MATH 240

Mon. October 14th, 2019


Identifying Subspaces

Let HH be a subset of a vector space VV. We have two main methods of determining if HH is a subspace of VV:


Null and Column Spaces

There are some “natural” subspaces that are worth examining:

Let AA be an m×nm\times n matrix. Recall that xAxx\rightarrow Ax is a linear transformation from Rn\R^n to Rm\R^m.

Null Space

Claim NulA={x:Ax=0x in Rn}\text{Nul}A=\{x:Ax=0\mid x\ \text{in} \ \R^n\} (the null space of AA) is a subspace of Rn\R^n:

  1. A0=0A0=0, so 00 is in NulA\text{Nul}A.
  2. If u,vu,v are in NulA\text{Nul}A, then u+vu+v is also in NulA\text{Nul}A.
    • Au=0Au=0
    • Av=0Av=0
    • A(u+v)=Au+Av=0+0=0A(u+v)=Au+Av=0+0=0
  3. If uu is in NulA\text{Nul}A and cc is any scalar, then cucu is also in NulA\text{Nul}A.
    • Au=0Au=0
    • A(cu)=c(Au)=c(0)=0A(cu)=c(Au)=c(0)=0

Thus NulA\text{Nul}A is a subspace of Rn\R^n.

xAxx\rightarrow Ax is one-to-one iff Ax=0Ax=0 has only the trivial solution. This is equivalent to NulA\text{Nul}A containing only the zero vector.

Column Space

Claim ColA=span{a1,...,an}\text{Col}A=\text{span}\{a_1,...,a_n\} where A=[a1,...,an]A=[a_1,...,a_n] (the column space of AA) is a subspace of Rm\R^m:

If vv is in Rm\R^m, then how would we determine if there is some xx in Rm\R^m such that Ax=vAx=v?
We’d determine if vv is in the span of the columns of AA, which is equivalent to vv being in ColA\text{Col}A. Thus ColA\text{Col}A can be thought of as the image of Rm\R^m under the mapping xAxx\rightarrow Ax.


Ex. Find a spanning set for NulA\text{Nul}A where A=(361171223124584)A=\begin{pmatrix} -3&6&-1&1&-7\\ 1&-2&2&3&-1\\ 2&-4&5&8&-4 \end{pmatrix} This is equivalent to solving Ax=0Ax=0.
(361170122310245840)(122310361170245840)(12231000510100001220)(122310001220000000)(120130001220000000)(x1x2x3x4x5)=(2x2+x43x5x22x4+2x5x4x5)=x2(21000)+x4(10210)+x5(30201)\begin{aligned} &\rightarrow \begin{pmatrix} -3&6&-1&1&-7&0\\ 1&-2&2&3&-1&0\\ 2&-4&5&8&-4&0 \end{pmatrix}\\ &\rightarrow \begin{pmatrix} 1&-2&2&3&-1&0\\ -3&6&-1&1&-7&0\\ 2&-4&5&8&-4&0 \end{pmatrix}\\ &\rightarrow \begin{pmatrix} 1&-2&2&3&-1&0\\ 0&0&5&10&-10&0\\ 0&0&1&2&-2&0 \end{pmatrix}\\ &\rightarrow \begin{pmatrix} 1&-2&2&3&-1&0\\ 0&0&1&2&-2&0\\ 0&0&0&0&0&0 \end{pmatrix}\\ &\rightarrow \begin{pmatrix} 1&-2&0&-1&3&0\\ 0&0&1&2&-2&0\\ 0&0&0&0&0&0 \end{pmatrix}\\\\ \begin{pmatrix} x_1\\x_2\\x_3\\x_4\\x_5 \end{pmatrix}&=\begin{pmatrix} 2x_2+x_4-3x_5\\x_2\\-2x_4+2x_5\\x_4\\x_5 \end{pmatrix}\\ &=x_2\begin{pmatrix} 2\\1\\0\\0\\0 \end{pmatrix}+x_4\begin{pmatrix} 1\\0\\-2\\1\\0 \end{pmatrix}+x_5\begin{pmatrix} -3\\0\\2\\0\\1 \end{pmatrix} \end{aligned}
Thus the null space is spanned by these three vectors.


Ex. Let T:R3R3T:\R^3\rightarrow \R^3 be T(xyz)=(xy0)T\begin{pmatrix} x\\y\\z \end{pmatrix}=\begin{pmatrix} x\\y\\0 \end{pmatrix}

The null space of the standard matrix of TT is {(00z)z in R}\{\begin{pmatrix} 0\\0\\z \end{pmatrix} \mid z \ \text{in} \ \R\}

The column space of the standard matrix of TT is the entire xyxy plane.