Statement
Suppose \(U_1\), \(U_2\), and \(W\) are subspaces of \(V\), such that:
\begin{equation} \begin{cases} V = U_1 \oplus W\\ V = U_2 \oplus W \end{cases} \end{equation}
Prove or give a counterexample that \(U_1=U_2\)
Intuition
The statement is not true. The definition of direct sums makes it such that, \(\forall v \in V\), there exists a unique representation of \(v\) with \(u_{1i}+w_{i} = v\) for \(u_{1j}\in U_1, w_{j} \in W\) as well as another unique representation \(u_{2i} + w_{i}=v\) for \(u_{2j} \in U_{2}, w_{j} \in W\).
However, the definition of direct sums doesn’t guarantee that the distinct unique representations are equivalent; although \(V\) can only be represented uniquely by EITHER a sum of \(U_1+W\) or \(U_2+W\), it does not mean that each \(v \in V\) itself has only one unique representation.
Counterexample
In constructing a counterexample, we turn to the fact that the sums of two variables creates a third free variable; therefore, we can figure two distinct ways of creating a third, final free variable that construct an equivalent space.
Constructing \(U_1\) as a subspace
We begin with constructing:
\begin{equation} U_1= \left\{\begin{pmatrix} x_1\\y_1\\2y_1 \end{pmatrix}, x_1,y_1 \in \mathbb{F} \right\} \end{equation}
By setting both free variables to \(0\), we construct the additive identity. Then:
\begin{equation} \lambda \begin{pmatrix} x_1 \\ y_1 \\ 2y_1 \end{pmatrix} = \begin{pmatrix} \lambda x_1 \\ \lambda y_1\\ 2(\lambda y_1) \end{pmatrix} \end{equation}
by multiplication in \(\mathbb{F}\), scalar multiplication, commutativity, and associativity. We can show closure under addition by inheriting the operation in \(\mathbb{F}\) as well as applying distributive to the factor of \(2\).
Therefore, we show that \(U_1\) is a subspace of \(\mathbb{F}^{3}\).
Constructing \(U_2\) as a subspace
Then, we construct:
\begin{equation} U_2=\left\{\begin{pmatrix} x_1 \\ y_1 \\ 0 \end{pmatrix}, x_1,y_1\in \mathbb{F} \right\} \end{equation}
We again have \(0\) by setting free variables to create the additive identity. Addition and scalar multiplication is closed by inheriting them from \(\mathbb{F}\) (and the fact that \(0\) is the additive inverse and therefore \(\lambda 0 = 0\)).
Therefore, \(U_2\) is a subspace as well in \(\mathbb{F}^{3}\).
Constructing \(W\) as a subspace
Finally, we have:
\begin{equation} W = \left\{\begin{pmatrix} 0 \\ 0 \\z_1 \end{pmatrix}, z_1\in \mathbb{F} \right\} \end{equation}
By setting \(z_1=0\), we have the additive identity. As with above, addition and scalar multiplication is closed through inheritance and that \(\lambda 0=0\).
Constructing Sum of Subsets
Let’s construct:
\begin{equation} U_1+W = V \end{equation}
Take \(u_1 \in U_1, w \in W\), attempting to construct a \(v\in V\), we have that:
\begin{equation} \begin{pmatrix} x_{1} \\ y_1 \\ 2y_1 \end{pmatrix} + \begin{pmatrix} 0 \\ 0 \\ z_1 \end{pmatrix} = \begin{pmatrix} x_1 \\ y_1 \\ 2y_1+z_1 \end{pmatrix} = \begin{pmatrix} a \\ b \\ c \end{pmatrix} \end{equation}
Constructing Direct Sum
For all vectors in \(\mathbb{F}^{3}\), this is an equivalence with 3 free variables and 3 expressions—rendering each vector in \(\mathbb{F}^{3}\) to have a representation by \(U_1+W\). We can see this also with the unique \(0\) test:
We see that for:
\begin{equation} 0 \in U_1+W \end{equation}
To solve for some \(u_1 \in U, w \in W : u_1+w = 0\) we have that:
\begin{equation} \begin{pmatrix} x_{1} \\ y_1 \\ 2y_1 \end{pmatrix} + \begin{pmatrix} 0 \\ 0 \\ z_1 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix} \end{equation}
where the first vector is in \(U_1\) and the second is in \(W\). The first two expressions tell us that \(x_1=y_1=0\); the final equation requires that \(2y_1+z_1=0+z_1=0\Rightarrow z_1=0\) .
Therefore, the only way to write \(0\) is to take each element in the sum to \(0\) (i.e. in this case \(u_1=w=0 \implies u_1+w = 0\)), making the above a direct sum.
Therefore:
\begin{equation} U_1 \oplus W = V \end{equation}
In almost the same manner, we can show that:
\begin{equation} U_2\oplus W = V \end{equation}
That, for some \(u_2\in U_2, w \in W, v \in V\):
\begin{equation} \begin{pmatrix} x_1\\y_1\\0 \end{pmatrix} + \begin{pmatrix} 0 \\ 0 \\ z_1 \end{pmatrix} = \begin{pmatrix} x_1\\y_1\\z_1 \end{pmatrix} \end{equation}
for the first vector in \(U_2\), the second in \(W\). In fact, this is the statement made in example 1.41.
Creating the Counterexample
Finally, we have that:
\begin{equation} \left\{\begin{pmatrix} x_1 \\ y_1 \\ 2y_1 \end{pmatrix}: x_1,y_1 \in \mathbb{F}\right\} \neq\left\{\begin{pmatrix} x_1 \\ y_1 \\ 0 \end{pmatrix}: x_1,y_1 \in \mathbb{F}\right\} \end{equation}
\(\forall y_1 \neq 0\) in the first expression. Therefore, \(U_1 \neq U_2\), finishing the counterexample. \(\blacksquare\)