Question

(5) let ( v ) be a vector space and suppose that ( v = m oplus n ). show that for every ( mathbf{v} in v ) there exist unique vectors ( mathbf{m} in m ) and ( mathbf{n} in n ) such that ( mathbf{v} = mathbf{m} + mathbf{n} ). hint. it should be clear that the only thing you have to establish is the uniqueness of the vectors ( mathbf{m} ) and ( mathbf{n} ). to this end, suppose that a vector ( mathbf{v} ) in ( v ) can be written as ( mathbf{m}_1 + mathbf{n}_1 ) and it can also be written as ( mathbf{m}_2 + mathbf{n}_2 ) where ( mathbf{m}_1, mathbf{m}_2 in m ) and ( mathbf{n}_1, mathbf{n}_2 in n ). prove that ( mathbf{m}_1 = mathbf{m}_2 ) and ( mathbf{n}_1 = mathbf{n}_2 ).

Explanation:

Step1: Start with given vector expressions

We know that \(\mathbf{v}=\mathbf{m}_1+\mathbf{n}_1\) and \(\mathbf{v}=\mathbf{m}_2+\mathbf{n}_2\), where \(\mathbf{m}_1,\mathbf{m}_2\in M\) and \(\mathbf{n}_1,\mathbf{n}_2\in N\). Since both equal \(\mathbf{v}\), we can set them equal to each other: \(\mathbf{m}_1+\mathbf{n}_1=\mathbf{m}_2+\mathbf{n}_2\).

Step2: Rearrange the equation

Subtract \(\mathbf{m}_2\) and \(\mathbf{n}_2\) from both sides: \(\mathbf{m}_1 - \mathbf{m}_2+\mathbf{n}_1 - \mathbf{n}_2=\mathbf{0}\). Let \(\mathbf{m}=\mathbf{m}_1 - \mathbf{m}_2\) and \(\mathbf{n}=\mathbf{n}_1 - \mathbf{n}_2\). Then \(\mathbf{m}+\mathbf{n}=\mathbf{0}\), where \(\mathbf{m}\in M\) (because \(M\) is a subspace, so the difference of two vectors in \(M\) is in \(M\)) and \(\mathbf{n}\in N\) (similarly, \(N\) is a subspace, so the difference of two vectors in \(N\) is in \(N\)).

Step3: Use the definition of direct sum

Since \(V = M\oplus N\), we know that \(M\cap N=\{\mathbf{0}\}\). From \(\mathbf{m}+\mathbf{n}=\mathbf{0}\), we can rewrite it as \(\mathbf{m}=-\mathbf{n}\). But if \(\mathbf{m}\in M\) and \(-\mathbf{n}\in M\) (because \(M\) is a subspace, so it's closed under scalar multiplication), then \(-\mathbf{n}\in M\cap N\). Similarly, \(\mathbf{n}\in M\cap N\). Since \(M\cap N =\{\mathbf{0}\}\), we have \(\mathbf{n}=\mathbf{0}\) and \(\mathbf{m}=\mathbf{0}\).

Step4: Conclude the uniqueness

If \(\mathbf{m}=\mathbf{m}_1 - \mathbf{m}_2=\mathbf{0}\), then \(\mathbf{m}_1=\mathbf{m}_2\). If \(\mathbf{n}=\mathbf{n}_1 - \mathbf{n}_2=\mathbf{0}\), then \(\mathbf{n}_1=\mathbf{n}_2\).

Answer:

We have proven that if \(\mathbf{v}=\mathbf{m}_1+\mathbf{n}_1=\mathbf{m}_2+\mathbf{n}_2\) with \(\mathbf{m}_1,\mathbf{m}_2\in M\) and \(\mathbf{n}_1,\mathbf{n}_2\in N\), then \(\mathbf{m}_1 = \mathbf{m}_2\) and \(\mathbf{n}_1=\mathbf{n}_2\), showing the uniqueness of the decomposition in the direct sum \(V = M\oplus N\).