# Prove that the statements are equivalent

#### mathmari

##### Well-known member
MHB Site Helper
Hey!!

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 $\mathcal{B}_i:=(b_{i,1}, \ldots , b_{i, d_i})$.

1. Show that $\displaystyle{\sum_{i=1}^kU_i=\left \{\sum_{i=1}^ku_i\mid u_i\in U_i\right \}\leq_{\mathbb{K}}V}$.
2. 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 $\displaystyle{\dim_{\mathbb{K}}\sum_{i=1}^kU_i\leq \sum_{i=1}^kd_i}$.
3. Show that the following are equivalent:
1. 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 $u_1=\ldots =u_k=0_V$.
2. For all $2\leq m\leq k$ it holds that $\displaystyle{\left (\sum_{i=1}^{m-1}U_i\right )\cap U_m=\{0_V\}}$.
3. $\displaystyle{\dim_{\mathbb{K}}\left (\sum_{i=1}^kU_i\right )=\sum_{i=1}^kd_i}$.
4. $\mathcal{B}:=(b_{1,1}, \ldots ,b_{1,d_1}, b_{2,1}, \ldots , b_{k,d_k})$ 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 ?

#### mathmari

##### Well-known member
MHB Site Helper
At the question 3 we have to show thw following equivalences: $(1) \Rightarrow (2) \Rightarrow (3) \Rightarrow (4) \Rightarrow (1)$, right?

For $(1) \Rightarrow (2)$ :

Let $\displaystyle{x\in \left (\sum_{i=1}^{m-1}U_i\right )\cap U_m}$. That means that $\displaystyle{x=\sum_{i=1}^{m-1}u_i}$ and $x=-u_m$. Subtracting these equations we get $\displaystyle{x-x=\sum_{i=1}^{m-1}u_i-(-u_m)\Rightarrow \sum_{i=1}^{m-1}u_i+u_m=0\Rightarrow \sum_{i=1}^{m}u_i=0 }$.
From the assumption (1) we get that $u_i=0$ fr all $1\leq i\leq m$. Therefore we have that $\displaystyle{x=\sum_{i=1}^{m-1}u_i=0}$ and $x=-u_m=0$, which means that $\displaystyle{\left (\sum_{i=1}^{m-1}U_i\right )\cap U_m=\{0\}}$.

For $(2) \Rightarrow (3)$ :

We already have that $\displaystyle{\dim_{\mathbb{K}}\sum_{i=1}^kU_i\leq \sum_{i=1}^kd_i}$. From the assumption (2) do we have that the $\mathcal{B}:=(b{1,1}, \ldots , b_{1, d_1}, b_{2,1}, \ldots , b_{k, d_k})$ is linearly independent?

For $(3) \Rightarrow (4)$ :

From the exercise 2 and teh assumption (3) the result follows, or not?

For $(4) \Rightarrow (1)$ :

Do we use here the condition for linear independence?

#### GJA

##### Well-known member
MHB Math Scholar
Hi mathmari ,

I will use Hindu-Arabic numerals (i.e., 1-3) for the primary problem statements and Roman numerals (i.e., I-IV) for the subparts to problem 3.

(I) $\Rightarrow$ (II)
Your proof looks good, nicely done.

(II) $\Rightarrow$ (III)
As you've shown in question (2), $\mathcal{B}=(b_{1,1},\ldots, b_{1,d_{1}},\ldots, b_{k,1},\ldots, b_{k,d_{k}})$ is a spanning set for $\displaystyle\sum_{i=1}^{k}U_{i}.$ Hence, we only need to show that the assumption in (II) implies the linear independence of $\mathcal{B}.$ I will give a fairly complete proof sketch of this and wait to hear back from you if you have any questions about filling in some of the finer details.

By way of contradiction, suppose $\mathcal{B}$ is not a linearly independent set. Then there is a vector in $\mathcal{B}$ that can be expressed as a linear combination of the other vectors in $\mathcal{B}$. Without loss of generality, we can assume this vector is $b_{kd_{k}}$ and write $$b_{kd_{k}} = \displaystyle\sum_{i=1}^{k-1}\displaystyle\sum_{j=1}^{d_{i}}\lambda_{j}^{i}b_{ij} + \sum_{j=1}^{d_{k}-1}\lambda_{j}^{k}b_{kj}.$$ Subtraction gives $$b_{kd_{k}}-\sum_{j=1}^{d_{k}-1}\lambda_{j}^{k}b_{kj} = \displaystyle\sum_{i=1}^{k-1}\displaystyle\sum_{j=1}^{d_{i}}\lambda_{j}^{i}b_{ij}.$$ Hence, by assumption (II), $$b_{kd_{k}}-\sum_{j=1}^{d_{k}-1}\lambda_{j}^{k}b_{kj} = 0.$$ But this last equation contradicts the fact that $\mathcal{B}_{k}$ is a basis for $U_{k}$. Hence, $\mathcal{B}$ is a basis for $\displaystyle\sum_{i=1}^{k}U_{i}$ and $\text{dim}_{\mathbb{K}}\left( \displaystyle\sum_{i=1}^{k}U_{i}\right) = \displaystyle\sum_{i=1}^{k}d_{i}.$

(III) $\Rightarrow$ (IV)
You are correct here as well. A spanning set of the space with the same number of elements as the dimension of the space is a basis for the space.

(IV) $\Rightarrow$ (I)
Write $u_{i} = \displaystyle\sum_{j=1}^{d_{i}}\lambda_{j}^{i}b_{ij}.$ Then $$\sum_{i=1}^{k}u_{i} = 0\,\Longrightarrow\,\sum_{i=1}^{k}\sum_{j=1}^{d_{i}}\lambda_{j}^{i}b_{ij} = 0.$$ Since $\mathcal{B}$ is a basis for $\displaystyle\sum_{i=1}^{k}U_{i}$, it follows that all the $\lambda_{j}^{i}$'s must be zero.

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