MATH 240

Fri. October 11th, 2019


Subspaces

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

  1. The zero vector 0 is in HH.
  2. If u,vu,v are in HH, then u+vu+v is in HH. (Closure under addition)
  3. If cc is any scalar and uu is any vector in HH, then cucu is in HH. (Closure under scalar multiplication)

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


Is R2\R^2 a subspace of R3\R^3? No, because elements of R2\R^2 are in the form (ab)\begin{pmatrix} a\\b \end{pmatrix}, and elements of R3\R^3 are in the form (cde)\begin{pmatrix} c\\d\\e \end{pmatrix}, which cannot be equal to each other; therefore R2\R^2 is not even a subset of R3\R^3, let alone a subspace. However, you can find subspaces within R3\R^3 that resemble R2\R^2.


Ex. Verify that H={(st0) s,tR}H=\{\begin{pmatrix} s\\t\\0 \end{pmatrix}\mid \ s,t \in \R\} is a subspace of R3\R^3.

  1. Note HH is a subset of R3\R^3.
  2. When s=t=0s=t=0, we get the zero vector (000)\begin{pmatrix} 0\\0\\0 \end{pmatrix} in HH.
  3. Let u=(s1t10), v=(s2t20)u=\begin{pmatrix} s_1\\t_1\\0 \end{pmatrix},\ v=\begin{pmatrix} s_2\\t_2\\0 \end{pmatrix} be in HH. u+v=(s1+s2t1+t20)u+v=\begin{pmatrix} s_1+s_2\\t_1+t_2\\0 \end{pmatrix} so u+vu+v is in HH.
  4. Let cc be any real number. cu=c(s1t10)=(cs1ct10)cu=c\begin{pmatrix} s_1\\t_1\\0 \end{pmatrix}=\begin{pmatrix} cs_1\\ct_1\\0 \end{pmatrix} so cucu is in HH.

Thus HH is a subspace of R3\R^3.


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

  1. Any polynomial can be differentiated as many times as you want, so PP is a subset of C(R)C^\infin(\R).
  2. f(x)=0f(x)=0 for all xx is a polynomial, so 00 is in PP.
  3. The sum of two polynomials is also a polynomial, so PP is closed under addition.
  4. The product of a scalar and any polynomial is also a polynomial, so PP is closed under scalar multiplication.

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


Generating Subspaces

Recall our example from before of H={(st0) s,tR}H=\{\begin{pmatrix} s\\t\\0 \end{pmatrix}\mid \ s,t \in \R\} being a subspace of R3\R^3.
H={s(100)+t(010) s,tR}=span{e1,e2}.\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 VV will always be a subspace of VV.


Ex. Let HH be the set of all vectors of the form (a3bbaab)\begin{pmatrix} a-3b\\b-a\\a\\b \end{pmatrix} Show that HH is a subspace of R4\R^4.

(a3bbaab)=a(1110)+b(3101)\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=span{(1110),(3101)}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 R4\R^4 is a subspace of R4\R^4, then HH is a subspace of R4\R^4.