Linear Algebra: Subspaces

Sums of Subspaces

Suppose $U_1, \ldots U_n$ are subspaces of a vector space $V$. The sum of these subspaces, denoted $U_1 + \ldots + U_n$ is the set of all possible sums of the elements in $U_1, \ldots, U_n$. In set builder notation

$$U_1 + \ldots + U_n = \left\{ u_1 + \ldots + u_n : u_1 \in U_1, \ldots u_n \in U_n \right\}$$

If you want to be fancy, you can use the sigma sum notation that all the grad students love

$$\displaystyle\sum\limits_{i=1}^{n} U_i = \left\{ \displaystyle\sum\limits_{i=1}^{n} u_i : u_1 \in U_1, \ldots u_n \in U_n\right\}$$

Although that's rather excessive.

The subspace sum operation is not exactly the most interesting character in the linear algebra saga, but its presence helps fills out the script.


  1. Let $U = \left\{ [u, 0, 0, 0] : u \in \mathbb{R} \right\}$ and let $V = \left\{ [0,0,v,0]: v \in \mathbb{R} \right\}$. Compute $U + V$.

    $ U + V = \left\{ [u + 0, 0+0, 0+v, 0+0] : u, v \in \mathbb{R} \right\} \\ U + V = \left\{ [u, 0, v, 0] : u, v \in \mathbb{R} \right\} \\ $

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  2. Let $U= \left\{ [a, b, a, b] : a, b \in \mathbb{C} \right\}$ and let $V = \left\{ [c, c, c, d] : c, d \in \mathbb{C} \right\}$. Compute $U+V$.

    $ U + V = \left\{ [a + c, b + c, a + c, b + d] : a, b, c, d \in \mathbb{C} \right\} \\ $

    Since $a$, $b$, $c$, and $d$ are all arbitrary complex numbers, their sums are also arbitrary complex numbers. All we need to do is keep track of the unique combinations, which allows us to simplify the expression:

    $ U + V = \left\{ [a, b, a, c] : a, b, c \in \mathbb{C} \right\} \\ $

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  3. Show that the sum of two subspaces is itself a subspace.

    Let $U$ and $W$ be subspaces of a vector space $V$ over a field $F$. By definition of the sum of subspaces, $U + W = \left\{ u + w : u \in U, w \in W \right\}$. In order to qualify as a subspace, this sum must meet the three conditions outlined in the first problem.

    Additive Identity: Because $U$ and $W$ are subspaces, $0 \in U$ and $0 \in W$. By the definition of subspace, $0 + 0 \in Q$, which of course simplifies to $0 \in W$.

    Closed under addition: Let $u, w \in U + W$. By definition of a sum of subspaces, we can write $u = a + b$, where $a \in U$ and $b \in W$, and we can likewise write $w = c + d$, where $c \in U$ and $d \in W$. Then $u + w = (a + b) + (c + d)$. By commutativity and associativity of vector addition, $u + w = (a + c) + (b + d)$. Since $(a + c) \in U$ and $(b + d) \in W$, we conclude that $u + w \in U+W$.

    Closed under scalar multiplication: Let $v \in U + W$ and let $c \in F$. By definition of the sum of subspaces, $v = a + b$, where $a \in U$ and $b \in W$. Since $U$ is closed under scalar multiplication, $ca \in U$, and likewise $cb \in W$. Therefore $ca + cb \in U + W$. By distributivity, $c(a+b) \in U+W$, which implies that $cv \in C+W$.

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  4. Let $U_1, \ldots, U_n$ be subspaces of a vector space $V$, and let $W=\displaystyle\sum\limits_{i=1}^{n} U_i$. Show that $U_1, \ldots, U_n$ are each contained within $W$.

    By definition of a subspace, $W = \left\{ u_1 + \ldots + u_n : u_1 \in U_1, \ldots, u_n \in U_n \right\}$. Let $u_k \in U_k$, where $k$ is an integer between $1$ and $n$, and let $u_i = 0$ for each $u_i \in U_i$ where $i \neq k$. Then $0 + \ldots + u_k + \ldots + 0 = u_k \in W$. Therefore $U_k \in W$.

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  5. Let $U_1, \ldots, U_n$ be subspaces of a vector space $V$. Show that $W=\displaystyle\sum\limits_{i}^{n} U_i$ is the smallest subspace containing all the elements of $U_1, \ldots, U_n$.

    By the above proof, each of $U_1, \ldots, U_n$ is contained within $W$.

    Conversely, each subspace of $V$ containing all the elements of $U_1, \ldots, U_n$ must contain $W$, since $V$ is closed under vector addition.

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