Linear Algebra/Definition and Examples of Vector Spaces/Solutions

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Solutions[edit]

Problem 1

Name the zero vector for each of these vector spaces.

  1. The space of degree three polynomials under the natural operations
  2. The space of  2 \! \times \! 4 matrices
  3. The space  \{f:[0,1]\to\mathbb{R}\,\big|\, f\text{ is continuous}\}
  4. The space of real-valued functions of one natural number variable
Answer
  1.  0+0x+0x^2+0x^3
  2.  \begin{pmatrix}
0  &0  &0  &0  \\
0  &0  &0  &0
\end{pmatrix}
  3. The constant function  f(x)=0
  4. The constant function  f(n)=0
This exercise is recommended for all readers.
Problem 2

Find the additive inverse, in the vector space, of the vector.

  1. In  \mathcal{P}_3 , the vector  -3-2x+x^2 .
  2. In the space  2 \! \times \! 2 ,
    
\begin{pmatrix}
1  &-1  \\
0  &3
\end{pmatrix}.
  3. In  \{ae^x+be^{-x}\,\big|\, a,b\in\mathbb{R}\} , the space of functions of the real variable  x under the natural operations, the vector  3e^x-2e^{-x} .
Answer
  1.  3+2x-x^2
  2.  \begin{pmatrix}
-1  &+1  \\
0  &-3
\end{pmatrix}
  3.  -3e^x+2e^{-x}
This exercise is recommended for all readers.
Problem 3

Show that each of these is a vector space.

  1. The set of linear polynomials  \mathcal{P}_1=\{a_0+a_1x\,\big|\, a_0,a_1\in\mathbb{R}\} under the usual polynomial addition and scalar multiplication operations.
  2. The set of  2 \! \times \! 2 matrices with real entries under the usual matrix operations.
  3. The set of three-component row vectors with their usual operations.
  4. The set
    
L=\{\begin{pmatrix} x \\ y \\ z \\ w \end{pmatrix}\in\mathbb{R}^4\,\big|\, x+y-z+w=0\}
    under the operations inherited from \mathbb{R}^4.
Answer

Most of the conditions are easy to check; use Example 1.3 as a guide. Here are some comments.

  1. This is just like Example 1.3; the zero element is  0+0x .
  2. The zero element of this space is the 2 \! \times \! 2 matrix of zeroes.
  3. The zero element is the vector of zeroes.
  4. Closure of addition involves noting that the sum
    
\begin{pmatrix} x_1 \\ y_1 \\ z_1 \\ w_1 \end{pmatrix}
+\begin{pmatrix} x_2 \\ y_2 \\ z_2 \\ w_2 \end{pmatrix}
=
\begin{pmatrix} x_1+x_2 \\ y_1+y_2 \\ z_1+z_2 \\ w_1+w_2 \end{pmatrix}
    is in  L because  (x_1+x_2)+(y_1+y_2)-(z_1+z_2)+(w_1+w_2) =(x_1+y_1-z_1+w_1)+(x_2+y_2-z_2+w_2)=0+0 . Closure of scalar multiplication is similar. Note that the zero element, the vector of zeroes, is in L.
This exercise is recommended for all readers.
Problem 4

Show that each of these is not a vector space. (Hint. Start by listing two members of each set.)

  1. Under the operations inherited from  \mathbb{R}^3 , this set
    
\{\begin{pmatrix} x \\ y \\ z \end{pmatrix}\in\mathbb{R}^3\,\big|\, x+y+z=1\}
  2. Under the operations inherited from  \mathbb{R}^3 , this set
    
\{\begin{pmatrix} x \\ y \\ z \end{pmatrix}\in\mathbb{R}^3\,\big|\, x^2+y^2+z^2=1\}
  3. Under the usual matrix operations,
    
\{\begin{pmatrix}
a  &1  \\
b  &c
\end{pmatrix} \,\big|\, a,b,c\in\mathbb{R}\}
  4. Under the usual polynomial operations,
    
\{a_0+a_1x+a_2x^2\,\big|\, a_0,a_1,a_2\in\mathbb{R}^+\}
    where \mathbb{R}^+ is the set of reals greater than zero
  5. Under the inherited operations,
    
\{\begin{pmatrix} x \\ y \end{pmatrix}\in\mathbb{R}^2\,\big|\,
x+3y=4 \text{ and } 2x-y=3 \text{ and } 6x+4y=10\}
Answer

In each item the set is called  Q . For some items, there are other correct ways to show that Q is not a vector space.

  1. It is not closed under addition; it fails to meet condition 1.
    
\begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix},
\begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix}\in Q
\qquad
\begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix}\not\in Q
  2. It is not closed under addition.
    
\begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix},
\begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix}\in Q
\qquad
\begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix}\not\in Q
  3. It is not closed under addition.
    
\begin{pmatrix}
0  &1  \\
0  &0
\end{pmatrix},
\,
\begin{pmatrix}
1  &1  \\
0  &0
\end{pmatrix}\in Q
\qquad
\begin{pmatrix}
1  &2  \\
0  &0
\end{pmatrix}\not\in Q
  4. It is not closed under scalar multiplication.
    
1+1x+1x^2\in Q
\qquad
-1\cdot(1+1x+1x^2)\not\in Q
  5. It is empty, violating condition 4.
Problem 5

Define addition and scalar multiplication operations to make the complex numbers a vector space over  \mathbb{R} .

Answer

The usual operations  (v_0+v_1i)+(w_0+w_1i)=(v_0+w_0)+(v_1+w_1)i and  r(v_0+v_1i)=(rv_0)+(rv_1)i suffice. The check is easy.

This exercise is recommended for all readers.
Problem 6

Is the set of rational numbers a vector space over  \mathbb{R} under the usual addition and scalar multiplication operations?

Answer

No, it is not closed under scalar multiplication since, e.g.,  \pi\cdot 1 is not a rational number.

Problem 7

Show that the set of linear combinations of the variables  x,y,z is a vector space under the natural addition and scalar multiplication operations.

Answer

The natural operations are  (v_1x+v_2y+v_3z)+(w_1x+w_2y+w_3z)=(v_1+w_1)x+(v_2+w_2)y+(v_3+w_3)z and  r\cdot(v_1x+v_2y+v_3z)=(rv_1)x+(rv_2)y+(rv_3)z . The check that this is a vector space is easy; use Example 1.3 as a guide.

Problem 8

Prove that this is not a vector space: the set of two-tall column vectors with real entries subject to these operations.


\begin{pmatrix} x_1 \\ y_1 \end{pmatrix}
+\begin{pmatrix} x_2 \\ y_2 \end{pmatrix}
=\begin{pmatrix} x_1-x_2 \\ y_1-y_2 \end{pmatrix}
\qquad
r\cdot\begin{pmatrix} x \\ y \end{pmatrix}
=\begin{pmatrix} rx \\ ry \end{pmatrix}
Answer

The "+" operation is not commutative (that is, condition 2 is not met); producing two members of the set witnessing this assertion is easy.

Problem 9

Prove or disprove that  \mathbb{R}^3 is a vector space under these operations.

  1. 
\begin{pmatrix} x_1 \\ y_1 \\ z_1 \end{pmatrix}
+\begin{pmatrix} x_2 \\ y_2 \\ z_2 \end{pmatrix}
=\begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}
\quad\text{and}\quad
r\begin{pmatrix} x \\ y \\ z \end{pmatrix}
=\begin{pmatrix} rx \\ ry \\ rz \end{pmatrix}
  2. 
\begin{pmatrix} x_1 \\ y_1 \\ z_1 \end{pmatrix}
+\begin{pmatrix} x_2 \\ y_2 \\ z_2 \end{pmatrix}
=\begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}
\quad\text{and}\quad
r\begin{pmatrix} x \\ y \\ z \end{pmatrix}
=\begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}
Answer
  1. It is not a vector space.
    
(1+1)\cdot\begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}\neq
\begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}
+\begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}
  2. It is not a vector space.
    
1\cdot\begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}\neq\begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}
This exercise is recommended for all readers.
Problem 10

For each, decide if it is a vector space; the intended operations are the natural ones.

  1. The diagonal  2 \! \times \! 2 matrices
    
\{\begin{pmatrix}
a  &0  \\
0  &b
\end{pmatrix}\,\big|\, a,b\in\mathbb{R}\}
  2. This set of  2 \! \times \! 2 matrices
    
\{\begin{pmatrix}
x    &x+y  \\
x+y  &y
\end{pmatrix}\,\big|\, x,y\in\mathbb{R}\}
  3. This set
    
\{\begin{pmatrix} x \\ y \\ z \\ w \end{pmatrix}\in\mathbb{R}^4
\,\big|\, x+y+w=1\}
  4. The set of functions  \{f:\mathbb{R}\to\mathbb{R}\,\big|\, df/dx+2f=0\}
  5. The set of functions  \{f:\mathbb{R}\to\mathbb{R}\,\big|\, df/dx+2f=1\}
Answer

For each "yes" answer, you must give a check of all the conditions given in the definition of a vector space. For each "no" answer, give a specific example of the failure of one of the conditions.

  1. Yes.
  2. Yes.
  3. No, it is not closed under addition. The vector of all 1/4's, when added to itself, makes a nonmember.
  4. Yes.
  5. No,  f(x)=e^{-2x}+(1/2) is in the set but  2\cdot f is not (that is, condition 6 fails).
This exercise is recommended for all readers.
Problem 11

Prove or disprove that this is a vector space: the real-valued functions  f of one real variable such that  f(7)=0 .

Answer

It is a vector space. Most conditions of the definition of vector space are routine; we here check only closure. For addition,  (f_1+f_2)\,(7)=f_1(7)+f_2(7)=0+0=0 . For scalar multiplication,  (r\cdot f)\,(7)=rf(7)=r0=0 .

This exercise is recommended for all readers.
Problem 12

Show that the set  \mathbb{R}^+ of positive reals is a vector space when " x+y " is interpreted to mean the product of  x and  y (so that  2+3 is  6 ), and " r\cdot x " is interpreted as the  r -th power of  x .

Answer

We check Definition 1.1.

First, closure under "+" holds because the product of two positive reals is a positive real. The second condition is satisfied because real multiplication commutes. Similarly, as real multiplication associates, the third checks. For the fourth condition, observe that multiplying a number by  1\in\mathbb{R}^+ won't change the number. Fifth, any positive real has a reciprocal that is a positive real.

The sixth, closure under " \cdot ", holds because any power of a positive real is a positive real. The seventh condition is just the rule that  v^{r+s} equals the product of  v^r and  v^s . The eight condition says that  (vw)^r=v^rw^r . The ninth condition asserts that  (v^r)^s=v^{rs} . The final condition says that  v^1=v .

Problem 13

Is  \{(x,y)\,\big|\, x,y\in\mathbb{R}\} a vector space under these operations?

  1.  (x_1,y_1)+(x_2,y_2)=(x_1+x_2,y_1+y_2) and  r\cdot (x,y)=(rx,y)
  2.  (x_1,y_1)+(x_2,y_2)=(x_1+x_2,y_1+y_2) and  r\cdot (x,y)=(rx,0)
Answer
  1. No:  1\cdot(0,1)+1\cdot(0,1)\neq (1+1)\cdot(0,1) .
  2. No; the same calculation as the prior answer shows a contition in the definition of a vector space that is violated. Another example of a violation of the conditions for a vector space is that  1\cdot (0,1)\neq (0,1) .
Problem 14

Prove or disprove that this is a vector space: the set of polynomials of degree greater than or equal to two, along with the zero polynomial.

Answer

It is not a vector space since it is not closed under addition, as  (x^2)+(1+x-x^2) is not in the set.

Problem 15

At this point "the same" is only an intuition, but nonetheless for each vector space identify the  k for which the space is "the same" as  \mathbb{R}^k .

  1. The  2 \! \times \! 3 matrices under the usual operations
  2. The  n \! \times \! m matrices (under their usual operations)
  3. This set of  2 \! \times \! 2 matrices
    
\{\begin{pmatrix}
a &0 \\
b &c
\end{pmatrix} \,\big|\, a,b,c\in\mathbb{R}\}
  4. This set of  2 \! \times \! 2 matrices
    
\{\begin{pmatrix}
a  &0  \\
b  &c
\end{pmatrix} \,\big|\, a+b+c=0\}
Answer
  1.  6
  2.  nm
  3.  3
  4. To see that the answer is  2 , rewrite it as
    
\{\begin{pmatrix}
a  &0  \\
b  &-a-b
\end{pmatrix} \,\big|\, a,b\in\mathbb{R}\}
    so that there are two parameters.
This exercise is recommended for all readers.
Problem 16

Using  \vec{+} to represent vector addition and  \,\vec{\cdot}\, for scalar multiplication, restate the definition of vector space.

Answer

A vector space (over  \mathbb{R} ) consists of a set  V along with two operations " \mathbin{\vec{+}} " and " \mathbin{\vec{\cdot}} " subject to these conditions. Where  \vec{v},\vec{w}\in V ,

  1. their vector sum  \vec{v}\mathbin{\vec{+}}\vec{w} is an element of  V . If  \vec{u},\vec{v},\vec{w}\in V then
  2.  \vec{v}\mathbin{\vec{+}}\vec{w}=\vec{w}\mathbin{\vec{+}}\vec{v} and
  3.  (\vec{v}\mathbin{\vec{+}}\vec{w})\mathbin{\vec{+}}\vec{u}=\vec{v}\mathbin{\vec{+}}(\vec{w}\mathbin{\vec{+}}\vec{u}) .
  4. There is a zero vector  \vec{0}\in V such that  \vec{v}\mathbin{\vec{+}}\vec{0}=\vec{v}\, for all  \vec{v}\in V.
  5. Each  \vec{v}\in V has an additive inverse  \vec{w}\in V such that  \vec{w}\mathbin{\vec{+}}\vec{v}=\vec{0} . If  r,s are scalars, that is, members of  \mathbb{R} ), and  \vec{v},\vec{w}\in V then
  6. each scalar multiple  r\cdot\vec{v} is in  V . If  r,s\in\mathbb{R} and  \vec{v},\vec{w}\in V then
  7.  (r+ s)\cdot\vec{v}=r\cdot\vec{v}\mathbin{\vec{+}} s\cdot\vec{v} , and
  8.  r\mathbin{\vec{\cdot}} (\vec{v}+\vec{w})=r\mathbin{\vec{\cdot}}\vec{v}+r\mathbin{\vec{\cdot}}\vec{w} , and
  9.  (rs)\mathbin{\vec{\cdot}} \vec{v} =r\mathbin{\vec{\cdot}} (s\mathbin{\vec{\cdot}}\vec{v}) , and
  10.  1\mathbin{\vec{\cdot}} \vec{v}=\vec{v} .
This exercise is recommended for all readers.
Problem 17

Prove these.

  1. Any vector is the additive inverse of the additive inverse of itself.
  2. Vector addition left-cancels: if  \vec{v},\vec{s},\vec{t}\in V then  \vec{v}+\vec{s}=\vec{v}+\vec{t}\, implies that  \vec{s}=\vec{t} .
Answer
  1. Let  V be a vector space, assume that  \vec{v}\in V , and assume that  \vec{w}\in V is the additive inverse of \vec{v} so that  \vec{w}+\vec{v}=\vec{0} . Because addition is commutative,  \vec{0}=\vec{w}+\vec{v}=\vec{v}+\vec{w} , so therefore  \vec{v} is also the additive inverse of  \vec{w} .
  2. Let  V be a vector space and suppose  \vec{v},\vec{s},\vec{t}\in V . The additive inverse of  \vec{v} is  -\vec{v} so  \vec{v}+\vec{s}=\vec{v}+\vec{t} gives that  -\vec{v}+\vec{v}+\vec{s}=-\vec{v}+\vec{v}+\vec{t} , which says that  \vec{0}+\vec{s}=\vec{0}+\vec{t} and so  \vec{s}=\vec{t} .
Problem 18

The definition of vector spaces does not explicitly say that  \vec{0}+\vec{v}=\vec{v} (it instead says that  \vec{v}+\vec{0}=\vec{v} ). Show that it must nonetheless hold in any vector space.

Answer

Addition is commutative, so in any vector space, for any vector  \vec{v} we have that  \vec{v}=\vec{v}+\vec{0}=\vec{0}+\vec{v} .

This exercise is recommended for all readers.
Problem 19

Prove or disprove that this is a vector space: the set of all matrices, under the usual operations.

Answer

It is not a vector space since addition of two matrices of unequal sizes is not defined, and thus the set fails to satisfy the closure condition.

Problem 20

In a vector space every element has an additive inverse. Can some elements have two or more?

Answer

Each element of a vector space has one and only one additive inverse.

For, let  V be a vector space and suppose that  \vec{v}\in V . If  \vec{w}_1,\vec{w}_2\in V are both additive inverses of  \vec{v} then consider  \vec{w}_1+\vec{v}+\vec{w}_2 . On the one hand, we have that it equals \vec{w}_1+(\vec{v}+\vec{w}_2)= \vec{w}_1+\vec{0}=\vec{w}_1. On the other hand we have that it equals (\vec{w}_1+\vec{v})+\vec{w}_2= \vec{0}+\vec{w}_2=\vec{w}_2. Therefore, \vec{w}_1=\vec{w}_2.

Problem 21
  1. Prove that every point, line, or plane thru the origin in  \mathbb{R}^3 is a vector space under the inherited operations.
  2. What if it doesn't contain the origin?
Answer
  1. Every such set has the form  \{r\cdot\vec{v}+s\cdot\vec{w}\,\big|\, r,s\in\mathbb{R}\} where either or both of  \vec{v},\vec{w} may be  \vec{0} . With the inherited operations, closure of addition  (r_1\vec{v}+s_1\vec{w})+(r_2\vec{v}+s_2\vec{w})=(r_1+r_2)\vec{v}+(s_1+s_2)\vec{w} and scalar multiplication  c(r\vec{v}+s\vec{w})=(cr)\vec{v}+(cs)\vec{w} are easy. The other conditions are also routine.
  2. No such set can be a vector space under the inherited operations because it does not have a zero element.
This exercise is recommended for all readers.
Problem 22

Using the idea of a vector space we can easily reprove that the solution set of a homogeneous linear system has either one element or infinitely many elements. Assume that  \vec{v}\in V is not  \vec{0} .

  1. Prove that  r\cdot\vec{v}=\vec{0} if and only if  r=0 .
  2. Prove that  r_1\cdot\vec{v}=r_2\cdot\vec{v} if and only if  r_1=r_2 .
  3. Prove that any nontrivial vector space is infinite.
  4. Use the fact that a nonempty solution set of a homogeneous linear system is a vector space to draw the conclusion.
Answer

Assume that  \vec{v}\in V is not  \vec{0} .

  1. One direction of the if and only if is clear: if r=0 then r\cdot\vec{v}=\vec{0}. For the other way, let  r be a nonzero scalar. If  r\vec{v}=\vec{0} then  (1/r)\cdot r\vec{v}=(1/r)\cdot \vec{0} shows that \vec{v}=\vec{0}, contrary to the assumption.
  2. Where  r_1,r_2 are scalars,  r_1\vec{v}=r_2\vec{v}\, holds if and only if  (r_1-r_2)\vec{v}=\vec{0} . By the prior item, then  r_1-r_2=0 .
  3. A nontrivial space has a vector  \vec{v}\neq\vec{0} . Consider the set  \{k\cdot\vec{v}\,\big|\, k\in\mathbb{R}\} . By the prior item this set is infinite.
  4. The solution set is either trivial, or nontrivial. In the second case, it is infinite.
Problem 23

Is this a vector space under the natural operations: the real-valued functions of one real variable that are differentiable?

Answer

Yes. A theorem of first semester calculus says that a sum of differentiable functions is differentiable and that  (f+g)^\prime=f^\prime+g^\prime , and that a multiple of a differentiable function is differentiable and that  (r\cdot f)^\prime=r\,f^\prime .

Problem 24

A vector space over the complex numbers \mathbb{C} has the same definition as a vector space over the reals except that scalars are drawn from  \mathbb{C} instead of from  \mathbb{R} . Show that each of these is a vector space over the complex numbers. (Recall how complex numbers add and multiply:  (a_0+a_1i)+(b_0+b_1i)=(a_0+b_0)+(a_1+b_1)i and  (a_0+a_1i)(b_0+b_1i)=(a_0b_0-a_1b_1)+(a_0b_1+a_1b_0)i .)

  1. The set of degree two polynomials with complex coefficients
  2. This set
    
\{\begin{pmatrix}
0  &a  \\
b  &0
\end{pmatrix}\,\big|\, a,b\in\mathbb{C}\text{ and }
a+b=0+0i \}
Answer

The check is routine. Note that "1" is  1+0i and the zero elements are these.

  1.  (0+0i)+(0+0i)x+(0+0i)x^2
  2.  \begin{pmatrix}
0+0i  &0+0i  \\
0+0i  &0+0i
\end{pmatrix}
Problem 25

Name a property shared by all of the  \mathbb{R}^n 's but not listed as a requirement for a vector space.

Answer

Notably absent from the definition of a vector space is a distance measure.

This exercise is recommended for all readers.
Problem 26
    • Prove that a sum of four vectors  \vec{v}_1,\ldots,\vec{v}_4\in V can be associated in any way without changing the result.
      \begin{array}{rl}
((\vec{v}_1+\vec{v}_2)+\vec{v}_3)+\vec{v}_4
&=(\vec{v}_1+(\vec{v}_2+\vec{v}_3))+\vec{v}_4 \\
&=(\vec{v}_1+\vec{v}_2)+(\vec{v}_3+\vec{v}_4) \\
&=\vec{v}_1+((\vec{v}_2+\vec{v}_3)+\vec{v}_4) \\
&=\vec{v}_1+(\vec{v}_2+(\vec{v}_3+\vec{v}_4))
\end{array}
      This allows us to simply write " \vec{v}_1+\vec{v}_2+\vec{v}_3+\vec{v}_4 " without ambiguity.
    • Prove that any two ways of associating a sum of any number of vectors give the same sum. (Hint. Use induction on the number of vectors.)
      Answer
      1. A small rearrangement does the trick.
        \begin{array}{rl}
(\vec{v}_1+(\vec{v}_2+\vec{v}_3))+\vec{v}_4
&=((\vec{v}_1+\vec{v}_2)+\vec{v}_3)+\vec{v}_4 \\
&=(\vec{v}_1+\vec{v}_2)+(\vec{v}_3+\vec{v}_4) \\
&=\vec{v}_1+(\vec{v}_2+(\vec{v}_3+\vec{v}_4)) \\
&=\vec{v}_1+((\vec{v}_2+\vec{v}_3)+\vec{v}_4)
\end{array}
        Each equality above follows from the associativity of three vectors that is given as a condition in the definition of a vector space. For instance, the second "=" applies the rule (\vec{w}_1+\vec{w}_2)+\vec{w}_3=\vec{w}_1+(\vec{w}_2+\vec{w}_3) by taking \vec{w}_1 to be \vec{v}_1+\vec{v}_2, taking \vec{w}_2 to be \vec{v}_3, and taking \vec{w}_3 to be \vec{v}_4.
      2. The base case for induction is the three vector case. This case  \vec{v}_1+(\vec{v}_2+\vec{v}_3)=(\vec{v}_1+\vec{v}_2)+\vec{v}_3 is required of any triple of vectors by the definition of a vector space. For the inductive step, assume that any two sums of three vectors, any two sums of four vectors, ..., any two sums of k vectors are equal no matter how the sums are parenthesized. We will show that any sum of  k+1 vectors equals this one  ((\cdots((\vec{v}_1+\vec{v}_2)+\vec{v}_3)+\cdots)+\vec{v}_k)+\vec{v}_{k+1} . Any parenthesized sum has an outermost " + ". Assume that it lies between  \vec{v}_m and  \vec{v}_{m+1} so the sum looks like this.
        
(\cdots\,(\vec{v}_1+(\cdots+\vec{v}_m))\,\cdots)+(\cdots\,(\vec{v}_{m+1}+(\cdots+\vec{v}_{k+1}))\,\cdots)
        The second half involves fewer than k+1 additions, so by the inductive hypothesis we can re-parenthesize it so that it reads left to right from the inside out, and in particular, so that its outermost "+" occurs right before \vec{v}_{k+1}.
        
=(\cdots\,(\vec{v}_1+(\,\cdots\,+\vec{v}_m))\,\cdots)+(\cdots(\vec{v}_{m+1}+(\cdots+\vec{v}_{k})\cdots)+\vec{v}_{k+1})
        Apply the associativity of the sum of three things
        
=((\,\cdots\,(\vec{v}_1+(\cdots+\vec{v}_m)\,\cdots\,)
+(\,\cdots\,(\vec{v}_{m+1}+(\cdots\,\vec{v}_k))\cdots)
+\vec{v}_{k+1}
        and finish by applying the inductive hypothesis inside these outermost parenthesis.
    Problem 27

    For any vector space, a subset that is itself a vector space under the inherited operations (e.g., a plane through the origin inside of  \mathbb{R}^3 ) is a subspace.

    1. Show that  \{a_0+a_1x+a_2x^2\,\big|\, a_0+a_1+a_2=0\} is a subspace of the vector space of degree two polynomials.
    2. Show that this is a subspace of the  2 \! \times \! 2 matrices.
      
\{\begin{pmatrix}
a  &b  \\
c  &0
\end{pmatrix} \,\big|\, a+b=0\}
    3. Show that a nonempty subset  S of a real vector space is a subspace if and only if it is closed under linear combinations of pairs of vectors: whenever  c_1,c_2\in\mathbb{R} and  \vec{s}_1,\vec{s}_2\in S then the combination  c_1\vec{v}_1+c_2\vec{v}_2 is in  S .
    Answer
    1. We outline the check of the conditions from Definition 1.1. Additive closure holds because if  a_0+a_1+a_2=0 and  b_0+b_1+b_2=0 then
      
(a_0+a_1x+a_2x^2)+(b_0+b_1x+b_2x^2)=
(a_0+b_0)+(a_1+b_1)x+(a_2+b_2)x^2
      is in the set since  (a_0+b_0)+(a_1+b_1)+(a_2+b_2)=(a_0+a_1+a_2)+(b_0+b_1+b_2) is zero. The second through fifth conditions are easy. Closure under scalar multiplication holds because if  a_0+a_1+a_2=0 then
      
r\cdot(a_0+a_1x+a_2x^2)=
(ra_0)+(ra_1)x+(ra_2)x^2
      is in the set as  ra_0+ra_1+ra_2=r(a_0+a_1+a_2) is zero. The remaining conditions here are also easy.
    2. This is similar to the prior answer.
    3. Call the vector space  V . We have two implications: left to right, if  S is a subspace then it is closed under linear combinations of pairs of vectors and, right to left, if a nonempty subset is closed under linear combinations of pairs of vectors then it is a subspace. The left to right implication is easy; we here sketch the other one by assuming  S is nonempty and closed, and checking the conditions of Definition 1.1. First, to show closure under addition, if  \vec{s}_1,\vec{s}_2\in S then  \vec{s}_1+\vec{s}_2\in S as  \vec{s}_1+\vec{s}_2=1\cdot\vec{s}_1+1\cdot\vec{s}_2 . Second, for any  \vec{s}_1,\vec{s}_2\in S , because addition is inherited from  V , the sum  \vec{s}_1+\vec{s}_2 in  S equals the sum  \vec{s}_1+\vec{s}_2 in  V and that equals the sum  \vec{s}_2+\vec{s}_1 in  V and that in turn equals the sum  \vec{s}_2+\vec{s}_1 in  S . The argument for the third condition is similar to that for the second. For the fourth, suppose that  \vec{s} is in the nonempty set  S and note that  0\cdot\vec{s}=\vec{0}\in S ; showing that the  \vec{0} of  V acts under the inherited operations as the additive identity of  S is easy. The fifth condition is satisfied because for any  \vec{s}\in S closure under linear combinations shows that the vector  0\cdot\vec{0}+(-1)\cdot\vec{s} is in  S ; showing that it is the additive inverse of  \vec{s} under the inherited operations is routine. The proofs for the remaining conditions are similar.