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Let \(W_{1}\) and \(W_{2}\) be subspaces of a vector space \(V\). Prove that \(V\) is the direct sum of \(W_{1}\) and \(W_{2}\) if and only if each vector in \(V\) can be uniquely written as \(x_{1}+x_{2}\), where \(x_{1} \in \mathrm{W}_{1}\) and $x_{2} \in \mathrm{W}_{2}$.

Short Answer

Expert verified
In summary, we have shown that a vector space V is the direct sum of two subspaces W鈧 and W鈧 if and only if each vector in V can be uniquely written as x鈧+x鈧, where x鈧 鈭 W鈧 and x鈧 鈭 W鈧. We proved this by first showing that the direct sum condition implies the unique representation, and then showing that the unique representation implies the direct sum condition.

Step by step solution

01

The definition of direct sum

Recall that a vector space V is the direct sum of two subspaces W鈧 and W鈧, denoted by V = W鈧 鈯 W鈧, if both of the following conditions are satisfied: 1. \(W_{1} \cap W_{2} = \{0\}\), that is, the intersection of the spaces contains only the zero vector. 2. \(W_{1} + W_{2} = V\), that is, every vector in V can be written as the sum of vectors in W鈧 and W鈧.
02

First implication: If V = W鈧 鈯 W鈧, then each vector in V can be uniquely written as x鈧+x鈧

Let's assume V is the direct sum of W鈧 and W鈧, that is, V = W鈧 鈯 W鈧. By definition, this means that every vector in V can be written as the sum of vectors in W鈧 and W鈧. Let v 鈭 V, then there exist unique vectors x鈧 鈭 W鈧 and x鈧 鈭 W鈧 such that v=x鈧+x鈧. Suppose there is a second representation \(v= x_{1}' + x_{2}'\) with \(x_{1}' \in W_{1}\) and \(x_{2}' \in W_{2}\). Now, we have \(x_{1} + x_{2} = v = x_{1}' + x_{2}'\), which can be rewritten as \(x_{1} - x_{1}' = x_{2}' - x_{2}\). Notice that both \(x_{1} - x_{1}'\) and \(x_{2}' - x_{2}\) belong to the intersection \(W_{1} \cap W_{2}\). Since V = W鈧 鈯 W鈧, the intersection \(W_{1} \cap W_{2}\) is only the zero vector. Therefore, we have \(x_{1} - x_{1}' = x_{2}' - x_{2} = 0\), which implies that \(x_{1} = x_{1}'\) and \(x_{2} = x_{2}'\). Therefore, each vector in V can be uniquely written as x鈧+x鈧 when V = W鈧 鈯 W鈧.
03

Second implication: If each vector in V can be uniquely written as x鈧+x鈧, then V = W鈧 鈯 W鈧

Now let's assume that each vector in V can be uniquely written as x鈧+x鈧, where x鈧 鈭 W鈧 and x鈧 鈭 W鈧. We need to show that V = W鈧 鈯 W鈧. In this case, we know that \(W_{1} + W_{2} = V\) since every vector in V can be written as the sum of vectors in W鈧 and W鈧. We also need to show that \(W_{1} \cap W_{2} = \{0\}\). Let w 鈭 \(W_{1} \cap W_{2}\), then w can be uniquely written as x鈧+x鈧 with x鈧 鈭 W鈧 and x鈧 鈭 W鈧. But w is also in W鈧, hence w=x鈧+x鈧=0+x鈧=x鈧. Since w 鈭 W鈧, we have W鈧 鈭 W鈧 containing zero vector. If there is any other non-zero vector w also belongs to \(W_{1} \cap W_{2}\), then it would contradict the assumption that each vector in V can be uniquely written as x鈧+x鈧. Therefore, \(W_{1} \cap W_{2} = \{0\}\), and V = W鈧 鈯 W鈧. In conclusion, V is the direct sum of W鈧 and W鈧 if and only if each vector in V can be uniquely written as x鈧+x鈧, where x鈧 鈭 W鈧 and x鈧 鈭 W鈧.

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