Linear Algebra/Subspaces and Spanning sets/Solutions

Solutions
{{TextBox|1= Find a set to span the given subspace of the given space. (Hint.  Parametrize each.)  the $$ xz $$-plane in $$ \mathbb{R}^3 $$  $$ \{\begin{pmatrix} x \\ y \\ z \end{pmatrix}\,\big|\, 3x+2y+z=0\} $$ in $$ \mathbb{R}^3 $$  $$ \{\begin{pmatrix} x \\ y \\ z \\ w \end{pmatrix}\,\big|\, 2x+y+w=0 \text{ and } y+2z=0\} $$ in $$ \mathbb{R}^4 $$  $$ \{a_0+a_1x+a_2x^2+a_3x^3\,\big|\, a_0+a_1=0 \text{ and } a_2-a_3=0\} $$ in $$ \mathcal{P}_3 $$  The set $$ \mathcal{P}_4 $$ in the space $$ \mathcal{P}_4 $$  $$ \mathcal{M}_{2 \! \times \! 2} $$ in $$ \mathcal{M}_{2 \! \times \! 2} $$ 
 * Problem 7:
 * Answer:

Each answer given is only one out of many possible.  We can parametrize in this way



\{\begin{pmatrix} x \\ 0 \\ z \end{pmatrix}\,\big|\, x,z\in\mathbb{R}\} =\{x\begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix} +z\begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix}\,\big|\, x,z\in\mathbb{R}\} $$

giving this for a spanning set.



\{\begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix},\begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix}\} $$

 As a way to solve it, is to express $$x$$ as $$-(2/3) y - (1/3) z$$ to get this parametrization.



\{\begin{pmatrix} -(2/3)y - (1/3)z \\ y \\ z \end{pmatrix}\,\big|\, y,z\in\mathbb{R}\} =\{ y\begin{pmatrix} -2/3 \\ 1 \\ 0 \end{pmatrix} +z\begin{pmatrix} -1/3 \\ 0 \\ 1 \end{pmatrix} \,\big|\, y,z \in \mathbb{R} \} $$

and get this as a spanning set.



\{\begin{pmatrix} -2/3 \\ 1 \\ 0 \end{pmatrix},\begin{pmatrix} -1/3 \\ 0 \\ 1 \end{pmatrix}\} $$

 Here as a way to go, is to notice that $$x$$ and $$z$$ depend on $$y$$ and $$w$$, but not each other, suggesting to choose $$y$$ and $$w$$ as free variables. Then re-expressing, $$x = -(1/2)y - (1/2)w$$ and $$z = -(1/2)y$$ to obtain this parametrization.



\{\begin{pmatrix}-(1/2)y -(1/2)w \\ y \\ -(1/2)y \\ w\end{pmatrix}\,\big|\, y,w\in\mathbb{R}\} =\{ y\begin{pmatrix}-1/2 \\ 1 \\ -1/2 \\ 0\end{pmatrix} +w\begin{pmatrix}-1/2 \\ 0 \\ 0 \\ 1\end{pmatrix} \,\big|\, y,w\in\mathbb{R}\} $$

and for a possible spanning set we have.



\{\begin{pmatrix}-1/2 \\ 1 \\ -1/2 \\ 0\end{pmatrix}, \begin{pmatrix}-1/2 \\ 0 \\ 0 \\ 1\end{pmatrix}\} $$

 Again, as a way to do it, re-express $$a_0 = -a_1$$ and $$a_2 = a_3$$ to get parametrization.



\{-a_1+a_1x+a_3x^2+a_3x^3\,\big|\, a_1,a_3\in\mathbb{R}\} =\{ a_1(-1+x) + a_3(x^2+x^3)\,\big|\, a_1,a_3\in\mathbb{R}\} \} $$

which gives this for a spanning set.


 * $$\{-1+x,\, x^2+x^3\}$$

 As there are no restrictions, we can get a spaning set right away, skipping parametrization.


 * $$\{1,\, x,\, x^2,\, x^3,\, x^4\}$$

 For same reason as the previous subproblem, we can skip parametrization and go to a spaning set immediatedly.



\begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}, \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}, \begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix}, \begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix} \}$$

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