Recall that the intersection $U\cap V$ is the set of elements that are both elements of $U$ and $V$.
In the set theoretical notation, we have
\[U \cap V=\{x \mid x\in U \text{ and } x\in V\}.\]

Proof.

To prove that the intersection $U\cap V$ is a subspace of $\R^n$, we check the following subspace criteria:

The zero vector $\mathbf{0}$ of $\R^n$ is in $U \cap V$.

For all $\mathbf{x}, \mathbf{y}\in U \cap V$, the sum $\mathbf{x}+\mathbf{y}\in U \cap V$.

For all $\mathbf{x}\in U \cap V$ and $r\in \R$, we have $r\mathbf{x}\in U \cap V$.

As $U$ and $V$ are subspaces of $\R^n$, the zero vector $\mathbf{0}$ is in both $U$ and $V$.
Hence the zero vector $\mathbf{0}\in \R^n$ lies in the intersection $U \cap V$.
So condition 1 is met.

Suppose that $\mathbf{x}, \mathbf{y} \in U \cap V$.
This implies that $\mathbf{x}$ is a vector in $U$ as well as a vector in $V$.
Similarly, $\mathbf{y}$ is a vector in $U$ as well as a vector in $V$.

Since $U$ is a subspace and $\mathbf{x}$ and $\mathbf{y}$ are both vectors in $U$, their sum $\mathbf{x}+\mathbf{y}$ is in $U$.
Similarly, since $V$ is a subspace and $\mathbf{x}$ and $\mathbf{y}$ are both vectors in $V$, their sum $\mathbf{x}+\mathbf{y}\in V$.

Therefore the sum $\mathbf{x}+\mathbf{y}$ is a vector in both $U$ and $V$.
Hence $\mathbf{x}+\mathbf{y}\in U \cap V$.
Thus condition 2 is met.

To verify condition 3, let $\mathbf{x}\in U \cap V$ and $r\in \R$.
As $\mathbf{x}\in U \cap V$, the vector $\mathbf{x}$ lies in both $U$ and $V$.
Since both $U$ and $V$ are subspaces, the scalar multiplication is closed in $U$ and $V$, respectively.

Thus $r\mathbf{x}\in U$ and $r\mathbf{x}\in V$.
It follows that $r\mathbf{x}\in U\cap V$.

This proves condition 3, and hence the intersection $U\cap V$ is a subspace of $\R^n$.

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We fix a nonzero vector $\mathbf{a}$ in $\R^3$ and define a map $T:\R^3\to \R^3$ by \[T(\mathbf{v})=\mathbf{a}\times \mathbf{v}\] for all $\mathbf{v}\in...