Education: Exercises: Subspace and Spanning sets


Exercises and Solutions:
Problem 1

Which of these subsets of the vector space of

2 \! \times \! 2

matrices are subspaces under the inherited operations? For each one that is a subspace, parametrize its description. For each that is not, give a condition that fails.

$\{\begin{pmatrix} a &0 \\ 0 &b \end{pmatrix} \,\big|\, a,b\in\mathbb{R}\}$
$\{\begin{pmatrix} a &0 \\ 0 &b \end{pmatrix} \,\big|\, a+b=0\}$
$\{\begin{pmatrix} a &0 \\ 0 &b \end{pmatrix} \,\big|\, a+b=5\}$
$\{\begin{pmatrix} a &c \\ 0 &b \end{pmatrix} \,\big|\, a+b=0, c\in\mathbb{R}\}$

Answer

By Lemma 2.9, to see if each subset of

\mathcal{M}_{2 \! \times \! 2}

is a subspace, we need only check if it is nonempty and closed.

Yes, it is easily checked to be nonempty and closed. This is a parametrization.
$\{a\begin{pmatrix} 1 &0 \\ 0 &0 \end{pmatrix} +b\begin{pmatrix} 0 &0 \\ 0 &1 \end{pmatrix} \,\big|\, a,b\in\mathbb{R}\}$
By the way, the parametrization also shows that it is a subspace, it is given as the span of the two-matrix set, and any span is a subspace.
Yes; it is easily checked to be nonempty and closed. Alternatively, as mentioned in the prior answer, the existence of a parametrization shows that it is a subspace. For the parametrization, the condition $a+b=0$ can be rewritten as $a=-b$ . Then we have this.
$\{\begin{pmatrix} -b &0 \\ 0 &b \end{pmatrix} \,\big|\, b\in\mathbb{R}\} =\{b\begin{pmatrix} -1 &0 \\ 0 &1 \end{pmatrix} \,\big|\, b\in\mathbb{R}\}$
No. It is not closed under addition. For instance,
$\begin{pmatrix} 5 &0 \\ 0 &0 \end{pmatrix} +\begin{pmatrix} 5 &0 \\ 0 &0 \end{pmatrix} =\begin{pmatrix} 10 &0 \\ 0 &0 \end{pmatrix}$
is not in the set. (This set is also not closed under scalar multiplication, for instance, it does not contain the zero matrix.)
Yes.
$\{b\begin{pmatrix} -1 &0 \\ 0 &1 \end{pmatrix} +c\begin{pmatrix} 0 &1 \\ 0 &0 \end{pmatrix} \,\big|\, b,c\in\mathbb{R}\}$

Problem 2

Is this a subspace of $\mathcal{P}_2$ : $\{a_0+a_1x+a_2x^2\,\big|\, a_0+2a_1+a_2=4\}$ ? If it is then parametrize its description.

Answer

No, it is not closed. In particular, it is not closed under scalar multiplication because it does not contain the zero polynomial.
Problem 3

Decide if the vector lies in the span of the set, inside of the space.
1. $\begin{pmatrix} 2 \\ 0 \\ 1 \end{pmatrix}$ , $\{\begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}, \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix} \}$ , in $\mathbb{R}^3$
2. $x-x^3$ , $\{x^2,2x+x^2,x+x^3\}$ , in $\mathcal{P}_3$
3. $\begin{pmatrix} 0 &1 \\ 4 &2 \end{pmatrix}$ , $\{\begin{pmatrix} 1 &0 \\ 1 &1 \end{pmatrix}, \begin{pmatrix} 2 &0 \\ 2 &3 \end{pmatrix} \}$ , in $\mathcal{M}_{2 \! \times \! 2}$
Answer
1. Yes, solving the linear system arising from
  $r_1\begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}+r_2\begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix} =\begin{pmatrix} 2 \\ 0 \\ 1 \end{pmatrix}$
  $r_1=2$ and $r_2=1$ .
2. Yes; the linear system arising from $r_1(x^2)+r_2(2x+x^2)+r_3(x+x^3)=x-x^3$
  $\begin{array}{*{3}{rc}r} & &2r_2 &+ &r_3 &= &1 \\ r_1 &+ &r_2 & & &= &0 \\ & & & &r_3 &= &-1 \end{array}$
  gives that $-1(x^2)+1(2x+x^2)-1(x+x^3)=x-x^3$ .
3. No; any combination of the two given matrices has a zero in the upper right.
Problem 4

Which of these are members of the span $[\{\cos^2x,\sin^2x\} ]$ in the vector space of real-valued functions of one real variable?
1. $f(x)=1$
2. $f(x)=3+x^2$
3. $f(x)=\sin x$
4. $f(x)=\cos (2x)$
Answer
1. Yes; it is in that span since $1\cdot\cos^2x+1\cdot\sin^2x=f(x)$ .
2. No, since $r_1\cos^2x+r_2\sin^2x=3+x^2$ has no scalar solutions that work for all $x$ . For instance, setting $x$ to be $0$ and $\pi$ gives the two equations $r_1\cdot 1+r_2\cdot 0=3$ and $r_1\cdot 1+r_2\cdot 0=3+\pi^2$ , which are not consistent with each other.
3. No; consider what happens on setting $x$ to be $\pi/2$ and $3\pi/2$ .
4. Yes, $\cos (2x)=1\cdot\cos^2(x)-1\cdot\sin^2(x)$ .
Problem 5

Which of these sets spans $\mathbb{R}^3$ ? That is, which of these sets has the property that any three-tall vector can be expressed as a suitable linear combination of the set's elements?
1. $\{ \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}, \begin{pmatrix} 0 \\ 2 \\ 0 \end{pmatrix}, \begin{pmatrix} 0 \\ 0 \\ 3 \end{pmatrix} \}$
2. $\{ \begin{pmatrix} 2 \\ 0 \\ 1 \end{pmatrix}, \begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix}, \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix} \}$
3. $\{ \begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix}, \begin{pmatrix} 3 \\ 0 \\ 0 \end{pmatrix} \}$
4. $\{ \begin{pmatrix} 1 \\ 0 \\ 1 \end{pmatrix}, \begin{pmatrix} 3 \\ 1 \\ 0 \end{pmatrix}, \begin{pmatrix} -1 \\ 0 \\ 0 \end{pmatrix}, \begin{pmatrix} 2 \\ 1 \\ 5 \end{pmatrix} \}$
5. $\{ \begin{pmatrix} 2 \\ 1 \\ 1 \end{pmatrix}, \begin{pmatrix} 3 \\ 0 \\ 1 \end{pmatrix}, \begin{pmatrix} 5 \\ 1 \\ 2 \end{pmatrix}, \begin{pmatrix} 6 \\ 0 \\ 2 \end{pmatrix} \}$
Answer
1. Yes, for any $x,y,z\in\mathbb{R}$ this equation
  $r_1\begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix} +r_2\begin{pmatrix} 0 \\ 2 \\ 0 \end{pmatrix} +r_3\begin{pmatrix} 0 \\ 0 \\ 3 \end{pmatrix} =\begin{pmatrix} x \\ y \\ z \end{pmatrix}$
  has the solution $r_1=x$ , $r_2=y/2$ , and $r_3=z/3$ .
2. Yes, the equation
  $r_1\begin{pmatrix} 2 \\ 0 \\ 1 \end{pmatrix} +r_2\begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix} +r_3\begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix} =\begin{pmatrix} x \\ y \\ z \end{pmatrix}$
  gives rise to this
  $\begin{array}{*{3}{rc}r} 2r_1 &+ &r_2 & & &= &x \\ & &r_2 & & &= &y \\ r_1 & & &+ &r_3 &= &z \\ \end{array} \;\xrightarrow[]{-(1/2)\rho_1+\rho_3}\;\;\xrightarrow[]{(1/2)\rho_2+\rho_3}\; \begin{array}{*{3}{rc}r} 2r_1 &+ &r_2 & & &= &x \\ & &r_2 & & &= &y \\ & & & &r_3 &= &-(1/2)x+(1/2)y+z \\ \end{array}$
  so that, given any $x$ , $y$ , and $z$ , we can compute that $r_3=(-1/2)x+(1/2)y+z$ , $r_2=y$ , and $r_1=(1/2)x-(1/2)y$ .
3. No. In particular, the vector
  $\begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix}$
  cannot be gotten as a linear combination since the two given vectors both have a third component of zero.
4. Yes. The equation
  $r_1\begin{pmatrix} 1 \\ 0 \\ 1 \end{pmatrix} +r_2\begin{pmatrix} 3 \\ 1 \\ 0 \end{pmatrix} +r_3\begin{pmatrix} -1\\ 0 \\ 0 \end{pmatrix} +r_4\begin{pmatrix} 2 \\ 1 \\ 5 \end{pmatrix} =\begin{pmatrix} x \\ y \\ z \end{pmatrix}$
  leads to this reduction.
  $\left(\begin{array}{*{4}{c}|c} 1 &3 &-1 &2 &x \\ 0 &1 &0 &1 &y \\ 1 &0 &0 &5 &z \end{array}\right) \xrightarrow[]{-\rho_1+\rho_3}\;\xrightarrow[]{3\rho_2+\rho_3} \left(\begin{array}{*{4}{c}|c} 1 &3 &-1 &2 &x \\ 0 &1 &0 &1 &y \\ 0 &0 &1 &6 &-x+3y+z \end{array}\right)$
  We have infinitely many solutions. We can, for example, set $r_4$ to be zero and solve for $r_3$ , $r_2$ , and $r_1$ in terms of $x$ , $y$ , and $z$ by the usual methods of back-substitution.
5. No. The equation
  $r_1\begin{pmatrix} 2 \\ 1 \\ 1 \end{pmatrix} +r_2\begin{pmatrix} 3 \\ 0 \\ 1 \end{pmatrix} +r_3\begin{pmatrix} 5 \\ 1 \\ 2 \end{pmatrix} +r_4\begin{pmatrix} 6 \\ 0 \\ 2 \end{pmatrix} =\begin{pmatrix} x \\ y \\ z \end{pmatrix}$
  leads to this reduction.
  $\left(\begin{array}{*{4}{c}|c} 2 &3 &5 &6 &x \\ 1 &0 &1 &0 &y \\ 1 &1 &2 &2 &z \end{array}\right) \xrightarrow[-(1/2)\rho_1+\rho_3]{-(1/2)\rho_1+\rho_2} \;\xrightarrow[]{-(1/3)\rho_2+\rho_3} \left(\begin{array}{*{4}{c}|c} 2 &3 &5 &6 &x \\ 0 &-3/2 &-3/2 &-3 &-(1/2)x+y \\ 0 &0 &0 &0 &-(1/3)x-(1/3)y+z \end{array}\right)$
  This shows that not every three-tall vector can be so expressed. Only the vectors satisfying the restriction that $-(1/3)x-(1/3)y+z=0$ are in the span. (To see that any such vector is indeed expressible, take $r_3$ and $r_4$ to be zero and solve for $r_1$ and $r_2$ in terms of $x$ , $y$ , and $z$ by back-substitution.)
Problem 6

Parametrize each subspace's description. Then express each subspace as a span.
1. The subset $\{\begin{pmatrix} a &b &c \end{pmatrix}\,\big|\, a-c=0\}$ of the three-wide row vectors
2. This subset of $\mathcal{M}_{2 \! \times \! 2}$
  $\{\begin{pmatrix} a &b \\ c &d \end{pmatrix} \,\big|\, a+d=0\}$
3. This subset of $\mathcal{M}_{2 \! \times \! 2}$
  $\{\begin{pmatrix} a &b \\ c &d \end{pmatrix} \,\big|\, 2a-c-d=0 \text{ and } a+3b=0 \}$
4. The subset $\{a+bx+cx^3\,\big|\, a-2b+c=0\}$ of $\mathcal{P}_3$
5. The subset of $\mathcal{P}_2$ of quadratic polynomials $p$ such that $p(7)=0$
Answer
1. $\{\begin{pmatrix} c &b &c \end{pmatrix}\,\big|\, b,c\in\mathbb{R}\} =\{b\begin{pmatrix} 0 &1 &0 \end{pmatrix}+c\begin{pmatrix} 1 &0 &1 \end{pmatrix} \,\big|\, b,c\in\mathbb{R}\}$ The obvious choice for the set that spans is $\{\begin{pmatrix} 0 &1 &0 \end{pmatrix},\begin{pmatrix} 1 &0 &1 \end{pmatrix}\}$ .
2. $\{\begin{pmatrix} -d &b \\ c &d \end{pmatrix} \,\big|\, b,c,d\in\mathbb{R}\} =\{b\begin{pmatrix} 0 &1 \\ 0 &0 \end{pmatrix} +c\begin{pmatrix} 0 &0 \\ 1 &0 \end{pmatrix} +d\begin{pmatrix} -1 &0 \\ 0 &1 \end{pmatrix} \,\big|\, b,c,d\in\mathbb{R}\}$ One set that spans this space consists of those three matrices.
3. The system
  $\begin{array}{*{4}{rc}r} a &+ &3b & & & & &= &0 \\ 2a & & & &-c &- &d &= &0 \end{array}$
  gives $b=-(c+d)/6$ and $a=(c+d)/2$ . So one description is this.
  $\{c\begin{pmatrix} 1/2 &-1/6 \\ 1 &0 \end{pmatrix} +d\begin{pmatrix} 1/2 &-1/6 \\ 0 &1 \end{pmatrix} \,\big|\, c,d\in\mathbb{R}\}$
  That shows that a set spanning this subspace consists of those two matrices.
4. The $a=2b-c$ gives $\{(2b-c)+bx+cx^3 \,\big|\, b,c\in\mathbb{R}\} =\{b(2+x)+c(-1+x^3) \,\big|\, b,c\in\mathbb{R}\}$ . So the subspace is the span of the set $\{2+x, -1+x^3\}$ .
5. The set $\{a+bx+cx^2\,\big|\, a+7b+49c=0\}$ parametrized as $\{b(-7+x)+c(-49+x^2)\,\big|\, b,c\in\mathbb{R}\}$ has the spanning set $\{-7+x,-49+x^2\}$
Problem 7

Find a set to span the given subspace of the given space. (Hint. Parametrize each.)
1. the $xz$ -plane in $\mathbb{R}^3$
2. $\{\begin{pmatrix} x \\ y \\ z \end{pmatrix}\,\big|\, 3x+2y+z=0\}$ in $\mathbb{R}^3$
3. $\{\begin{pmatrix} x \\ y \\ z \\ w \end{pmatrix}\,\big|\, 2x+y+w=0 \text{ and } y+2z=0\}$ in $\mathbb{R}^4$
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 $\mathcal{P}_3$
5. The set $\mathcal{P}_4$ in the space $\mathcal{P}_4$
6. $\mathcal{M}_{2 \! \times \! 2}$ in $\mathcal{M}_{2 \! \times \! 2}$
Answer

Each answer given is only one out of many possible.
1. 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.
  $\{\begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix},\begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix}\}$
  
  Problem 8
  
  Parametrize it with
  
  $\{y\begin{pmatrix} -2/3 \\ 1 \\ 0 \end{pmatrix}+z\begin{pmatrix} -1/3 \\ 0 \\ 1 \end{pmatrix} \,\big|\, y,z\in\mathbb{R} \}$ to get $\{\begin{pmatrix} -2/3 \\ 1 \\ 0 \end{pmatrix},\begin{pmatrix} -1/3 \\ 0 \\ 1 \end{pmatrix} \}$ .
  
  $\{\begin{pmatrix} 1 \\ -2 \\ 1 \\ 0 \end{pmatrix}, \begin{pmatrix} -1/2 \\ 0 \\ 0 \\ 1 \end{pmatrix} \}$
  
  Parametrize the description as $\{-a_1+a_1x+a_3x^2+a_3x^3\,\big|\, a_1,a_3\in\mathbb{R} \}$ to get $\{-1+x,x^2+x^3\}.$
  
  $\{1,x,x^2,x^3,x^4\}$
  
  $\{ \begin{pmatrix} 1 &0 \\ 0 &0 \end{pmatrix}, \begin{pmatrix} 0 &1 \\ 0 &0 \end{pmatrix}, \begin{pmatrix} 0 &0 \\ 1 &0 \end{pmatrix}, \begin{pmatrix} 0 &0 \\ 0 &1 \end{pmatrix} \}$
  
  Problem 9
  
  Is $\mathbb{R}^2$ a subspace of $\mathbb{R}^3$ ?
  
  Answer
  
  Technically, no. Subspaces of $\mathbb{R}^3$ are sets of three-tall vectors, while $\mathbb{R}^2$ is a set of two-tall vectors. Clearly though, $\mathbb{R}^2$ is "just like" this subspace of $\mathbb{R}^3$ .
  
  $\{\begin{pmatrix} x \\ y \\ 0 \end{pmatrix}\,\big|\, x,y\in\mathbb{R}\}$
  
  Problem 10
  
  Decide if each is a subspace of the vector space of real-valued functions of one real variable.
  
  The even functions $\{f:\mathbb{R}\to \mathbb{R} \,\big|\, f(-x)=f(x) \text{ for all } x\}$ . For example, two members of this set are $f_1(x)=x^2$ and $f_2(x)=\cos (x)$ .
  
  The odd functions $\{f:\mathbb{R}\to \mathbb{R} \,\big|\, f(-x)=-f(x) \text{ for all } x\}$ . Two members are $f_3(x)=x^3$ and $f_4(x)=\sin(x)$ .
  
  Answer
  
  Of course, the addition and scalar multiplication operations are the ones inherited from the enclosing space.
  
  This is a subspace. It is not empty as it contains at least the two example functions given. It is closed because if $f_1,f_2$ are even and $c_1,c_2$ are scalars then we have this.
  $(c_1f_1+c_2f_2)\,(-x) =c_1\,f_1(-x)+c_2\,f_2(-x) =c_1\,f_1(x)+c_2\,f_2(x) =(c_1f_1+c_2f_2)\,(x)$
  
  This is also a subspace; the check is similar to the prior one
  
  Problem 11
  
  Example 2.16 says that for any vector $\vec{v}$ that is an element of a vector space $V$ , the set $\{r\cdot\vec{v}\,\big|\, r\in\mathbb{R}\}$ is a subspace of $V$ . (This is of course, simply the span of the singleton set $\{\vec{v}\}$ .) Must any such subspace be a proper subspace, or can it be improper?
  
  Answer
  
  It can be improper. If $\vec{v}=\vec{0}$ then this is a trivial subspace. At the opposite extreme, if the vector space is $\mathbb{R}^1$ and $\vec{v}\neq\vec{0}\,$ then the subspace is all of $\mathbb{R}^1$ .
  
  Problem 12
  
  An example following the definition of a vector space shows that the solution set of a homogeneous linear system is a vector space. In the terminology of this subsection, it is a subspace of $\mathbb{R}^n$ where the system has $n$ variables. What about a non-homogeneous linear system; do its solutions form a subspace (under the inherited operations)?
  
  Answer
  
  No, such a set is not closed. For one thing, it does not contain the zero vector.
  
  Problem 13
  
  Example 2.19 shows that $\mathbb{R}^3$ has infinitely many subspaces. Does every nontrivial space have infinitely many subspaces?
  
  Answer
  
  No. The only subspaces of $\mathbb{R}^1$ are the space itself and its trivial subspace. Any subspace $S$ of $\mathbb{R}$ that contains a nonzero member $\vec{v}$ must contain the set of all of its scalar multiples $\{r\cdot\vec{v}\,\big|\, r\in\mathbb{R}\}$ . But this set is all of $\mathbb{R}$
  
  Problem 14
  
  Show that each vector space has only one trivial subspace.
  
  Answer
  
  An exercise in the prior subsection shows that every vector space has only one zero vector (that is, there is only one vector that is the additive identity element of the space). But a trivial space has only one element and that element must be this (unique) zero vector.
  
  Problem 15
  
  All of the subspaces that we've seen use zero in their description in some way. For example, the subspace in Example 2.3 consists of all the vectors from $\mathbb{R}^2$ with a second component of zero. In contrast, the collection of vectors from $\mathbb{R}^2$ with a second component of one does not form a subspace (it is not closed under scalar multiplication). Another example is Example 2.2, where the condition on the vectors is that the three components add to zero. If the condition were that the three components add to one then it would not be a subspace (again, it would fail to be closed). This exercise shows that a reliance on zero is not strictly necessary. Consider the set
  
  $\{\begin{pmatrix} x \\ y \\ z \end{pmatrix}\,\big|\, x+y+z=1\}$
  
  under these operations.
  
  $\begin{pmatrix} x_1 \\ y_1 \\ z_1 \end{pmatrix}+\begin{pmatrix} x_2 \\ y_2 \\ z_2 \end{pmatrix} =\begin{pmatrix} x_1+x_2-1 \\ y_1+y_2 \\ z_1+z_2 \end{pmatrix} \qquad r\begin{pmatrix} x \\ y \\ z \end{pmatrix}=\begin{pmatrix} rx-r+1 \\ ry \\ rz \end{pmatrix}$
  
  Show that it is not a subspace of $\mathbb{R}^3$ . (Hint. See Example 2.5).
  
  Show that it is a vector space. Note that by the prior item, Lemma 2.9 can not apply.
  
  Show that any subspace of $\mathbb{R}^3$ must pass through the origin, and so any subspace of $\mathbb{R}^3$ must involve zero in its description. Does the converse hold? Does any subset of $\mathbb{R}^3$ that contains the origin become a subspace when given the inherited operations?
  
  Answer
  
  It is not a subspace because these are not the inherited operations. For one thing, in this space,
  $0\cdot\begin{pmatrix} x \\ y \\ z \end{pmatrix}=\begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}$
  while this does not, of course, hold in $\mathbb{R}^3$ .
  
  We can combine the argument showing closure under addition with the argument showing closure under scalar multiplication into one single argument showing closure under linear combinations of two vectors. If $r_1,r_2,x_1,x_2,y_1,y_2,z_1,z_2$ are in $\mathbb{R}$ then
  $r_1\begin{pmatrix} x_1 \\ y_1 \\ z_1 \end{pmatrix} +r_2\begin{pmatrix} x_2 \\ y_2 \\ z_2 \end{pmatrix} =\begin{pmatrix} r_1x_1-r_1+1 \\ r_1y_1 \\ r_1z_1 \end{pmatrix} +\begin{pmatrix} r_2x_2-r_2+1 \\ r_2y_2 \\ r_2z_2 \end{pmatrix} =\begin{pmatrix} r_1x_1-r_1+r_2x_2-r_2+1 \\ r_1y_1+r_2y_2 \\ r_1z_1+r_2z_2 \end{pmatrix}$
  (note that the definition of addition in this space is that the first components combine as $(r_1x_1-r_1+1)+(r_2x_2-r_2+1)-1$ , so the first component of the last vector does not say " $+2$ "). Adding the three components of the last vector gives $r_1(x_1-1+y_1+z_1)+r_2(x_2-1+y_2+z_2)+1=r_1\cdot0+r_2\cdot0+1=1$ . Most of the other checks of the conditions are easy (although the oddness of the operations keeps them from being routine). Commutativity of addition goes like this.
  $\begin{pmatrix} x_1 \\ y_1 \\ z_1 \end{pmatrix}+\begin{pmatrix} x_2 \\ y_2 \\ z_2 \end{pmatrix} =\begin{pmatrix} x_1+x_2-1 \\ y_1+y_2 \\ z_1+z_2 \end{pmatrix} =\begin{pmatrix} x_2+x_1-1 \\ y_2+y_1 \\ z_2+z_1 \end{pmatrix} =\begin{pmatrix} x_2 \\ y_2 \\ z_2 \end{pmatrix}+\begin{pmatrix} x_1 \\ y_1 \\ z_1 \end{pmatrix}$
  Associativity of addition has
  $(\begin{pmatrix} x_1 \\ y_1 \\ z_1 \end{pmatrix}+\begin{pmatrix} x_2 \\ y_2 \\ z_2 \end{pmatrix}) +\begin{pmatrix} x_3 \\ y_3 \\ z_3 \end{pmatrix} =\begin{pmatrix} (x_1+x_2-1)+x_3-1 \\ (y_1+y_2)+y_3 \\ (z_1+z_2)+z_3 \end{pmatrix}$
  while
  $\begin{pmatrix} x_1 \\ y_1 \\ z_1 \end{pmatrix} +(\begin{pmatrix} x_2 \\ y_2 \\ z_2 \end{pmatrix}+\begin{pmatrix} x_3 \\ y_3 \\ z_3 \end{pmatrix}) =\begin{pmatrix} x_1+(x_2+x_3-1)-1 \\ y_1+(y_2+y_3) \\ z_1+(z_2+z_3) \end{pmatrix}$
  and they are equal. The identity element with respect to this addition operation works this way
  $\begin{pmatrix} x \\ y \\ z \end{pmatrix}+\begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix} =\begin{pmatrix} x+1-1 \\ y+0 \\ z+0 \end{pmatrix} =\begin{pmatrix} x \\ y \\ z \end{pmatrix}$
  and the additive inverse is similar.
  $\begin{pmatrix} x \\ y \\ z \end{pmatrix}+\begin{pmatrix} -x+2 \\ -y \\ -z \end{pmatrix} =\begin{pmatrix} x+(-x+2)-1 \\ y-y \\ z-z \end{pmatrix} =\begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}$
  The conditions on scalar multiplication are also easy. For the first condition,
  $(r+s)\begin{pmatrix} x \\ y \\ z \end{pmatrix} =\begin{pmatrix} (r+s)x-(r+s)+1 \\ (r+s)y \\ (r+s)z \end{pmatrix}$
  while
  $r\begin{pmatrix} x \\ y \\ z \end{pmatrix}+s\begin{pmatrix} x \\ y \\ z \end{pmatrix} =\begin{pmatrix} rx-r+1 \\ ry \\ rz \end{pmatrix}+\begin{pmatrix} sx-s+1 \\ sy \\ sz \end{pmatrix} =\begin{pmatrix} (rx-r+1)+(sx-s+1)-1 \\ ry+sy \\ rz+sz \end{pmatrix}$
  and the two are equal. The second condition compares
  $r\cdot(\begin{pmatrix} x_1 \\ y_1 \\ z_1 \end{pmatrix}+\begin{pmatrix} x_2 \\ y_2 \\ z_2 \end{pmatrix}) =r\cdot\begin{pmatrix} x_1+x_2-1 \\ y_1+y_2 \\ z_1+z_2 \end{pmatrix} =\begin{pmatrix} r(x_1+x_2-1)-r+1 \\ r(y_1+y_2) \\ r(z_1+z_2) \end{pmatrix}$
  with
  $r\begin{pmatrix} x_1 \\ y_1 \\ z_1 \end{pmatrix}+r\begin{pmatrix} x_2 \\ y_2 \\ z_2 \end{pmatrix} =\begin{pmatrix} rx_1-r+1 \\ ry_1 \\ rz_1 \end{pmatrix} +\begin{pmatrix} rx_2-r+1 \\ ry_2 \\ rz_2 \end{pmatrix} =\begin{pmatrix} (rx_1-r+1)+(rx_2-r+1)-1 \\ ry_1+ry_2 \\ rz_1+rz_2 \end{pmatrix}$
  and they are equal. For the third condition,
  $(rs)\begin{pmatrix} x \\ y \\ z \end{pmatrix} =\begin{pmatrix} rsx-rs+1 \\ rsy \\ rsz \end{pmatrix}$
  while
  $r(s\begin{pmatrix} x \\ y \\ z \end{pmatrix}) =r(\begin{pmatrix} sx-s+1 \\ sy \\ sz \end{pmatrix}) =\begin{pmatrix} r(sx-s+1)-r+1 \\ rsy \\ rsz \end{pmatrix}$
  and the two are equal. For scalar multiplication by $1$ we have this.
  $1\cdot\begin{pmatrix} x \\ y \\ z \end{pmatrix} =\begin{pmatrix} 1x-1+1 \\ 1y \\ 1z \end{pmatrix} =\begin{pmatrix} x \\ y \\ z \end{pmatrix}$
  Thus all the conditions on a vector space are met by these two operations. Remark. A way to understand this vector space is to think of it as the plane in $\mathbb{R}^3$
  $P=\{\begin{pmatrix} x \\ y \\ z \end{pmatrix}\,\big|\, x+y+z=0\}$
  displaced away from the origin by $1$ along the $x$ -axis. Then addition becomes: to add two members of this space,
  $\begin{pmatrix} x_1 \\ y_1 \\ z_1 \end{pmatrix},\;\begin{pmatrix} x_2 \\ y_2 \\ z_2 \end{pmatrix}$
  (such that $x_1+y_1+z_1=1$ and $x_2+y_2+z_2=1$ ) move them back by $1$ to place them in $P$ and add as usual,
  $\begin{pmatrix} x_1-1 \\ y_1 \\ z_1 \end{pmatrix}+\begin{pmatrix} x_2-1 \\ y_2 \\ z_2 \end{pmatrix} =\begin{pmatrix} x_1+x_2-2 \\ y_1+y_2 \\ z_1+z_2 \end{pmatrix} \qquad\text{(in }P\text{)}$
  and then move the result back out by $1$ along the $x$ -axis.
  $\begin{pmatrix} x_1+x_2-1 \\ y_1+y_2 \\ z_1+z_2 \end{pmatrix}.$
  Scalar multiplication is similar.
  
  For the subspace to be closed under the inherited scalar multiplication, where $\vec{v}$ is a member of that subspace,
  $0\cdot\vec{v}=\begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}$
  must also be a member. The converse does not hold. Here is a subset of $\mathbb{R}^3$ that contains the origin
  $\{\begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix},\begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}\}$
  (this subset has only two elements) but is not a subspace.

18/09/2014

Exercises: Subspace and Spanning sets

Exercises and Solutions: