Show that a subset W of a vector space V is a subspace of V if and only if the following condition holds: If u and v are any vectors in W and a and b are any scalars, then au + bv is in W.
To show that a subset $W$ of a vector space $V$ is a subspace of $V$ if and only if the condition holds, we need to prove two things:
1. If $W$ is a subspace of $V$, then for any vectors $u, v \in W$ and scalars $a, b$, the vector $au + bv$ is also in $W$.
2. If for any vectors $u, v \in W$ and scalars $a, b$, the vector $au + bv$ is in $W$, then $W$ is a subspace of $V$.
Let's prove these two statements one by one.
1. If $W$ is a subspace of $V$, then for any vectors $u, v \in W$ and scalars $a, b$, the vector $au + bv$ is also in $W":
To prove this, we need to show that $W$ satisfies the three conditions of being a subspace:
a) The zero vector 0 must belong to $W$. Since $W$ is a subspace of $V$, it contains the zero vector.
b) $W$ must be closed under addition. Let $u, v \in W$. Since $W$ is a subspace, $u + v$ must also be in $W$.
c) $W$ must be closed under scalar multiplication. Let $u \in W$ and $a$ be a scalar. Since $W$ is a subspace, $au$ must also be in $W$.
Now, let's consider the vector $au + bv$. Since $W$ is closed under addition and scalar multiplication, both $au$ and $bv$ are in $W$. Therefore, their sum $au + bv$ is also in $W$. Thus, $W$ satisfies the condition if it is a subspace of $V$.
2. If for any vectors $u, v \in W$ and scalars $a, b$, the vector $au + bv$ is in $W$, then $W$ is a subspace of $V":
To prove this, we need to show that $W$ satisfies the three conditions of being a subspace:
a) The zero vector 0 belongs to $W$. Since $W$ is closed under scalar multiplication, choosing $u = v = 0$ and $a = b = 0$ gives $au + bv = 0$, which implies that the zero vector is in $W$.
b) $W$ is closed under addition. Let $u, v \in W$. We need to show that $u + v$ is also in $W$. By choosing $a = b = 1$, we have $au + bv = u + v$ which is in $W$.
c) $W$ is closed under scalar multiplication. Let $u \in W$ and $a$ be any scalar. To show that $au \in W$, we choose $v = 0$ and $b = 0$. This gives $au + bv = au + 0 = au$, which is in $W$.
Since $W$ satisfies all three conditions of being a subspace of $V$, we can conclude that $W$ is indeed a subspace of $V$.
Therefore, we have shown that a subset $W$ of a vector space $V$ is a subspace of $V$ if and only if the condition holds: "If $u$ and $v$ are any vectors in $W$ and $a$ and $b$ are any scalars, then $au + bv$ is in $W$."