You signed in with another tab or window. Reload to refresh your session.You signed out in another tab or window. Reload to refresh your session.You switched accounts on another tab or window. Reload to refresh your session.Dismiss alert
A subspace of a vector space $V$ is a subset $U$ of $V$ that is in itself a vector space.
Properties of a subspace $U$:
Contains the zero vector
Closed under vector addition
Closed under scalar multiplication
An example of something that is not a subspace is ${(x_1, x_2, 1) \in \mathbb R^3 \mid x, y \in \mathbb R}$. This is clearly not a subspace of $\mathbb R^3$ since neither property above is satisfied.
Example: $U = {(x, 3x) \mid x \in \mathbb F^2}$. Verifying that this is a subspace is trivial and is left as an exercise for the reader.
Note that any vector space $V$ has at least two (trivial) subspaces: the empty set, and $V$ itself. Analogous to subrings I would say.
Sums
Let $U_1, \ldots, U_m$ be subspaces of $V$. Their sum, $U_1 + \ldots U_m$, is the set of all possible linear combinations (since subspaces are closed under scalar multiplication): ${u_1 + \ldots + u_m \mid u_i \in U_i}$. The sum of subspaces is a subspace of $V$ in itself.
If any element in $V$ can be written as a combination of vectors in $U_i$ (so $v = u_1 + \ldots + v_m$ for any $v \in V$), then we say that $V = U_1 + \ldots + u_m$.
Example: $U_1 = {(x, y, 0) \mid x, y \in \mathbb R}$, $U_2 = {(0, 0, z) \mid z\in \mathbb R}$, $U_3 = {(0, y, y) \mid y \in \mathbb R}$. Clearly, $U_1 + U_2 + U_3 = V$. We can write the zero vector in more than one way: either as the sum of zero vectors, or as $(0, a, 0) + (0, 0, a) + (0, -a, -a)$ (for any $a \in \mathbb R$). From this, we deduce that one of the subspaces is superfluous. We'll look into this more in the next section.
Direct sums
If any element $v \in V$ can be written uniquely as a sum of vectors in $U_1, \ldots, U_m$, then we say that $V$ is the direct sum of $U_1, \ldots, U_m$, which we write as $V = \oplus U_1 + \ldots + U_m$.
Examples:
$U_1 = {(x, y, 0)}$, $U_2 = {(0, 0, z)}$; then $U_1 \oplus U_2 = \mathbb F^3$.
Proposition 1.8: To check that a vector space $V$ is the direct sum of subspaces $U_1, \ldots, U_m$, we need to check that the following conditions are satisfied:
$V$ is the sum (i.e. any vector in $V$ can be written as a sum of vectors in the $U$s etc)
The zero element can be written uniquely as a sum of vectors in the $U$s ($0 + \ldots + 0$)
This proposition will be useful for homework questions.
Proof: ($\to$) Assume $V$ is the direct sum. Then 1) comes for free. 2) is also kind of free.
($\rightarrow$) Assume that 1) and 2) hold. Then we need to prove that $V$ is the direct sum (such that there is a unique decomposition for any element $v \in V$). We can prove this by contradiction. Assume that there are two distinct ways of writing an element $v \in V$, as $v = u_1 + \ldots + u_m$ and $v = v_1 + \ldots + v_m$, where $u_i \neq v_i$ for at least one $i$. Consider $-v$ (which we know is an element of $V$, by property 5 of vector spaces). $-v + v = 0$, since the zero vector is the additive identity. So then we have two distinct decompositions for the zero vector: $0 = -v + u_1 + \ldots + u_m$, and $0 = -v + v_1 + \ldots + v_m$. This contradicts condition 2) above. $\blacksquare$1
Footnotes
This differs somewhat from the proof shown in class, which I believe was not a proof by contradiction. I don't think he finished the proof though so who knows. ↩