Problem 1762

### Part (a)
1. Given: $A$ is a $3 \times 3$ real matrix such that the vectors $Au$ and $u$ are orthogonal for every column vector $u \in \mathbb{R}^3$.
2. Orthogonality Condition: For any $u \in \mathbb{R}^3$, we have:
$$\langle Au, u \rangle = 0$$
where $\langle \cdot, \cdot \rangle$ denotes the dot product.
3. Matrix Representation: Let $A = (a_{ij})$ and $u = (u_1, u_2, u_3)^T$.
4. Dot Product Expansion: The condition $\langle Au, u \rangle = 0$ can be written as:
$$(Au)^T u = 0$$
Expanding this, we get:
$$u^T A^T u = 0$$
5. Component Form: For any $u = (u_1, u_2, u_3)^T$, this becomes:
$$\sum_{i=1}^3 \sum_{j=1}^3 a_{ij} u_i u_j = 0$$
6. Choosing Basis Vectors: Let $u$ be the standard basis vectors $e_i$ where $e_i$ has 1 in the $i$-th position and 0 elsewhere.
- For $u = e_1$, we get $a_{11} = 0$.
- For $u = e_2$, we get $a_{22} = 0$.
- For $u = e_3$, we get $a_{33} = 0$.
7. Off-Diagonal Elements: Consider $u = e_i + e_j$ for $i \neq j$:
- For $u = e_1 + e_2$, we get:
$$a_{11} + a_{12} + a_{21} + a_{22} = 0 \implies a_{12} + a_{21} = 0 \implies a_{21} = -a_{12}$$
- Similarly, for other combinations, we get:
$$a_{13} + a_{31} = 0 \implies a_{31} = -a_{13}$$
$$a_{23} + a_{32} = 0 \implies a_{32} = -a_{23}$$
8. Conclusion: The matrix $A$ is skew-symmetric, i.e., $A^T = -A$.

$\blacksquare$

### Part (b)
1. Given: We need to show that there exists a vector $v \in \mathbb{R}^3$ such that $Au = v \times u$ for every $u \in \mathbb{R}^3$.
2. Skew-Symmetric Matrix Representation: From part (a), we know $A$ is skew-symmetric. Thus, $A$ can be written as:
$$A = \begin{pmatrix} 0 & -a_{12} & -a_{13} \\ a_{12} & 0 & -a_{23} \\ a_{13} & a_{23} & 0 \end{pmatrix}$$
3. **Vector $v$**: Define $v$ as:
$$v = \begin{pmatrix} -a_{23} \\ a_{13} \\ -a_{12} \end{pmatrix}$$
4. Cross Product: The cross product $v \times u$ for $u = (u_1, u_2, u_3)^T$ is:
$$v \times u = \begin{pmatrix} -a_{23} \\ a_{13} \\ -a_{12} \end{pmatrix} \times \begin{pmatrix} u_1 \\ u_2 \\ u_3 \end{pmatrix} = \begin{pmatrix} a_{13} u_3 - (-a_{12}) u_2 \\ -a_{23} u_1 - (-a_{13}) u_3 \\ -a_{12} u_2 - a_{23} u_1 \end{pmatrix}$$
Simplifying, we get:
$$v \times u = \begin{pmatrix} a_{13} u_3 + a_{12} u_2 \\ -a_{23} u_1 + a_{13} u_3 \\ -a_{12} u_1 - a_{23} u_2 \end{pmatrix}$$
5. Matrix-Vector Multiplication: Now, compute $Au$:
$$Au = \begin{pmatrix} 0 & -a_{12} & -a_{13} \\ a_{12} & 0 & -a_{23} \\ a_{13} & a_{23} & 0 \end{pmatrix} \begin{pmatrix} u_1 \\ u_2 \\ u_3 \end{pmatrix} = \begin{pmatrix} -a_{12} u_2 - a_{13} u_3 \\ a_{12} u_1 - a_{23} u_3 \\ a_{13} u_1 + a_{23} u_2 \end{pmatrix}$$
6. Comparison: Comparing the components of $Au$ and $v \times u$, we see that:
$$Au = v \times u$$
Thus, the vector $v$ defined as $v = (-a_{23}, a_{13}, -a_{12})^T$ satisfies the required condition.

$\blacksquare$