-coding task) Since matrix \mathbf{A} is symmetric, let us do the spectral decomposition of it.
That is, you should write \mathbf{A} in the following form:
\mathbf{A} = \mathbf{Q} \mathbf{\Lambda} \mathbf{Q}^\top ,
where
\mathbf{Q} \in \Bbb R^{2 \times 2} is an orthonormal matrix.
\mathbf{\Lambda} = \begin{bmatrix} \lambda_0 & 0 \\ 0 & \lambda_1 \end{bmatrix} \in \Bbb R^{2 \times 2} is a diagnomal matrix with \lambda_0 > \lambda_1
In this task, you need to compute \mathbf{Q} . Reasoning is required.
USAAIO
March 28, 2025, 5:09am
2
\color{green}{\text{### WRITE YOUR SOLUTION HERE ###}}
To compute an eigenvector \mathbf{x}_i assocaited with the eigenvalue \lambda_i for i \in \left\{ 0, 1 \right\} , we solve the following equation:
\begin{bmatrix}
1 - \lambda_i & 1 \\
1 & - \lambda_i
\end{bmatrix}
\begin{bmatrix}
x_{i0} \\
x_{i1}
\end{bmatrix}
= \begin{bmatrix}
0 \\
0
\end{bmatrix} .
Since the determintant of the 2-by-2 matrix on the L.H.S. is 0, the rank of this matrix is 1. Thus, it is sufficient to only solve the following equation:
\begin{bmatrix}
1 & - \lambda_i
\end{bmatrix}
\begin{bmatrix}
x_{i0} \\
x_{i1}
\end{bmatrix}
= \begin{bmatrix}
0 \\
0
\end{bmatrix} .
One solution is
\mathbf{x}_i = \begin{bmatrix}
x_{i0} \\
x_{i1}
\end{bmatrix}
= \begin{bmatrix}
\lambda_i \\
1
\end{bmatrix} .
Thus, the $i$th orthonormal vector is
\begin{align*}
\mathbf{q}_i
& = \frac{\mathbf{x}_i}{|| \mathbf{x}_i ||_2} \\
& = \frac{1}{\sqrt{1 + \lambda_i^2}}
\begin{bmatrix}
\lambda_i \\
1
\end{bmatrix} .
\end{align*}
Therefore,
\begin{align*}
\mathbf{Q}
= \boxed{\begin{bmatrix}
\frac{\lambda_0}{\sqrt{1 + \lambda_0^2}}
&
\frac{\lambda_1}{\sqrt{1 + \lambda_1^2}} \\
\frac{1}{\sqrt{1 + \lambda_0^2}}
&
\frac{1}{\sqrt{1 + \lambda_1^2}}
\end{bmatrix} }.
\end{align*}
\color{red}{\text{""" END OF THIS PART """}}