Question

for non-trivial subspaces U and W of a finite-dimensional vector space V, define
U+W := {u+w | u element of U and w element of W}. Prove that U+W is a subspace of V

Answer 1

Conditions

for non-trivial subspaces $U$ and $W$ of a finite-dimensional vector space $V$, define $U+W := \{u+w \mid u \text{ element of } U \text{ and } w \text{ element of } W\}$. Prove that $U+W$ is a subspace of $V$.

Solution

Consider a finite-dimensional vector space $V$ with a field $P$.

U,W – subspaces of V

If we want prove, that $U+W$ is a subspace of V, we must use the Subspace criterion. It claims, that:

Non-trivial set $X \in V$ is a subspace of $V$ if and only if, when:

1. $\forall x, y \in X, x + y \in X$

2. $\forall a \in P, \forall x \in X, ax \in X$

First we must prove, that $U+W$ is in V.

Let's $V$ is a n-dimensional space. Fix a random basis $(v_{1}, v_{2}, \ldots, v_{n}) \in V$.

As $U$ is a subset of V, then:

$$\forall u \in U, u = a_{1}v_{1} + a_{2}v_{2} + \dots + a_{n}v_{n}, a_{1}, \dots, a_{n} \in P$$

As $W$ is a subset of V, then:

$$\forall w \in W, w = b_{1}v_{1} + b_{2}v_{2} + \dots + b_{n}v_{n}, b_{1}, \dots, b_{n} \in P$$ $$\begin{array}{l} u + v = a_{1}v_{1} + a_{2}v_{2} + \dots + a_{n}v_{n} + b_{1}v_{1} + b_{2}v_{2} + \dots + b_{n}v_{n}, b_{1}, \dots, b_{n} \\ = (a_{1} + b_{1})v_{1} + (a_{2} + b_{2})v_{2} + \dots + (a_{n} + b_{n})v_{n} \end{array}$$

So, $u+w$ is an element of $V$ for each $u$ and $w$, so $U+W$ is in V.

The first condition of the Subspace Criterion is obvious:

$\forall u_{1}, u_{2} \in U, u_{1} + u_{2} \in U$ (as $U$ is a subspace of $V$ and the criterion for $U$ is correct)

$\forall w_{1}, w_{2} \in W, w_{1} + w_{2} \in W$ (as $W$ is a subspace of $V$ and the criterion for $W$ is correct)

$\forall u_{1}, u_{2} \in U, \forall w_{1}, w_{2} \in W, u_{1} + w_{1} + u_{2} + w_{2}$ is in $U+W$, because $u_{1} + u_{2} \in U$ and $w_{1} + w_{2} \in W$.

The second condition of the Subspace Criterion is obvious too:

As $U$ and $W$ are subspaces, then the $2^{\text{nd}}$ condition is correct for them:

$\forall a \in P, \forall u \in U, au \in U$

$\forall b \in P, \forall w \in W, bw \in W$

For U+W:

$\forall c \in P, q \in (U + W)$

Consider c(U+W):

$$c(U + W) = cu + cw, u \in U, w \in W$$

As we know,

$\forall a \in P, \forall u \in U, au \in U,$ especially, for $a = c$

$\forall b \in P, \forall w \in W, bw \in W,$ especially, for $b = c$

As $cu \in U, cw \in W$, than $cu + cw \in U + W$ (by the definition of U+W)

Then the $2^{\text{nd}}$ condition is correct.

Q.E.D.