# Direct Sum: Vector Spaces

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1. Feb 29, 2016

### Kevin_H

During lecture, the professor gave us a theorem he wants us to prove on our own before he goes over the theorem in lecture.

Theorem: Let $V_1, V_2, ... V_n$ be subspaces of a vector space $V$. Then the following statements are equivalent.
1. $W=\sum V_i$ is a direct sum.
2. Decomposition of the zero vector is unique.
3. $V_i\cap\sum_{i\neq j}V_j =\{0\}$ for $i = 1, 2, ..., n$
4. dim$W$ = $\sum$dim$V_i$
What I understand:
• Definition of Basis
• Dimensional Formula
• Definition of Direct Sum

My Attempt: $1 \rightarrow 2 \rightarrow 3 \rightarrow 4 \rightarrow 1$

$1 \rightarrow 2$
1 state $W=\sum V_i$ is a direct sum. Then by definition there is a unique decomposition for $\alpha \in W$ such that $\alpha = \alpha_1 + \alpha_2 + ... + \alpha_n$ where $\alpha_i \in V_i$ for $i = 1, 2, ..., n.$ Let $\alpha = 0$, then it is necessary obvious $\alpha_i = 0$ for all $i$.

$2 \rightarrow 3$
2 states there is a unique decomposition for $0 = \alpha_1 + ... + \alpha_n$ where $\alpha_i \in V_i$ for $i = 1, 2, ..., n$. Suppose there exists $x_i \neq 0 \in V_i \cap \sum_{j\neq i}V_j$. Then $x_i = \sum_{j\neq i} x_j$ for some $x_j \in V_j$, hence $x_i - \sum_{j\neq i} x_j = 0$. Since $x_i \neq 0$, then $x_j$ can not be all zero. This contradicts the fact $0 = \alpha_1 + ... + \alpha_n$ is the unique decomposition of the zero vector. Therefore $V_i \cap \sum_{j\neq i}V_j = \{0\}$.

$3 \rightarrow 4$
3 states $V_i\cap\sum_{i\neq j}V_j =\{0\}$ for $i = 1, 2, ..., n$. This implies dim($V_i\cap\sum_{i\neq j}V_j$) = $0$. Now by direct application of the dimensional formula, which states dim($X+Y$) = dim($X$) + dim($Y$) - dim($X\cap Y$). Then

\begin{eqnarray*}
\text{dim}(V_1+(V_2 + ... + V_n)) & = & \text{dim}(V_1) + \text{dim}(V_2 + (V_3 + ... + V_n)) - \text{dim}(V_1 \cap \sum_{2}^nV_j)\\
& = & \text{dim}(V_1) + \text{dim}(V_2) + \text{dim}(V_3 + (V_4 +... + V_n)) - \text{dim}(V_2 \cap \sum_{3}^nV_j)\\
\end{eqnarray*}
repeatedly applying the dimensional formula to dim($V_i + V_{i + 1} + ... + V_{n}$) yields
\begin{eqnarray*}
\text{dim}(V_1+(V_2 + ... + V_n)) & = & \text{dim}(V_1) + \text{dim}(V_2) + ... + \text{dim}(V_n)\\
& = & \sum_{i = 1}^n\text{dim}(V_i)\\
\end{eqnarray*}
Where $W = \sum_{i = 1}^n(V_i)$

$4 \rightarrow 1$
4 states dim$W$ = $\sum$dim$V_i$. By direct consequence of the dimensional formula, we know $W = \sum_{i=1}^nV_i = \{\alpha = \alpha_1 + \alpha_2 + ... + \alpha_n \in V: \alpha_i \in V_i \text{for } i = 1,..., n\}$. We seek to show $\forall \alpha \in W$, there exists a unique decomposition. By hypothesis, dim($W) = m$ and dim($V_i) = m_i$ where $m = \sum_{i = 1}^nm_i$. Now, each $V_i$ has a basis $\Lambda_i$ with $m_i$ linearly independent vectors. Since $\alpha_i \in V_i$, there exists a unique linear combination $\alpha_i = \sum_{k=1}^{m_i}c_{i,k}\beta_{i,k}$ where $c_{i,k}$ is a scalar in the field and $\beta_{i,k} \in \Lambda_i$. Thus $\alpha \in W$ can be written as
\begin{eqnarray*}
\alpha & = & \alpha_1 + \alpha+2 + ... + \alpha_n\\
& = & (\sum_{k=1}^{m_1}c_{1,k}\beta_{1,k}) + (\sum_{k=1}^{m_2}c_{2,k}\beta_{2,k}) + ... + (\sum_{k=1}^{m_n}c_{n,k}\beta_{n,k})
\end{eqnarray*}
It follows by hypothesis that $\alpha$ is composed of $m = m_1 + ... + m_n$ linearly independent vectors. Thus $\alpha$ is indeed a unique decomposition $\alpha = \alpha_1 + \alpha_2 + ... + \alpha_n$ where $\alpha_i \in V_i$ for $i = 1, 2, ..., n$; therefore, $W = \sum_{i = 1}^nV_i$ is a direct sum.

Since $1 \rightarrow 2 \rightarrow 3 \rightarrow 4 \rightarrow 1$, then all statements are equivalent.

_________________________

Now I feel like my proof overall, especially $4 \rightarrow 1$, could be improved upon. I wanted to ask if you all have any suggestions on how I can do to make the proof better? Are there any logical errors? Is there an alternative way to prove this? I appreciate any feedback or criticism. Thank You for your time and have a wonderful day.

2. Feb 29, 2016

### andrewkirk

It looks pretty good.
However $4\to 1$ is missing a piece. When you write
that statement is not supported by the assumption of (4), which is simply a statement about dimensions and says nothing (directly) about the relationships between the subspaces $V_i$. We know by supposition that the vectors in each set $\Lambda_i\equiv\{\beta_{i,1},...,\beta_{i,m_i}\}$ are mutually independent, but not that the vectors in $\Lambda_i$ are independent of those in $\Lambda_j$ for $i\neq j$.

I wonder whether the contrapositive might be an easier way to prove this. That is, prove that $\neg 1\to\neg 4$. If you assume the sum is not direct it should be easy enough to identify a nonzero vector in the intersection of two subspaces which, by the dimensional formula, will entail that the dimension of the sum of subspaces is less than the sum of dimensions.