Lecture notes from lecture #3 of MATH 236. Material covered: some properties of vector spaces, subspaces, sums and direct sums.
1Properties of vector spaces¶
 A vector space $V$ has a unique additive identity element, the zero vector ($\vec 0 \in V$).
 Any element in $V$ has a unique additive inverse.
 $0\vec v = 0$ for any $\vec v \in V$.
 $a \vec 0 = \vec 0$ for any $a \in \mathbb F$.
 $(1)\vec v = \vec v$ for any $\vec v \in V$.
Proof of 4:
$$\begin{align} a\cdot \vec 0 & = a \cdot (\vec 0 + \vec 0) \tag{by property 1} \\ & = a \cdot \vec 0 + a \cdot \vec 0 \tag{by distributivity} \\ \therefore & a\cdot \vec 0 = \vec 0 \tag{by property 1}\end{align}$$
The proofs for the others are left as exercises.
2Subspaces¶
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 nontrivial vector space $V$ (i.e., a vector space that is not $\{0\}$) has at least two (distinct) subspaces: the set consisting of only the zero vector ($\{0\}$), and $V$ itself. The trivial vector space $\{0\}$ has only one distinct subspace. The empty set is of course not a subspace.
3Sums¶
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.
3.1Direct 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 $V = U_1 \oplus \ldots \oplus U_m$.
Examples:
 $U_1 = \{(x, y, 0)\}$, $U_2 = \{(0, 0, z)\}$; then $U_1 \oplus U_2 = \mathbb F^3$.
 $U_e(\mathbb F) = a_0 + a_2x^2 + \ldots + a_{2m} + x^{2m}$ (even polynomials), $U_o(\mathbb F) = a_1 + a_3x^3 + \ldots + a_{2m+1} x^{2m+1}$ (odd polynomials); then $U_e(\mathbb F) \oplus U_o(\mathbb F) \mathbb P(\mathbb F)$ (the polynomial vector space).
3.1.1Verifying direct sums¶
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.
($\leftarrow$) 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}

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. ↩