Vector Subspaces Examples 2

Vector Subspaces Examples 2

Recall from the Vector Subspaces page that a subset $U$ of the subspace $V$ is said to be a vector subspace of $V$ if $U$ contains the zero vector of $V$ and is closed under both addition and scalar multiplication defined on $V$.

We also saw an important lemma which says that if $V$ is a vector space, then a subset $U$ of $V$ is a subspace of $V$ if and only if $(ax + by) \in U$ for all $x, y \in U$ and for all $a, b \in \mathbb{F}$.

We will now look at some more examples and non-examples of vector subspaces.

Example 1

Determine whether or not the subset $U$ of all differentiable real-valued functions $f$ on the interval $[-2, 2]$ where $f'(1) = 2f(-1)$ is a subspace of $C[-2, 2]$.

Let $f(x) = 0$ be the zero function. Then $f'(x) = 0$ and so $f'(1) = 0$ and $2f(-1) = 0$ so $f'(1) = 2f(-1)$. Thus the zero function is in $U$.

Now let $f(x), p(x) \in U$, and let $a, b \in \mathbb{R}$. Then we have that:

\begin{align} \quad (af + bp)'(1) = af'(1) + bp'(1) = 2af(-1) + 2bp(-1) = 2(af + bp)(-1) \end{align}

Therefore $(af(x) + bp(x)) \in U$, so indeed $U$ is a subspace of $C[-2, 2]$.

Example 2

Determine whether or not the subset $U$ of continuous real-valued functions such that $\int_0^1 f(x) \: dx = 1$ is a subspace of $C[0, 1]$.

Consider the zero function $f(x) = 0$. We have that:

\begin{align} \quad \int_0^1 f(x) \: dx = \int_0^1 0 \: dx = 0 \neq 1 \end{align}

Therefore the zero function is not in $U$ so $U$ is not a subspace of $C[0, 1]$.

Example 3

Show that if $U_1$ and $U_2$ are subspaces of $V$ then $U_1 \cap U_2$ is also a subspace of $V$.

We will prove this as a theorem later on. For now, suppose that $U_1$ and $U_2$ are subspaces of $V$. Then $0 \in U_1$ and $0 \in U_2$ which implies that $0 \in U_1 \cap U_2$.

Let $x, y \in U_1 \cap U_2$. Then any scalar multiple of $x$, say $ax \in U_1 \cap U_2$. Furthermore, any scalar multiple of $y$, say $by \in U_1 \cap U_2$. Thus since $U_1$ and $U_2$ are subspaces of $V$ we must also have that $(ax + by) \in U_1$ and $(ax + by) \in U_2$ so $(ax + by) \in U_1 \cap U_2$.

Therefore $U_1 \cap U_2$ is a subspace of $V$.

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