MAT223 Lecture 18

• 1
  • Assum $T:{\mathbb{R}}^2\to {\mathbb{R}}^2$ is a linear transformation and $T(\overrightarrow{v})=\left[\begin{array}{cc}2& 0\\ 0& 1\end{array}\right]\left[\begin{array}{cc}0& 1\\ -1& 0\end{array}\right]\overrightarrow{v}$
  • What does $T$ do?
    • $\left[\begin{array}{cc}0& 2\\ -1& 0\end{array}\right]$
    • ${\overrightarrow{e}}_1=-{\overrightarrow{e}}_2$
    • ${\overrightarrow{e}}_2=2{\overrightarrow{e}}_1$
    • Axes are flipped.
    • Then ${\overrightarrow{e}}_1$ is stretched by a factor of 2 and ${\overrightarrow{e}}_2$ is stretched by a factor of -1.
  • Answer was:
    • Rotated clockwise by 90 degrees and then stretched by a factor of 2 in the horizontal direction and a factor of 1 in the vertical direction.
  • So my intuition was correct.
  • Algorithmically: $T(\overrightarrow{v})=[\text{S}][\text{R}]\overrightarrow{v}$
  • So it's $S\circ R$ so $R$ is applied first and then $S$ is applied to the result.
• 1
  • Subspaces
  • Is $V=\left[\begin{array}{c}2x-2y+3z\\ 5x-3y+z\\ x-3y\end{array}\right]|x,y,z\in \mathbb{R}$ a subspace of ${\mathbb{R}}^3$?
    • $V$ is a subspace of ${\mathbb{R}}^3$ if it is closed under addition and scalar multiplication.
    • Let ${\overrightarrow{v}}_1=\left[\begin{array}{c}2x_1-2y_1+3z_1\\ 5x_1-3y_1+z_1\\ x_1-3y_1\end{array}\right]$ and ${\overrightarrow{v}}_2=\left[\begin{array}{c}2x_2-2y_2+3z_2\\ 5x_2-3y_2+z_2\\ x_2-3y_2\end{array}\right]$ be two vectors in $V$.
    • Then ${\overrightarrow{v}}_1+{\overrightarrow{v}}_2=\left[\begin{array}{c}2(x_1+x_2)-2(y_1+y_2)+3(z_1+z_2)\\ 5(x_1+x_2)-3(y_1+y_2)+(z_1+z_2)\\ (x_1+x_2)-3(y_1+y_2)\end{array}\right]$ which is in $V$.
    • Let $c\in \mathbb{R}$ be a scalar and $\overrightarrow{v}=\left[\begin{array}{c}2x-2y+3z\\ 5x-3y+z\\ x-3y\end{array}\right]$ be a vector in $V$.
    • Then $c\overrightarrow{v}=\left[\begin{array}{c}2(cx)-2(cy)+3(cz)\\ 5(cx)-3(cy)+(cz)\\ (cx)-3(cy)\end{array}\right]$ which is in $V$.
    • Therefore, $V$ is a subspace of ${\mathbb{R}}^3$.
  • Pattern I see, if there's a constant. Then it might not be a subspace. If it's quadratic or higher, it might not be. If it's simply linear then it could be.
  • Types of sets that are always subspaces:
    • $\mathrm{Im}(T_A)=\mathrm{Im}(A)=A\overrightarrow{x}|\overrightarrow{x}\in {\mathbb{R}}^n$
      • Or $\overrightarrow{b}|A\overrightarrow{x}=\overrightarrow{b}$ has at least $1$ solution.
    • $\mathrm{null}(A)=\overrightarrow{x}\in {\mathbb{R}}^n|A\overrightarrow{x}=\overrightarrow{0}$
      • $\overrightarrow{x}\in {\mathbb{R}}^n$ is the domain of $T_A$
  • $\mathrm{Im}\left[\begin{array}{ccc}3& -2& 3\\ 5& -3& 1\\ 1& -3& 0\end{array}\right]$
  • $\mathrm{Im}(T)$ where $T\left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\left[\begin{array}{c}2z-2y+3z\\ 5x-3y+z\\ x-3y\end{array}\right]$
  • An eigenspace is a null space of $A-\lambda I$ where $\lambda$ is an eigenvalue of $A$.
  • Example:
    • $T=\left[\begin{array}{c}x\\ y\\ z\end{array}\right]\in {\mathbb{R}}^3|2x-3z=y+1$
    • $\overrightarrow{0}⟹0=1$
    • So $T$ is not a subspace of ${\mathbb{R}}^3$.
  • $S=\left[\begin{array}{c}x\\ y\\ z\end{array}\right]\in {\mathbb{R}}^3|\{\begin{array}{l}2x+y=2z-y\\ 3z-y=y+x\end{array}$
  • Then you can do $\mathrm{null}\left[\begin{array}{ccc}2& 2& -2\\ -1& -2& 3\end{array}\right]$
• 3
  • Is $\left[\begin{array}{c}1\\ 0\\ 1\end{array}\right]\in \mathrm{span}\{\left[\begin{array}{c}2\\ 1\\ 3\end{array}\right],\left[\begin{array}{c}5\\ 1\\ 6\end{array}\right]\}$
  • Yes
  • $\overrightarrow{x}\in \mathrm{span}\{\overrightarrow{u},\overrightarrow{v}\}⟺\overrightarrow{x}=a\overrightarrow{u}+b\overrightarrow{v}$
  • $\left[\begin{array}{ccc}2& 5& 1\\ 1& 1& 0\\ 3& 6& 1\end{array}\right]$
• 4
  • $U\subseteq {\mathbb{R}}^3$
  • Which can justify $U$ is a subspace of ${\mathbb{R}}^3$?
    • If $\overrightarrow{0},{\overrightarrow{e}}_1,{\overrightarrow{e}}_2,\left[\begin{array}{c}1\\ 1\\ 0\end{array}\right]$ are in u
      • I don't know why we have that last vector.
    • For every $\overrightarrow{u},\overrightarrow{v}\in U$ we have $\overrightarrow{u}+\overrightarrow{v}\in U$
      • This is the definition of a subspace.
    • $U=\mathrm{span}\{{\overrightarrow{e}}_1,{\overrightarrow{e}}_2\}$
      • Span is the set of all linear combinations of ${\overrightarrow{e}}_1$ and ${\overrightarrow{e}}_2$.
      • So that would be a stronger condition than just saying ${\overrightarrow{e}}_1$ and ${\overrightarrow{e}}_2$ are in $U$.
    • $U=\mathrm{null}\left[\begin{array}{ccc}1& 2& 3\\ 4& 5& 6\end{array}\right]$
      • Null space is the set of all vectors $\overrightarrow{x}$ such that $A\overrightarrow{x}=\overrightarrow{0}$.
    • $u=\{\overrightarrow{0}\}$
      • Trivial subspace.
  • Spans are always subspaces
  • Null spaces are always subspaces
  • Images are always subspaces
  • Subspaces are either finite (one vector) or infinite. Similar to how matrices can have 1 or infinite solutions.
• Which of the following on their own can be used to say $U$ is not a subspace of ${\mathbb{R}}^2$
  • ${\overrightarrow{e}}_1,{\overrightarrow{e}}_2\in U$ but $⟨-2,3⟩\notin U$
    • Well it's a linear combination of ${\overrightarrow{e}}_1$ and ${\overrightarrow{e}}_2$ so it should be in $U$ if ${\overrightarrow{e}}_1$ and ${\overrightarrow{e}}_2$ are in $U$.
  • $⟨2,-1⟩\in U$ but $⟨-2,1⟩\notin U$
    • This means $\overrightarrow{v}\in U,-\overrightarrow{v}\notin U$.
    • It should be in there, but it's not. So $U$ is not a subspace.
  • $\overrightarrow{0}\notin U$
    • This is the definition of a subspace. So if $\overrightarrow{0}$ is not in $U$ then $U$ is not a subspace.
  • $⟨2,-1⟩\in U$ but $2{\overrightarrow{e}}_1,-{\overrightarrow{e}}_2\notin U$
    • Doesn't really say anything, because maybe the spanning vectors are not basis vectors. So maybe $⟨2,-1⟩$ is a linear combination of some other vectors in $U$.