let $u=(x_1,\ldots,x_n)$, $y=(y_1,\ldots,y_n)$ two vectors in $\mathbb{R}^n$. What is the vector belonging to the intersection of all the kernels of the matrices $\begin{pmatrix}x_jy_i&-y_iy_j\\ -x_ix_j&x_iy_j\end{pmatrix}$

Given two vectors $ u = (x_1, \ldots, x_n) $ and $ y = (y_1, \ldots, y_n) $ in $ \mathbb{R}^n $, we are asked to find a vector in the intersection of all the kernels of the matrices:

\[
A_{ij} = \begin{pmatrix} x_j y_i & -y_i y_j \\ -x_i x_j & x_i y_j \end{pmatrix}
\]

for all $ i, j = 1, \ldots, n $.

### Step 1: Determine the kernel of a single matrix $ A_{ij} $
The kernel of the matrix $ A_{ij} $ consists of vectors $ v = \begin{pmatrix} v_1 \\ v_2 \end{pmatrix} $ such that:

\[
A_{ij} v = \begin{pmatrix} x_j y_i & -y_i y_j \\ -x_i x_j & x_i y_j \end{pmatrix} \begin{pmatrix} v_1 \\ v_2 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix}
\]

This gives the following system of equations:

\[
\begin{aligned}
x_j y_i v_1 - y_i y_j v_2 &= 0 \quad \text{(1)} \\
-x_i x_j v_1 + x_i y_j v_2 &= 0 \quad \text{(2)}
\end{aligned}
\]

### Step 2: Analyze the system
From equation (1):

\[
x_j y_i v_1 = y_i y_j v_2 \implies v_1 = \frac{y_j}{x_j} v_2
\]

From equation (2):

\[
-x_i x_j v_1 + x_i y_j v_2 = 0 \implies v_1 = \frac{y_j}{x_j} v_2
\]

Both equations give the same relationship:

\[
v_1 = \frac{y_j}{x_j} v_2
\]

This suggests that any vector $ v $ in the kernel of $ A_{ij} $ must satisfy this condition.

### Step 3: Generalize for all matrices $ A_{ij} $
To be in the intersection of all kernels of the matrices $ A_{ij} $, the vector $ v = \begin{pmatrix} v_1 \\ v_2 \end{pmatrix} $ must satisfy the condition for all $ i, j $. 

However, this implies that for all $ i, j $, the ratio $ \frac{v_1}{v_2} $ must equal $ \frac{y_j}{x_j} $, i.e.,

\[
\frac{v_1}{v_2} = \frac{y_j}{x_j} \quad \text{for all } j
\]

For this to hold true for all $ j $, it must be that $ \frac{y_j}{x_j} $ is the same for all $ j $. Thus, we require $ y_j = c x_j $ for some constant $ c $.

### Step 4: Solution
If $ y = cx $, then any vector of the form $ v = \begin{pmatrix} c \\ 1 \end{pmatrix} $ will belong to the intersection of all kernels. Therefore, the vector $ v = \begin{pmatrix} x_j \\ y_j \end{pmatrix} $ for any $ j $ will belong to the intersection.

Thus, the vector that belongs to the intersection of all the kernels is:

\[
\mathbf{0} = \begin{pmatrix} 0 \\ 0 \end{pmatrix} \quad \text{if } y_j \neq c x_j \text{ for any constant } c.
\]

If $ y_j = c x_j $ for all $ j $, then any vector $ v = \begin{pmatrix} c \\ 1 \end{pmatrix} $ (where $ c $ is such that $ y = c x $) will belong to the intersection.

Would you like to go over the details of this result or have any questions?

Here are five questions for further exploration:
1. What conditions on $ u $ and $ y $ ensure that the intersection of the kernels is non-trivial?
2. How would the solution change if $ u $ and $ y $ are linearly independent?
3. Can we interpret the geometric meaning of the intersection of these kernels in $ \mathbb{R}^2 $?
4. What would happen if $ u $ and $ y $ are in higher-dimensional spaces, such as $ \mathbb{R}^m $ and $ \mathbb{R}^n $ with $ m \neq n $?
5. How does the determinant of the matrix $ A_{ij} $ relate to the kernel?

**Tip:** When working with matrices, examining special cases and symmetries can often simplify the problem and reveal deeper insights.

Explore the intersection of kernels of matrices with detailed explanations and examples. Understand how vectors in the intersection are determined based on given vectors u and y in ℝ^n. Suitable for advanced undergraduate students studying linear algebra.

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