# Bases and Dimension

Let $V$ be some vector space over a field $F$ and let $W_1$ and $W_2$ be subspaces of $V$ .

1. Prove that $W_1 \cap W_2$ is also a subspace of $V$ .
2. Denote the set of all sums $w_1 + w_2$ where $w_1 \in W_1$ and $w_2 \in W_2$ by $W_1 + W_2$. Prove that $W_1 + W_2$ is a subspace of $V$.
3. If in addition it is known that both $W_1$ and $W_2$ are finite dimensional, prove that $\dim(W_1 + W_2) = \dim(W_1) + \dim(W_2) − \dim(W_1 \cap W_2)$.

Hint. You may want to start with a basis $\beta$ of $W_1 \cap W_2$ and extend it to a basis $\beta_1$ of $W_1$ and a basis $\beta_2$ of $W_2$.

• Please, try to write in TeX. – Sigur Dec 8 '12 at 19:45

For 1: To show that $W_1 \cap W_2$ is a subspace, use your definitions! Remember that a subset of a vector space is a subset if it is closed under linear combinations. So, try showing first that $0$ is in $W_1 \cap W_2$. Let $x_1, x_2 \in W_1 \cap W_2$. Try to argue that $x_1 + x_2 \in W_1 \cap W_2$. Remember, $W_1$ and $W_2$ are both subspaces, hence closed under addition. You also need to show that $W_1 \cap W_2$ is closed under scalar multiplication. Let $\alpha$ be a scalar and $x \in W_1 \cap W_2$. Try arguing that $\alpha x \in W_1 \cap W_2$. Once again, remember that $W_1$ and $W_2$ are both subspaces and that $x$ is in the intersection (so $x$ is in both $W_1$ and $W_2$).
For 2: Proceed in a similar way. Show that $W_1 + W_2$ is closed under scalar multiplication and addition. So, for example, let $z_1 = x_1 + y_1 \in W_1 + W_2$ and $z_2= x_2 + y_2 \in W_1 + W_2$. What is $z_1 + z_2$? Try arguing that this is in $W_1 + W_2$. Proceed similarly for scalar multiplication.
For 3: The hint should help you get started. Use the hint to get the bases and remember that the dimension is the number of vectors in a basis. So, start with a basis for $W_1\cap W_2$ and extend it to $\beta_1$ for a basis of $W_1$ and starting with $\beta$ again, extend it to a basis for $W_2$. Then, try thinking about what a basis for $W_1 + W_2$ would be. Note that since $W_1 +W_2$ is formed by sums of elements from $W_1$ and $W_2$, it will involve the basis elements from both of them, but you don't want to 'double count' the basis elements that are in both (that is, the elements of $\beta$ from $W_1 \cap W_2$).