Problem 1640

### Part (a)

1. **Eigenvalues of $Y$:**
Given $Y = \begin{pmatrix} 0_n & A \\ A^T & 0_{n+1} \end{pmatrix}$, we need to show that 0 is an eigenvalue of $Y$ and the other eigenvalues can be expressed as $\pm \sqrt{\lambda}$, where $\lambda$ is a nonnegative eigenvalue of $AA^T$.

2. Characteristic Polynomial:
To find the eigenvalues of $Y$, we consider the characteristic polynomial $\det(Y - \lambda I)$. We have:
$$Y - \lambda I = \begin{pmatrix} -\lambda I_n & A \\ A^T & -\lambda I_{n+1} \end{pmatrix}$$
The determinant of this block matrix can be computed using the Schur complement:
$$\det(Y - \lambda I) = \det(-\lambda I_n) \det(-\lambda I_{n+1} - A^T(-\lambda I_n)^{-1}A)$$
Simplifying, we get:
$$\det(Y - \lambda I) = (-\lambda)^n \det(-\lambda I_{n+1} + \frac{1}{\lambda} A^T A)$$
$$= (-\lambda)^n \det(-\lambda I_{n+1} + \frac{1}{\lambda} A^T A)$$
$$= (-\lambda)^n \det(-\lambda^2 I_{n+1} + A^T A)$$

3. **Eigenvalues of $AA^T$:**
Let $\lambda$ be an eigenvalue of $AA^T$. Then there exists a nonzero vector $v$ such that:
$$AA^T v = \lambda v$$
Since $A^T A$ and $AA^T$ have the same nonzero eigenvalues, the eigenvalues of $A^T A$ are also $\lambda$.

4. **Eigenvalues of $Y$:**
The eigenvalues of $Y$ are the solutions to:
$$\det(-\lambda^2 I_{n+1} + A^T A) = 0$$
This implies:
$$\lambda^2 = \text{eigenvalues of } A^T A$$
Therefore, the eigenvalues of $Y$ are $\pm \sqrt{\lambda}$ where $\lambda$ are the eigenvalues of $AA^T$. Additionally, since $Y$ is a $(2n+1) \times (2n+1)$ matrix, there is an extra eigenvalue 0.

$\blacksquare$

### Part (b)

1. **Given Matrix $A$:**
For $n = 3$, we have $A = (x_{i,j})$ where:
$$x_{i,j} = a_i \delta_{i,j} + a_j \delta_{4,j} - \frac{1}{a^2}(a_i^2 + a_4^2) a_j$$
where $1 \leq i \leq 3$ and $1 \leq j \leq 4$.

2. Distinct Eigenvalues:
We need to show that $Y$ has 7 distinct eigenvalues. According to the solution, $Y$ has 7 distinct eigenvalues if $\text{rank}(A) = 3$ and $AA^T$ has 3 distinct eigenvalues.

3. Symmetric Polynomials:
Let $b_i = a_i^2$ for $i = 1, \ldots, 4$. The characteristic polynomial of $AA^T$ is given by:
$$\chi_{AA^T}(x) = \sigma_1 x^3 - 2\sigma_2 x^2 + 3\sigma_3 x - 4\sigma_4$$
where $\sigma_i$ are the elementary symmetric polynomials of $b_i$.

4. Discriminant:
The discriminant $\delta$ of the characteristic polynomial is:
$$\delta = -432\sigma_1^2\sigma_4^2 + 432\sigma_1\sigma_2\sigma_3\sigma_4 - 108\sigma_1\sigma_3^3 - 128\sigma_2^3\sigma_4 + 36\sigma_2^2\sigma_3^2$$
We need to show that $\delta > 0$.

5. Homogeneous Polynomial:
Since $\delta$ is a homogeneous polynomial of degree 10 in $b_i$, we can assume $\sigma_2 = 1$. By solving the partial derivatives, we find that $\delta = 0$ is not reached, implying $\delta > 0$.

6. Conclusion:
Since $\delta > 0$, $AA^T$ has 3 distinct eigenvalues, and thus $Y$ has 7 distinct eigenvalues.

$\blacksquare$