Lecture 5 - Subspaces and Direct Sums

Review

Subspace

A subset $U$ of a vector space $V$ is a subspace of $V$ (which is over field $F$ ) if $U$ is a vector space itself (over the same field and using the same operations as those in $V$ ).

This is a pretty simple definition. It's just a smaller vector space inside of another one. For example:

Thus we require certain conditions on the subspace:

Subspace conditions

$U$ is a subspace of $V$ iff:

$\vec{0} \in U$ , so $U$ contains the additive identity
$u_{1} + u_{2} \in U$ for any two vectors $u_{1}, u_{2} \in U$ . This is closure under addition.
$α u \in U$ for any $α \in F, u \in U$ . This is closure under scalar multiplication.

Note that (1) says that $\vec{0} \in U$ . But this is really saying that $\vec{0} \in V$ must be the same $\vec{0}$ as the one from $U$ . But why is this guaranteed?!?!? Suppose that ${\vec{0}}_{u} \neq {\vec{0}}_{v}$ where both are still additive identities from their respective sets. Let $\vec{u} \in U$ be arbitrary, and notice that $\vec{u} + {\vec{0}}_{u} = \vec{u} = \vec{u} + {\vec{0}}_{v}$ . Cancel $\vec{u}$ from both sides and see that ${\vec{0}}_{u} = {\vec{0}}_{v}$ .

Another example:

Example

Let $U = {p (x) \in P_{6} (R) | p^{'} (1) = 0}$ . Is $U$ a subspace of $P_{6} (R)$ ?

Proof
It is because:

$p_{0} (x) = 0$ so $p_{0}^{'} (x) = 0$ and $p_{0}^{'} (1) = 0$ so $p_{0} (x) \in U$ .
Let $p, q \in U$ be arbitrary, so $p^{'} (1) = 0 = q^{'} (1)$ . Then notice that $(p + q)^{'} (1) = (p^{'} + q^{'}) (1) = p^{'} (1) + q^{'} (1) = 0 + 0 = 0$ so $p + q \in U$ .
Let $α \in R$ and $p \in U$ be arbitrary, so $p^{'} (1) = 0$ . Then notice that $(α p)^{'} (1) = (α p^{'}) (1) = α p^{'} (1) = α 0 = 0$ so then $α p \in U$ .
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Direct Sums of Subspaces

Sums of Subspaces

Given subspaces $U_{1}, . . ., U_{n}$ of $V$ , then sum of $U_{1}, . . ., U_{n}$ , denoted by:

U_{1} + . . . + U_{n} = {{\vec{u}}_{1} + . . . + {\vec{u}}_{n} | {\vec{u}}_{i} \in U_{i}}

Is the smallest subspaces containing each $U_{1}, . . ., U_{n}$ .

Note that any ${\vec{u}}_{i}$ could be $\vec{0}$ , or even any linear combination of vectors from $U_{i}$ . Thus, this simple addition allows vector combinations between possibly different subspaces.

A direct sum goes one step further in saying that each vector in this space can be written uniquely in this way. When it's a direct sum we say:

U_{1} \oplus \dots \oplus U_{n}

Instead of the standard $+$ sign.