Vector Subspace Sums Examples 1

# Vector Subspace Sums Examples 1

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 examples regarding vector subspaces sums and vector subspace direct sums.

## Example 1

Let $V = \mathbb{F}^3$, and consider the subspaces $U_1 = \{ (x, 0, 0) \in \mathbb{F}^3 : x \in \mathbb{F} \}$, $U_2 = \{ (0, y, 0) \in \mathbb{F}^3 : y \in \mathbb{F} \}$, and $U_3 = \{ (0, 0, z) \in \mathbb{F}^3 : z \in \mathbb{F} \}$. Do these subspaces form a direct sum?

Yes, these vectors form a direct sum. Let $u_1 = (x, 0, 0)$, $u_2 = (0, y, 0)$ and $u_3 = (0, 0, z)$. Then their sum $u_1 + u_2 + u_3 = (x, 0, 0) + (0, y, 0) + (0, 0, z) = (x, y, z)$. Each coordinate in this sum is uniquely determined, for example, the first coordinate in $(x, y, z)$ is determined by only the vectors in $U_1$, the second coordinate is determined only by the vectors in $U_2$, and the third coordinate is determined only by the vectors in $U_3$.

So indeed these subspaces form a direct sum of $V$, that is $V = U_1 \oplus U_2 \oplus U_3$.

## Example 2

Let $V = \mathbb{F}^3$, and consider the subspaces $U_1 = \{ (x, y, 0) \in \mathbb{F}^3 : x \in \mathbb{F} \}$, $U_2 = \{ (0, y, 0) \in \mathbb{F}^3 : y \in \mathbb{F} \}$, and $U_3 = \{ (x, 0, z) \in \mathbb{F}^3 : z \in \mathbb{F} \}$. Do these subspaces form a direct sum?

We first note that these vectors clearly form a sum, that is $V = U_1 + U_2 + U_3$ since we can create any vector in $\mathbb{F}$ to be:

(2)
\begin{align} \underbrace{(x, y, z)}_{\in V} = \underbrace{(x, y, 0)}_{\in U_1} + \underbrace{(0, 0, 0)}_{\in U_2} + \underbrace{(0, 0, z)}_{\in U_3} \end{align}

However, the sum does not necessarily produce a unique way to write $(x, y, z)$. For example, consider the vector $(0, 0, 0)$ which can be written in a few different ways:

(3)
\begin{align} \underbrace{(0, 0, 0)}_{\in V} = \underbrace{(1, 1, 0)}_{\in U_1} + \underbrace{(0,-1, 0)}_{\in U_2} + \underbrace{(-1, 0, 0)}_{\in U_3} \\ \underbrace{(0, 0, 0)}_{\in V} = \underbrace{(0, 0, 0)}_{\in U_1} + \underbrace{(0, 0, 0)}_{\in U_2} + \underbrace{(0, 0, 0)}_{\in U_3} \\ \end{align}

Thus $(0, 0, 0)$ is not formed uniquely, and $V \neq U_1 \oplus U_2 \oplus U_3$.