NUS-MATH530 1.C Problem 23

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