Vector Subspace Sums

# Vector Subspace Sums

We will now look at an important definition regarding vector subspaces.

 Definition: Let $U_1, U_2, ..., U_m$ all be vector subspaces of the $\mathbb{F}$-vector space $V$. We define the Vector Sum of these subspaces $\sum_{i=1}^{m} = U_1 + U_2 + ... + U_m$ is defined to be the set of all possible sums $u_1 + u_2 + ... + u_m$ where each $u_i \in U_i$, that is $\sum_{i=1}^{m} U_i = U_1 + U_2 + ... + U_m := \{ u_1 + u_2 + ... + u_m : u_1 \in U_1, u_2 \in U_2, ..., u_m \in U_m \}$.

For example, consider the subspaces $U_1 = \{ (x, 0) : x \in \mathbb{F} \}$ and $U_2 = \{ (0, y) : y \in \mathbb{F} \}$ of $\mathbb{F}^2$. We denote the sum of these subspaces $U_1 + U_2 = \{ (x, 0) + (0, y) : x, y, \in \mathbb{F} \} = \{ (x, y) : x, y \in \mathbb{F} \}$, which is also a subspace of $\mathbb{F}^2$. In fact, the sum of two subspaces of a vector space $V$ will always be a subspace of $V$.

 Lemma 1: If $V$ is a vector space and $U_1, U_2, ..., U_m$ are subspaces of $V$ then the sum $U_1 + U_2 + ... + U_m$ is also a subspace of $V$.
• Proof: Since $U_1, U_2, ..., U_m$ are subspaces, we must show that their sum $\sum_{i=1}^{m} U_i$ contains the zero vector, is closed under addition, and is closed under multiplication.
• We note that $\sum_{i=1}^{m} U_i = U_1 + U_2 + ... + U_m := \{ u_1 + u_2 + ... + u_m : u_1 \in U_1, u_2 \in U_2, ..., u_m \in U_m \}$, so since each subspace $U_i$ contains the zero vector, then the zero vector is also contained in the sum by summing up the zero vector $m$-times.
• We also note that the sum is closed under addition since each vector in the sum $u_1 + u_2 + ... + u_m$ is closed under addition.
• Lastly we have that the sum is closed under scalar multiplication. That is for any $k$ we have that $ku_1 + ku_2 + ... + ku_m$ is in the sum, since each vector $ku_i \in U_i$. $\blacksquare$
 Proposition 1: The sum $\sum_{i=1}^{m} U_i$ is the smallest vector subspace containing all of the subspaces $U_1, U_2, ..., U_m$.
• Proof: Suppose that a smaller vector subspace existed by removing a finite number of elements from $\sum_{i=1}^{m} U_i$, none of which removed vectors are the zero vector. Then not all possible sums $u_1 + u_2 + ... + u_m$ could be formed.

We will now look at another type of sum known as a vector subspace direct sum.

# Vector Subspace Direct Sums

 Definition: Let $U_1, U_2, ..., U_m$ all be vector subspaces of the $\mathbb{F}$-vector space $V$. We define the Vector Direct Sum of these subspaces $\bigoplus_{i=1}^{m} = U_1 \oplus U_2 \oplus ... \oplus U_m = V$ is defined to be a sum of the subspaces $U_1, U_2, ..., U_m$ to which each element in $V$ can be uniquely written as $u_1 + u_2 + ... + u_m$ where $u_i \in U_i$ for $i = 1, 2, ..., m$.

One such example of a direct sum comes from the $\mathbb{F}$-vector space $\mathbb{R}^3$. Let $U_1 = \{ (x, 0, 0) : x \in \mathbb{R} \}$, $U_2 = \{ (0, y, 0) : y \in \mathbb{R} \}$ and $U_3 = \{ (0, 0, z) : z \in \mathbb{R} \}$, and suppose that we wanted to show that $U_1 \oplus U_2 \oplus U_3 = V$. Let $x \in U_1$, $y \in U_2$ and $z \in U_3$ such that $x = (x, 0, 0)$, $y = (0, y, 0)$ and $z = (0, 0, z)$. Then clearly $x + y + z = (x, y, z)$ is unique determined since the first coordinate is determined only by vectors in $U_1$, the second coordinate is determined only by vectors in $U_2$, and the third coordinate is determined only by vectors in $U_3$. Therefore $U_1 \oplus U_2 \oplus U_3 = V$.

It is important to note that not all sums are direct sums though. Consider the $\mathbb{F}$-vector space $\wp _5(\mathbb{F})$ which is defined to be the set of all polynomials whose degree is less than or equal to $5$ and whose coefficients are from $\mathbb{F}$, that is $\wp _5(\mathbb{F}) := \{ a_0 + a_1x + a_2x^2 + a_3x^3 + a_4x^4 + a_5x^5 : a_i \in \mathbb{F} \: i = 0, 1, 2, 3, 4, 5 \}$.

Now consider the vector subspace $\wp _4(\mathbb{F})$ (the $\mathbb{F}$-vector space containing all polynomials whose degree is less than or equal to $4$) and $U_2 = \{ ax^2 + bx^3 + cx^5 : a, b, c \in \mathbb{F} \}$.

We note that the polynomial $1 + x + x^2 + x^3 + x^4 + x^5$ is an element within the sum $U_1 + U_2$. Note that this element can be written as $(1 + x + x^4) + (x^2 + x^3 + x^5)$ where $(1 + x + x^4) \in U_1$ and $(x^2 + x^3 + x^5) \in U_2$. This polynomial can also be written as $(1 + x + x^2 + x^3 + x^4) + (x^5)$ where $(1 + x + x^2 + x^3 + x^4) \in U_1$ and $(x^5) \in U_2$. So each element in the sum $U_1 + U_2$ is not necessarily uniquely determined, and so $\wp _5(\mathbb{F})$ is not a direct sum of the subspaces $U_1$ and $U_2$, that is $\wp _5(\mathbb{F}) \neq U_1 \oplus U_2$.

We will now look at an important lemma to determine whether a sum of vector subspaces is a direct sum of a specific vector space.

 Lemma 2: Let $U_1, U_2, ..., U_m$ be vector subspaces of the $\mathbb{F}$-vector space $V$. Then these subspaces form a direct sum $\bigoplus_{i=1}^{m} U_i = V$ if and only if the sum of these subspaces is equal to $V$, that is $\sum_{i=1}^{m} U_i = V$ and when $\sum_{i=1}^{m} u_i = 0$ where $u_i \in U_i$ implies that $u_i = 0$ for every $i = 1, 2, ..., m$.
• Proof: $\Rightarrow$ Let $V = \bigoplus_{i=1}^{m} U_i$. Then it follows by the definition of the direct sum that $V = \sum_{i=1}^{m} U_i$ and suppose $\sum_{i=1}^{m} u_i = 0$ were $u_i \in U_i$ for every $i = 1, 2, ..., m$. Since the direct sum procures uniquely determined sums, then $0 = u_1 + u_2 + ... + u_m$ implies that $u_1 = u_2 = ... = u_m = 0$.
• $\Leftarrow$ Now suppose that $V = \sum_{i=1}^{m} U_i$ and $\sum_{i=1}^{m} u_i = 0$ and $u_1 = u_2 = ... = u_m = 0$. Now let a vector $v \in V$ and suppose that $v$ has two representations, that is $v = \sum_{i=1}^{m} u_i = \sum_{i=1}^{m} u_i'$ where $u_i, u_i' \in U_i$ for $i = 1, 2, ..., m$, and so $\sum_{i=1}^{m} u_i - \sum_{i=1}^{m} u_i' = \sum_{i=1}^{m} (u_i - u_i') = 0$. Therefore $u_i - u_i' = 0$ which implies that $u_i = u_i'$ for every $i = 1, 2, ..., m$ and so $V = \bigoplus_{i=1}^{m} U_i$. $\blacksquare$