Vector Subspace Sums Examples 2

# Vector Subspace Sums Examples 2

Recall from the Vector Subspace Sums page that if $U_1$, $U_2$, …, $U_m$ are all vector subspaces of the vector space $V$, then the sum of $U_1$, $U_2$, …, $U_m$ denoted $U_1 + U_2 + ... + U_m$ is defined to be the set of all possible sums $u_1 + u_2 + ... + u_m$ where $u_i \in U_i$ for each $i = 1, 2, ..., m$, that is:

(1)
\begin{align} \quad U_1 + U_2 + ... + U_m = \{ u_1 + u_2 + ... + u_m : u_i \in U_i \: \mathrm{for \: each \:} i = 1, 2, ..., m \} \end{align}

Furthermore, we said that $U_1$, $U_2$, …, $U_m$ form a direct sum of the vector space $V$ written $V = U_1 \oplus U_2 \oplus ... \oplus U_m$ if every vector $v \in V$ can be written uniquely as $v = u_1 + u_2 + ... + u_m$ where $u_i \in U_i$ for each $i = 1, 2, ..., m$.

We will now look at some more examples regarding vector subspaces sums and vector subspace direct sums.

## Example 1

Prove by giving a counterexample that if $U_1 + U_2 = U_1 + U_3$ then $U_2$ need not equal $U_3$.

Consider the following subspaces from $\mathbb{R}^2$:

(2)
\begin{align} U_1 = \{ (a, b) : a, b \in \mathbb{F} \} \\ U_2 = \{ (a, 0) : a \in \mathbb{F} \} \\ U_3 = \{ (0, b) : b \in \mathbb{F} \} \end{align}

It's not hard to verify that each of $U_1$, $U_2$ and $U_3$ are indeed subspaces of $\mathbb{R}^2$. Furthermore $U_1 + U_2 = \mathbb{R}^2$ since $U_1 = \mathbb{R}^2$, and $U_1 + U_3 = \mathbb{R}^2$ for the same reason. However, clearly $U_2 \neq U_3$. Take a vector $(a, 0) \in U_2$ where $a \neq 0$. Then $(a, 0) \not \in U_3$.

## Example 2

Consider the subspace $U = \{ (c, c, c, d) \in \mathbb{R}^4 : c, d \in \mathbb{R} \}$ of $\mathbb{R}^4$. Find a subspace $W$ of $\mathbb{R}^4$ such that $\mathbb{R}^4 = U \oplus W$.

Consider the vector space $W = \{ (0, a, b, 0) \in \mathbb{R}^4 : a, b \in \mathbb{R} \}$. Then for $(x, y, z, w) \in \mathbb{R}^4$ we have that:

(3)
\begin{align} \quad (x, y, z, w) = \underbrace{(x, x, x, y)}_{\in U} + \underbrace{(0, y - x, z - x, 0)}_{\in W} \end{align}

Clearly $U \cap W = \{ 0 \}$. To show this, consider a vector in $U \cap W$. Then we have that:

(4)
\begin{align} \quad (c, c, c, d) = (0, a, b, 0) \\ \quad (c, c, c, d) - (0, a, b, 0) = (0, 0, 0, 0) \\ \quad (c, c - a, c - b, d) = (0, 0, 0, 0) \end{align}

Note that then $c = 0$, $c = a$, $c = b$ and $d = 0$. Thus $a = b = c = d = 0$, which implies that the vector we were considering was the zero vector.