- Thread starter
- #1

- Apr 14, 2013

- 4,601

Let $\mathbb{K}$ be a field and let $V$ be a $\mathbb{K}$-vector space. Let $1\leq k\in \mathbb{N}$ and let $U_1, \ldots , U_k\leq_{\mathbb{K}}V$ be subspaces of $V$. Let $d_i:=\dim_{\mathbb{K}}U_i$ and $\mathcal{B}_i:=(b_{i,1}, \ldots , b_{i, d_i})$.

- Show that $\displaystyle{\sum_{i=1}^kU_i=\left \{\sum_{i=1}^ku_i\mid u_i\in U_i\right \}\leq_{\mathbb{K}}V}$.
- It holds that $\displaystyle{\sum_{i=1}^kU_i=\text{span}(b_{1,1}, \ldots ,b_{1,d_1}, b_{2,1}, \ldots , b_{k,d_k})}$ and that $\displaystyle{\dim_{\mathbb{K}}\sum_{i=1}^kU_i\leq \sum_{i=1}^kd_i}$.
- Show that the following are equivalent:
- Let $u_1\in U_1, \ldots , u_k\in U_k$ such that $\displaystyle{\sum_{i=1}^ku_i=0_V}$, it follows that $u_1=\ldots =u_k=0_V$.
- For all $2\leq m\leq k$ it holds that $\displaystyle{\left (\sum_{i=1}^{m-1}U_i\right )\cap U_m=\{0_V\}}$.
- $\displaystyle{\dim_{\mathbb{K}}\left (\sum_{i=1}^kU_i\right )=\sum_{i=1}^kd_i}$.
- $\mathcal{B}:=(b_{1,1}, \ldots ,b_{1,d_1}, b_{2,1}, \ldots , b_{k,d_k})$ is a basis of $\displaystyle{\sum_{i=1}^kU_i}$.

I have done the following:

For

**question 1**we show that the 3 conditions are satisfied:

Since $U_i$ are subspaces, it holds that $0\in U_i$. So, it holds that $\displaystyle{0=\sum_{i=1}^k0\in \sum_{i=1}^kU_i}$. Therefore, $\displaystyle{\sum_{i=1}^kU_i}$ is a non-empty set.

Let $\displaystyle{u=\sum_{i=1}^ku_i}$ and $\displaystyle{u'=\sum_{i=1}^ku_i'}$ be two elements of $\displaystyle{\sum_{i=1}^kU_i}$ (mit $u_i, u_i' \in U_i$ for all $i \in \{1, 2, \ldots , n\}$). Then we have that \begin{equation*}u+u'=\sum_{i=1}^ku_i+\sum_{i=1}^ku_i'=\sum_{i=1}^k(u_i+u_i')\in \sum_{i=1}^kU_i\end{equation*} We have that $u_i+u_i'\in U_i$, since $U_i$ is a subspace, for each $i \in \{1, 2, \ldots , n\}$.

Let $\lambda\in \mathbb{K}$ be a scalar and $\displaystyle{u=\sum_{i=1}^ku_i}$ an element of $\displaystyle{\sum_{i=1}^kU_i}$. Then we have that \begin{equation*}\lambda u=\lambda \sum_{i=1}^ku_i=\sum_{i=1}^k(\lambda u_i)\in \sum_{i=1}^kU_i\end{equation*} We have that $\lambda u_i\in U_i$, since $U_i$ is a subspace, for each $i \in \{1, 2, \ldots , n\}$.

Therefore $\displaystyle{\sum_{i=1}^kU_i}$ is a subspace of $V$.

For

**question 2**I have done the following:

$B:=(b{1,1}, \ldots , b_{1, d_1}, b_{2,1}, \ldots , b_{k, d_k})$ is a generating set of $\displaystyle{\sum_{i=1}^kU_i}$, since $B$ contains the generating sets of all $U_i$.

A linear independent subset of $\mathcal{B}$ (or $\mathcal{B}$ itself) is a basis of $\displaystyle{\sum_{i=1}^kU_i}$. This means that the number of elements of the basis is at most $\displaystyle{\sum_{i=1}^kd_i}$. This means that $\displaystyle{\dim_{\mathbb{K}}\sum_{i=1}^kU_i\leq \sum_{i=1}^kd_i}$.

Is everything correct so far?

Could you give me a hint for

**question 3**?