MATH 240

Fri. October 11th, 2019

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Subspaces

Definition. Let $V$ be a vector space and let $H\sube V$ ($H$ is a subset of $V$). We say that $H$ is a subspace of $V$ if all of the following are true:

1. The zero vector 0 is in $H$.
2. If $u,v$ are in $H$, then $u+v$ is in $H$. (Closure under addition)
3. If $c$ is any scalar and $u$ is any vector in $H$, then $cu$ is in $H$. (Closure under scalar multiplication)

$H=\{0\}$ (the set only containing the zero vector) is always a subset of any vector space $V$, and $V$ is always a subset of itself.

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Is $\R^2$ a subspace of $\R^3$? No, because elements of $\R^2$ are in the form $\begin{pmatrix} a\\b \end{pmatrix}$, and elements of $\R^3$ are in the form $\begin{pmatrix} c\\d\\e \end{pmatrix}$, which cannot be equal to each other; therefore $\R^2$ is not even a subset of $\R^3$, let alone a subspace. However, you can find subspaces within $\R^3$ that resemble $\R^2$.

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Ex. Verify that $H=\{\begin{pmatrix} s\\t\\0 \end{pmatrix}\mid \ s,t \in \R\}$ is a subspace of $\R^3$.

1. Note $H$ is a subset of $\R^3$.
2. When $s=t=0$, we get the zero vector $\begin{pmatrix} 0\\0\\0 \end{pmatrix}$ in $H$.
3. Let $u=\begin{pmatrix} s_1\\t_1\\0 \end{pmatrix},\ v=\begin{pmatrix} s_2\\t_2\\0 \end{pmatrix}$ be in $H$. $u+v=\begin{pmatrix} s_1+s_2\\t_1+t_2\\0 \end{pmatrix}$ so $u+v$ is in $H$.
4. Let $c$ be any real number. $cu=c\begin{pmatrix} s_1\\t_1\\0 \end{pmatrix}=\begin{pmatrix} cs_1\\ct_1\\0 \end{pmatrix}$ so $cu$ is in $H$.

Thus $H$ is a subspace of $\R^3$.

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Recall that $C^\infin(\R)$ is the set of infinitely differentiable functions on the reals.
Is $P$, the collection of all polynomials, a subspace of $C^\infin(\R)$?

1. Any polynomial can be differentiated as many times as you want, so $P$ is a subset of $C^\infin(\R)$.
2. $f(x)=0$ for all $x$ is a polynomial, so $0$ is in $P$.
3. The sum of two polynomials is also a polynomial, so $P$ is closed under addition.
4. The product of a scalar and any polynomial is also a polynomial, so $P$ is closed under scalar multiplication.

Thus $P$ is a subspace of $C^\infin(\R)$.

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Generating Subspaces

Recall our example from before of $H=\{\begin{pmatrix} s\\t\\0 \end{pmatrix}\mid \ s,t \in \R\}$ being a subspace of $\R^3$.
$\begin{aligned} H&=\{s\begin{pmatrix} 1\\0\\0 \end{pmatrix}+t\begin{pmatrix} 0\\1\\0 \end{pmatrix}\mid \ s,t \in \R\}\\ &=\text{span}\{e_1,e_2\}. \end{aligned}$

The span of a set of vectors in $V$ will always be a subspace of $V$.

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Ex. Let $H$ be the set of all vectors of the form $\begin{pmatrix} a-3b\\b-a\\a\\b \end{pmatrix}$ Show that $H$ is a subspace of $\R^4$.

$\begin{pmatrix} a-3b\\b-a\\a\\b \end{pmatrix}=a\begin{pmatrix} 1\\-1\\1\\0 \end{pmatrix}+b\begin{pmatrix} -3\\1\\0\\1 \end{pmatrix}$$H=\text{span}\{\begin{pmatrix} 1\\-1\\1\\0 \end{pmatrix},\begin{pmatrix} -3\\1\\0\\1 \end{pmatrix}\}$
Since the span of a collection of vectors from $\R^4$ is a subspace of $\R^4$, then $H$ is a subspace of $\R^4$.