MAT223 Lecture 20

• 1
  • Every plane in ${\mathbb{R}}^3$ through the origin is a subspace.
  • $\mathrm{null}(abc)=\{\left[\begin{array}{c}x\\ y\\ z\end{array}\right]\in {\mathbb{R}}^3|ax+by+cz=0\}$
• Activity 3
  • Subspaces of ${\mathbb{R}}^3$
  • Let $U_{A,B,C}=\{(x,y,z)\in {\mathbb{R}}^3|ax+by+cz=0\}$
  • So it's also a null space.
  • It's a plane through origin.
  • Show that $V_{A,B,C}=\{(x,y,z)\in {\mathbb{R}}^3|⟨x,y,z⟩=t⟨a,b,c⟩\}$ is a subspace of ${\mathbb{R}}^3$.
  • $x=tay=tbz=tc$
  • $t=\frac{x}{a}=\frac{y}{b}=\frac{z}{c}$
  • It's $\mathrm{span}\left\{\left[\begin{array}{c}a\\ b\\ c\end{array}\right]\right\}$
  • It's $\mathrm{Im}\left[\begin{array}{c}a\\ b\\ c\end{array}\right]$
• Are there any other subspaces of ${\mathbb{R}}^3$ that aren't lines or planes.
  • No, there are only 3 types of subspaces of ${\mathbb{R}}^3$: $\{0\}$, lines through the origin, planes through the origin, and ${\mathbb{R}}^3$ itself.
  • You can have $0,1,2,3$ dimensions as subspaces of ${\mathbb{R}}^3$.
  • $0=\overrightarrow{0}$
  • $1=\overrightarrow{L}(t)$
  • $2=\pi$
  • $3={\mathbb{R}}^3$
• Polls
• 1
  • Which of the subsets are linearly independent?
  • $\{⟨1,0,0⟩,⟨0,1,0⟩,⟨2,3,0⟩\}$
    • No, because $⟨2,3,0⟩$ is a linear combination of the first two vectors.
• 2
  • What's a basis for the $yz$ plane?
  • This is when $x=0$
  • We need independence AND span.
• 3
  • $U=\{(x,y,z):x+y+z=0\}$
  • Find a basis for $U$
  • $x=-y-z$
  • $y=-x-z$
  • $z=-y-x$
  • So we can write $U$ as $\{(-y-z,y,z):y,z\in \mathbb{R}\}$
  • So we can write $U$ as $\{(-t-s,t,s):t,s\in \mathbb{R}\}$
  • So we can write $U$ as $\{t(-1,1,0)+s(-1,0,1):t,s\in \mathbb{R}\}$
  • Dimension is now 2.
• 4
  • True or false
  • $U$ is a subspace
  • $S$ is linearly independent subset of $U$
  • $U=\mathrm{span}(S)$
  • I think false. Because $S$ as a subset may not span it.
• Activity 1
  • Linear Independence
  • 1
    • Why two non parallel vectors in ${\mathbb{R}}^3$ are linearly independent?
    • Non parallel means $\overrightarrow{v}\ne c\overrightarrow{w}$
    • So they automatically satisfy the definition of linear independence. Because one can't produce the other.
    • Linear Independence
      • $\overrightarrow{u}-k\overrightarrow{v}\ne 0$
      • $a\overrightarrow{u}+b\overrightarrow{v}=\overrightarrow{0}$
      • $a\overrightarrow{u}=-b\overrightarrow{v}$
      • $a\ne 0$
        • $u=-\frac{b}{a}\overrightarrow{v}$
          • $b\ne u$ and $u=k\overrightarrow{v}$
          • $b=0$ and $u=\overrightarrow{0}$
      • $a=0$
        • $\overrightarrow{0}=b\overrightarrow{v}$
  • 2
    • Why do two such vectors in ${\mathbb{R}}^3$ span a plane?
    • They span a plane because the linear combinations of these two nonparallel vectors can produce any point on that plane.
  • 3
    • What does it mean geometrically for a set of vectors to be linearly independent?
    • That they can define a unique plane or line in space without being able to be expressed as a linear combination of each other.
    • Independent means span is a plane, and they're a basis.
    • If they're dependent. It's a line, or if we have $\overrightarrow{u}=\overrightarrow{v}$ then it's a point.
  • 4
    • If we have $\{\left[\begin{array}{c}1\\ 0\\ 0\end{array}\right],\left[\begin{array}{c}0\\ 1\\ 0\end{array}\right],\left[\begin{array}{c}1\\ 2\\ 0\end{array}\right]\}$
      • Is any pair parallel?
      • Are these independent?
      • $\overrightarrow{c}=\overrightarrow{a}+2\overrightarrow{b}$
      • So no.
      • No trivial pairs are parallel.
      • So we basically just have $xy$ plane.