MAT223 Lecture 19

• Activity 1
  • What's a subspace of ${\mathbb{R}}^4$?
  • $U$
    • Null
    • $U=\mathrm{null}\left[\begin{array}{cccc}1& 6& -3& 1\\ 1& 1& -4& -1\end{array}\right]$
  • $V$
    • Span
    • $V=\{\left[\begin{array}{c}2s+t\\ 3s-t\\ 7s+4t\\ 2s+t\end{array}\right]|s,t,\in R\}$
    • $\mathrm{span}\{\left[\begin{array}{c}2\\ 3\\ 7\\ 2\end{array}\right],\left[\begin{array}{c}1\\ -1\\ 4\\ 1\end{array}\right]\}$
    • $\mathrm{Im}\left[\begin{array}{cc}2& 1\\ 3& -1\\ 7& 4\\ 2& 1\end{array}\right]$
      • $A\overrightarrow{x}$
      • $\left[\begin{array}{cc}2& 1\\ 3& -1\\ 7& 4\\ 2& 1\end{array}\right]\left[\begin{array}{c}s\\ t\end{array}\right]$
      • $\left[\begin{array}{c}2s+t\\ 3s-t\\ 7s+4t\\ 2s+t\end{array}\right]$
    • #tk why does the image of that matrix equal the span of those two vectors?
      • Image of the transformation is everything that can get hit in the codomain.
  • $X$
    • Null
• 2
• 3
  • Find spanning sets for $UVX$
  • $U$
    • $A\overrightarrow{x}=\overrightarrow{0}$
    • $A\overrightarrow{y}=\overrightarrow{0}$
    • $U=\{(x,y,z,w)|\begin{array}{c}x+6y=3z-w\\ y-4z=w-x\end{array}\}$
    • $U=\mathrm{null}\left[\begin{array}{cccc}1& 6& -3& 1\\ 1& 1& -4& -1\end{array}\right]$
    • $\left[\begin{array}{ccccc}1& 6& -3& -1& 0\\ 1& 1& -4& -1& 0\end{array}\right]$
    • $R_2=R_2-R_1$
    • $\left[\begin{array}{ccccc}1& 6& -3& -1& 0\\ 0& -5& -7& 0& 0\end{array}\right]$
    • $R_2=\frac{R_2}{-5}$
    • $\left[\begin{array}{ccccc}1& 6& -3& -1& 0\\ 0& 1& \frac{7}{5}& 0& 0\end{array}\right]$
    • $R_1=R_1-6R_2$
    • $-3-6\frac{7}{5}=-\frac{57}{5}$
    • $\left[\begin{array}{ccccc}1& 0& -\frac{57}{5}& -1& 0\\ 0& 1& \frac{7}{5}& 0& 0\end{array}\right]$
• Polls
• 7
  • The set $Q=\{\left[\begin{array}{c}x\\ y\end{array}\right]\in {\mathbb{R}}^2|(x,y)\text{ is in the first quadrant}\}$ is a subspace of ${\mathbb{R}}^2$
  • False because it's not closed under scalar multiplication. For example, if you take a vector in the first quadrant and multiply it by -1, you get a vector in the third quadrant, which is not in $Q$.
• Activity 2
  • $U=\mathrm{span}\{\overrightarrow{u},\overrightarrow{v},\overrightarrow{w}\}=\{a\overrightarrow{u}+b\overrightarrow{v}+c\overrightarrow{w}|a,b,c\in \mathbb{R}\}$
  • $=\mathrm{Im}\left[\begin{array}{ccc}\overrightarrow{u}& \overrightarrow{v}& \overrightarrow{w}\end{array}\right]=\mathrm{Im}(A)$
  • $\overrightarrow{u}=⟨1,1,1⟩$
  • $\overrightarrow{v}=⟨1,-1,2⟩$
  • $\overrightarrow{w}=⟨-1,5,8⟩$
  • I don't see any parallel vectors or such.
  • $\left[\begin{array}{ccc}1& 1& -1\\ 1& -1& 5\\ 1& 2& 8\end{array}\right]$
  • $R_2=R_2-R_1$
  • $\left[\begin{array}{ccc}1& 1& -1\\ 0& -2& 6\\ 1& 2& 8\end{array}\right]$
  • $R_3=R_3-R_1$
  • $\left[\begin{array}{ccc}1& 1& -1\\ 0& -2& 6\\ 0& 1& 9\end{array}\right]$
  • $R_3=R_3-\frac{1}{2}{R}_2$
  • $\left[\begin{array}{ccc}1& 1& -1\\ 0& -2& 6\\ 0& 0& 12\end{array}\right]$
  • $R_2=\frac{R_2}{-2}$
  • $\left[\begin{array}{ccc}1& 1& -1\\ 0& 1& -3\\ 0& 0& 12\end{array}\right]$
  • $R_3=\frac{R_3}{12}$
  • $R_1=R_1-R_2$
  • $\left[\begin{array}{ccc}1& 0& 2\\ 0& 1& -3\\ 0& 0& 1\end{array}\right]$
  • $R_2=R_2+3R_3$
  • $\left[\begin{array}{ccc}1& 0& 2\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]$
  • $detA=1$, it's invertible.
  • In principle this whole thing is possible. Depending on the subspace we get though. Not with these.
  • This is tied into the concept of linear dependence and independence.
  • If we have $A=n\times n$ and is invertible. It means that $A\overrightarrow{x}=\overrightarrow{y}$ for all $\overrightarrow{y}\in {\mathbb{R}}^n$
  • The image is all of ${\mathbb{R}}^3$ for our $A$
  • If let's say that $detA=0$, it's not invertible so one column is a linear combination of one of the others.