MATH 240

Mon. October 14th, 2019

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Identifying Subspaces

Let $H$ be a subset of a vector space $V$. We have two main methods of determining if $H$ is a subspace of $V$:

• Method I:
  • Determine that $0$ is in $H$
  • Determine that if $u,v$ are in $H$, then $u+v$ is also in $H$ (closed under addition)
  • Determine that if $u$ is in $H$ and $c$ is any scalar, then $cu$ is also in $H$ (closed under scalar multiplication)
• Method II:
  • Find a set of vectors in $V$ that span $H$.
  • "Theorem. The span of any collection of vectors from $V$ is a subspace of $V$."

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Null and Column Spaces

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

Let $A$ be an $m\times n$ matrix. Recall that $x\rightarrow Ax$ is a linear transformation from $\R^n$ to $\R^m$.

Null Space

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

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

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

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

Column Space

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

• $a_1,...,a_n$ are all vectors in $\R^m$, thus $\text{Col}A=\text{span}\{a_1,...,a_n\}$ is a subspace of $\R^m$ by the previously stated theorem.

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

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Ex. Find a spanning set for $\text{Nul}A$ where $A=\begin{pmatrix} -3&6&-1&1&-7\\ 1&-2&2&3&-1\\ 2&-4&5&8&-4 \end{pmatrix}$ This is equivalent to solving $Ax=0$.
$\begin{aligned} &\rightarrow \begin{pmatrix} -3&6&-1&1&-7&0\\ 1&-2&2&3&-1&0\\ 2&-4&5&8&-4&0 \end{pmatrix}\\ &\rightarrow \begin{pmatrix} 1&-2&2&3&-1&0\\ -3&6&-1&1&-7&0\\ 2&-4&5&8&-4&0 \end{pmatrix}\\ &\rightarrow \begin{pmatrix} 1&-2&2&3&-1&0\\ 0&0&5&10&-10&0\\ 0&0&1&2&-2&0 \end{pmatrix}\\ &\rightarrow \begin{pmatrix} 1&-2&2&3&-1&0\\ 0&0&1&2&-2&0\\ 0&0&0&0&0&0 \end{pmatrix}\\ &\rightarrow \begin{pmatrix} 1&-2&0&-1&3&0\\ 0&0&1&2&-2&0\\ 0&0&0&0&0&0 \end{pmatrix}\\\\ \begin{pmatrix} x_1\\x_2\\x_3\\x_4\\x_5 \end{pmatrix}&=\begin{pmatrix} 2x_2+x_4-3x_5\\x_2\\-2x_4+2x_5\\x_4\\x_5 \end{pmatrix}\\ &=x_2\begin{pmatrix} 2\\1\\0\\0\\0 \end{pmatrix}+x_4\begin{pmatrix} 1\\0\\-2\\1\\0 \end{pmatrix}+x_5\begin{pmatrix} -3\\0\\2\\0\\1 \end{pmatrix} \end{aligned}$
Thus the null space is spanned by these three vectors.

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Ex. Let $T:\R^3\rightarrow \R^3$ be $T\begin{pmatrix} x\\y\\z \end{pmatrix}=\begin{pmatrix} x\\y\\0 \end{pmatrix}$

The null space of the standard matrix of $T$ is $\{\begin{pmatrix} 0\\0\\z \end{pmatrix} \mid z \ \text{in} \ \R\}$

The column space of the standard matrix of $T$ is the entire $xy$ plane.