Problem 2 - Algebra

For subspaces $U$, $W$ of $n$-dimensional real vector space $V_n$, define $U \cap W$, $U+W$ by

$$U \cap W = \{ \boldsymbol{a} \; \vert \; \boldsymbol{a} \in U, \;\boldsymbol{a} \in W \}$$

$$U + W = \{ \boldsymbol{b} \; \vert \; \boldsymbol{b} = \boldsymbol{a}_1 + \boldsymbol{a}_2, \;\boldsymbol{a}_1 \in U, \; \boldsymbol{a}_2 \in W \}$$

1.

Show that $U \cap W$ and $U+W$ are real vector spaces.

2.

Suppose that $V_n$ is the 3-dimensional space with $(x,y,z)$-coordinate.

(a)

When $U$ is the $(x,y)$-plane and $W$ is the $(y,z)$-plane, whate are $U \cap W$, $U+W$ ? Also, give the dimensions of $U \cap W$ and $U+W$.

(b)

When $U$ is the $(x,y)$-plane and $W$ is the $z$-axis, whate are $U \cap W$, $U+W$ ? Also, give the dimensions of $U \cap W$ and $U+W$.

3.

Denote the dimension of real vector space $V$ by $\dim (V)$. Prove the following equality:

$$\dim (U \cap W) + \dim (U+W) = \dim(U) + \dim(W)$$

1.

$U \cap W$ and $U+W$ are they real vector spaces?

(a)

$U \cap W$ is a real vector space because:

i.

$U \cap W$ is an abelian group because:

a.

$U$ and $W$ are subspaces, therefore $\boldsymbol{0} \in U$ and $\boldsymbol{0} \in W$, therefore $\boldsymbol{0} \in U \cap W$;

b.

$\forall (\boldsymbol{x}, \boldsymbol{y}) \in (U \cap W)^2$, $\boldsymbol{x} \in U$ and $\boldsymbol{x} \in W$ and $\boldsymbol{y} \in U$ and $\boldsymbol{y} \in W$, therefore $\boldsymbol{x}+\boldsymbol{y} \in U$ and $\boldsymbol{x}+\boldsymbol{y} \in W$ (since $U$ and $W$ are vector spaces), therefore $\boldsymbol{x}+\boldsymbol{y} \in U\cap W$;

c.

$\forall \boldsymbol{x} \in U\cap W$, $\boldsymbol{x} \in U$ and $\boldsymbol{y} \in W$, therefore $-\boldsymbol{x} \in U$ and $-\boldsymbol{y} \in W$ (since $U$ and $W$ are vector spaces), therefore $-\boldsymbol{x} \in U\cap W$.

ii.

$(\forall (\lambda_1, \lambda_2) \in \mathbb{R}^2)(\forall (\boldsymbol{x},\boldsymbol{y} \in (U\cap W)^2))$ we have (since these properties are already true for both $U$ and $W$):

a.

$(\lambda_1 + \lambda_2)\boldsymbol{x} = \lambda_1 \boldsymbol{x} + \lambda_2 \boldsymbol{y}$;

b.

$\lambda_1 (\boldsymbol{x} + \boldsymbol{y}) = \lambda_1 \boldsymbol{x} + \lambda_1 \boldsymbol{y}$.

iii.

$(\forall \boldsymbol{x} \in U\cap W) 1 \boldsymbol{x}=\boldsymbol{x}$ (since this property is already true for both $U$ and $W$).

iv.

$(\forall (\lambda, \mu) \in \mathbb{R}^2)(\forall \boldsymbol{x} \in U\cap W)\lambda(\mu \boldsymbol{x} = (\lambda \mu)\boldsymbol{x}$ (since these property is already true for both $U$ and $W$).

(b)

$U+W$ is also a real vector space for the same reasons.

This is raw check of the definition of a vector space. It is better to apply the theorem saying that a set included in a vector space is a subspace iff it is stable for the addition and for the scalar multiplication.

2.

Can you explicit $U \cap W$ and $U+W$ and their dimensions in a practical case study?

(a)

i.

$U\cap W = \{ \boldsymbol{x} \; \vert \; \boldsymbol{x} \in ( \mathbb{R}\boldsy... ...mathbb{R}\boldsymbol{j}+ \mathbb{R}\boldsymbol{k})\} = \mathbb{R}\boldsymbol{j}$ because $(\boldsymbol{i}, \boldsymbol{j}, \boldsymbol{k})$ is a set of basis vectors. Therefore, $\dim (U\cap W) = 1$.

ii.

$U+W = \{ \boldsymbol{x} \; \vert \; (\exists (\boldsymbol{a}, \boldsymbol{b}) ... ...thbb{R}\boldsymbol{k})\; \boldsymbol{x}=\boldsymbol{a} + \boldsymbol{b}\} = V_n$. Therefore, $\dim (U+W) = 3$.

(b)

i.

$U \cap W = \{ \boldsymbol{0} \}$ and $\dim (U\cap W)=0$.

ii.

$U+W = V_n$ and $\dim (U+W) = 3$.

3.

Can you show that $\dim (U\cap W) + \dim (U+W) = \dim (U) + \dim (W)$?

Let us consider the application $f : U \times W \rightarrow V_n; (x,y) \mapsto x+y$. $\Im (f) = U+W$. $\ker (f) = \{ (x,y) \; \vert \; f(x,y)=x+y=0 \} = \{ x \; \vert \; (x \in U)\wedge (-x \in W) \} = \{ x \; \vert \; (x \in U)\wedge (x \in W)\} = U \cap W$ (remember that W is a vector space). The rank formula gives us the following equality:

$$\dim (U \times W) = \dim (U+W) + \dim (U\cap W).$$

Since we know that $\dim (U\times W) = \dim (U) + \dim(W)$, we have shown that the following equality holds:

$$\dim (U\cap W) + \dim (U+W) = \dim (U) + \dim (W).$$