4.18: Amira Vector Space Axioms

Learning Objectives

By the end of this section, students will be able to:

• Identify the zero element and the additive inverse in a given set when nonstandard operations are defined.
• Verify closure under addition and scalar multiplication using arbitrary elements.
• Explain how different operation rules affect the structure of a set and determine whether it forms a vector space.

Why we Should Study Vector Spaces?

Previously, we worked with matrices and solved systems of linear equations, finding solutions and solution sets in \(\mathbb{R}^n\). In those problems, we were allowed to add real numbers and multiply them by scalars. We also explored matrix addition and scalar multiplication, noting that these operations are only defined for matrices of the same size.

We discovered that square matrices play a special role in solving systems of equations, they follow consistent algebraic rules that allow us to perform operations like finding determinants and inverses.

What makes the set of all points in \(\mathbb{R}^n\) and the set of all square matrices of a given size so powerful is that addition and scalar multiplication always produce another element of the same set. This property, staying “inside the set” after performing these operations, is one of the key ideas behind vector spaces.

The diagram below shows the pitch control system of the Space Shuttle, adapted from Figure 3 in the NASA Technical Memorandum Space Shuttle Digital Flight Control System (1976). It illustrates how feedback signals are combined to maintain the desired pitch angle during ascent. The diagram includes several summing junctions, represented by the symbol \(\otimes\), where signals such as the commanded pitch and the measured pitch are added or subtracted. Each junction \(\otimes\)demonstrates the closure property of a vector space, meaning that when two valid signals are added, the result is another valid signal of the same type.

The system also includes blocks that multiply signals by constants, showing scalar multiplication. Together, the junctions \(\otimes\) and the scaling blocks illustrate how the collection of all control signals, including commands, sensor readings, and feedback corrections, forms a vector space of continuous functions. This example connects the abstract definition of a vector space to a real engineering system that depends on addition and scaling of signals to maintain stability and performance.

                        

Definition \(\PageIndex{1}\): Vector Space

A Vector Space is a set \(V\) of elements called vectors, having operations of addition and scalar multiplication defined on it that satisfy the following conditions (\(u\), \(v\), and \(w\) are arbitrary elements of \(V\), and \(c\) and \(d\) are scalars.)

Closure Axioms

1. The sum \(u+v\) exists and is an element of \(V\). (\(V\) is closed under addition.)
2. \(cu\) is an element of \(V\). (\(V\) is closed under multiplication.)

Addition Axioms

1. \(u+v=v+u\) (commutative property)
2. \(u+(v+w)=(u+v)+w\) (associative property)
3. There exists an element of \(V\), called a zero vector, denoted 0, such that \(u+0=u\)
4. For every element \(u\) of \(V\), there exists an element called a negative of \(u\), denoted \(−u\), such that \(u+(−u)=0\).

Scalar Multiplication Axioms

1. \(c(u+v)=cu+cv\)
2. \((c+d)u=cu+du\)
3. \(c(du)=(cd)u\)
4. \(1u=u\)

In the definition we define two operations; vector addition, denoted by \(+\) and scalar multiplication denoted by placing the scalar next to the vector. A vector space need not have usual operations, and for this reason the operations will always be given in the definition of the vector space. The below axioms for addition (written +) and scalar multiplication must hold for however addition and scalar multiplication are defined for the vector space.

It is important to note that we have seen much of this content before, in terms of \(\mathbb{R}^n\). We will prove in this section that \(\mathbb{R}^n\) is an example of a vector space and therefore all discussions in this chapter will pertain to \(\mathbb{R}^n\). While it may be useful to consider all concepts of this chapter in terms of \(\mathbb{R}^n\), it is also important to understand that these concepts apply to all vector spaces.

Points to Remember When Checking a Vector Space

• The definition requires that (\(u\), \(v\), and \(w\) be arbitrary (random) elements of the set implies that

  • We cannot use specific examples or numbers to verify the axioms.

  • The properties must hold for all elements in the set.

• When testing if a set is a vector space, always start by checking the closure properties:

  1. Closure under addition

  2. Closure under scalar multiplication

  • These two checks often give a quick answer about whether the set can be a vector space.

• The operations of addition and scalar multiplication might not be the usual ones.

  • Because of this, the zero element may also be different from the usual zero.

  • You may need to solve for the additive inverse based on the operations defined in the problem.

Below are examples and non-examples of vector spaces. It is crucial to focus on how randomness (using arbitrary elements) is applied when showing whether a set is a vector space. In every case, we must test the properties using general elements, not specific numbers, to verify that the axioms hold for all possible cases.

Example \(\PageIndex{1}\): \(\mathbb{R}^n\)

\(\mathbb{R}^n\), under the usual operations of vector addition and scalar multiplication, is a vector space.

Solution

To show that \(\mathbb{R}^n\) is a vector space, we need to show that the above axioms hold. Let \(\vec{x}, \vec{y}, \vec{z}\) be vectors in \(\mathbb{R}^n\). We first prove the axioms for vector addition.

• To show that \(\mathbb{R}^n\) is closed under addition, we must show that for two vectors in \(\mathbb{R}^n\) their sum is also in \(\mathbb{R}^n\). The sum \(\vec{x} + \vec{y}\) is given by: \[\left[ \begin{array}{c} x_1 \\ x_2 \\ \vdots \\ x_n \end{array}\right] + \left[ \begin{array}{c} y_1 \\ y_2 \\ \vdots \\ y_n \end{array}\right] = \left[ \begin{array}{c} x_1 + y_1 \\ x_2 + y_2 \\ \vdots \\ x_n + y_n \end{array}\right]\nonumber \] The sum is a vector with \(n\) entries, showing that it is in \(\mathbb{R}^n\). Hence \(\mathbb{R}^n\) is closed under vector addition.
• To show that addition is commutative, consider the following: \[\begin{aligned} \vec{x} + \vec{y} &= \left[ \begin{array}{c} x_1 \\ x_2 \\ \vdots \\ x_n \end{array} \right] + \left[ \begin{array}{c} y_1 \\ y_2 \\ \vdots \\ y_n \end{array} \right] \\ &= \left[ \begin{array}{c} x_1 + y_1 \\ x_2 + y_2 \\ \vdots \\ x_n + y_n \end{array} \right] \\ &= \left[ \begin{array}{c} y_1 + x_1 \\ y_2 + x_2 \\ \vdots \\ y_n + x_n \end{array} \right] \\ &= \left[ \begin{array}{c} y_1 \\ y_2 \\ \vdots \\ y_n \end{array} \right] + \left[ \begin{array}{c} x_1 \\ x_2 \\ \vdots \\ x_n \end{array} \right] \\ &= \vec{y} + \vec{x}\end{aligned}\] Hence addition of vectors in \(\mathbb{R}^n\) is commutative.
• We will show that addition of vectors in \(\mathbb{R}^n\) is associative in a similar way. \[\begin{aligned} (\vec{x} + \vec{y}) + \vec{z} &= \left( \left[ \begin{array}{c} x_1 \\ x_2 \\ \vdots \\ x_n \end{array} \right] + \left[ \begin{array}{c} y_1 \\ y_2 \\ \vdots \\ y_n \end{array} \right] \right) + \left[ \begin{array}{c} z_1 \\ z_2 \\ \vdots \\ z_n \end{array} \right] \\ &= \left[ \begin{array}{c} x_1 + y_1 \\ x_2 + y_2 \\ \vdots \\ x_n + y_n \end{array} \right] + \left[ \begin{array}{c} z_1 \\ z_2 \\ \vdots \\ z_n \end{array} \right] \\ &= \left[ \begin{array}{c} (x_1 + y_1) + z_1 \\ (x_2 + y_2) + z_2\\ \vdots \\ (x_n + y_n) + z_n \end{array} \right] \\ &= \left[ \begin{array}{c} x_1 + (y_1 + z_1) \\ x_2 + (y_2 + z_2)\\ \vdots \\ x_n + (y_n + z_n) \end{array} \right ) \\ &= \left ( \begin{array}{c} x_1 \\ x_2 \\ \vdots \\ x_n \end{array} \right] + \left[ \begin{array}{c} y_1 + z_1 \\ y_2 + z_2\\ \vdots \\ y_n + z_n \end{array} \right] \\ &= \left[ \begin{array}{c} x_1 \\ x_2 \\ \vdots \\ x_n \end{array} \right] + \left( \left[ \begin{array}{c} y_1 \\ y_2 \\ \vdots \\ y_n \end{array} \right] + \left[ \begin{array}{c} z_1 \\ z_2 \\ \vdots \\ z_n \end{array} \right] \right) \\ &= \vec{x} + \left( \vec{y} + \vec{z} \right)\end{aligned}\]Hence addition of vectors is associative.
• Next, we show the existence of an additive identity. Let \(\vec{0} = \left[ \begin{array}{c} 0 \\ 0 \\ \vdots \\ 0 \end{array} \right].\) \[\begin{aligned}\vec{x}+\vec{0}&=\left[\begin{array}{c}x_1\\x_2\\ \vdots \\x_n\end{array}\right]+\left[\begin{array}{c}0\\0\\ \vdots \\0\end{array}\right] \\ &=\left[\begin{array}{c}x_1+0 \\ x_2+0 \\ \vdots \\ x_n+0\end{array}\right] \\ &=\left[\begin{array}{c}x_1\\x_2\\ \vdots \\ x_n\end{array}\right] \\ &=\vec{x}\end{aligned}\] Hence the zero vector \(\vec{0}\) is an additive identity.
• Next, we prove the existence of an additive inverse. Let \(-\vec{x} = \left ( \begin{array}{c} -x_1 \\ -x_2 \\ \vdots \\ -x_n \end{array}\right )\). \[\begin{aligned}\vec{x}+(-\vec{x})&=\left[\begin{array}{c}x_1\\x_2\\ \vdots\\x_n\end{array}\right]+\left[\begin{array}{c}-x_1 \\ -x_2\\ \vdots \\ -x_n\end{array}\right] \\ &=\left[\begin{array}{c}x_1-x_1 \\ x_2-x_2 \\ \vdots \\ x_n-x_n\end{array}\right] \\ &=\left[\begin{array}{c}0\\0\\ \vdots\\0\end{array}\right] \\ &=\vec{0}\end{aligned}\] Hence \(-\vec{x}\) is an additive inverse.

We now need to prove the axioms related to scalar multiplication. Let \(a,b\) be real numbers and let \(\vec{x}, \vec{y}\) be vectors in \(\mathbb{R}^n\).

• We first show that \(\mathbb{R}^n\) is closed under scalar multiplication. To do so, we show that \(a\vec{x}\) is also a vector with \(n\) entries. \[a\vec{x} = a\left[ \begin{array}{c} x_1 \\ x_2 \\ \vdots \\ x_n \end{array} \right] = \left[ \begin{array}{c} ax_1 \\ ax_2 \\ \vdots \\ ax_n \end{array} \right]\nonumber \] The vector \(a\vec{x}\) is again a vector with \(n\) entries, showing that \(\mathbb{R}^n\) is closed under scalar multiplication.
• We wish to show that \(a (\vec{x} + \vec{y}) = a\vec{x} + a\vec{y}\). \[\begin{aligned}a(\vec{x}+\vec{y})&=a\left(\left[\begin{array}{c}x_1\\x_2\\ \vdots\\x_n\end{array}\right]+\left[\begin{array}{c}x_1\\x_2\\ \vdots x_n\end{array}\right]\right)\\ &=a\left[\begin{array}{c}x_1+y_1 \\ x_2+y_2 \\ \vdots \\ x_n+y_n\end{array}\right] \\ &=\left[\begin{array}{c}a(x_1+y_1) \\ a(x_2+y_2) \\ \vdots \\ a(x_n+y_n)\end{array}\right] \\ &=\left[\begin{array}{c}ax_1+ay_1 \\ ax_2+ay_2 \\ \vdots \\ ax_n+ay_n\end{array}\right] \\ &=\left[\begin{array}{c}ax_1 \\ ax_2 \\ \vdots \\ ax_n\end{array}\right] +\left[\begin{array}{c}ay_1\\ay_2\\ \vdots\\ay_n\end{array}\right] \\ &=a\vec{x}+a\vec{y}\end{aligned}\]
• Next, we wish to show that \((a+b)\vec{x} = a\vec{x} + b\vec{x}\). 
  
  \[
  \begin{aligned}
  (a+b)\vec{x}
  &= (a+b)\left[\begin{array}{c}x_1\\x_2\\ \vdots\\x_n\end{array}\right] \\[4pt]
  &= \left[\begin{array}{c}(a+b)x_1 \\ (a+b)x_2 \\ \vdots \\ (a+b)x_n\end{array}\right] \\[4pt]
  &= \left[\begin{array}{c}ax_1+bx_1 \\ ax_2+bx_2 \\ \vdots \\ ax_n+bx_n\end{array}\right] \\[4pt]
  &= \left[\begin{array}{c}ax_1 \\ ax_2 \\ \vdots \\ ax_n\end{array}\right]
  + \left[\begin{array}{c}bx_1 \\ bx_2 \\ \vdots \\ bx_n\end{array}\right] \\[4pt]
  &= a\vec{x} + b\vec{x}
  \end{aligned}
  \]
• We wish to show that \(a(b\vec{x}) = (ab) \vec{x}\). \[\begin{aligned} a(b\vec{x}) &= a\left(b \left[ \begin{array}{c} x_1 \\ x_2 \\ \vdots \\ x_n \end{array} \right] \right) \\ &= a\left( \left[ \begin{array}{c} bx_1 \\ bx_2 \\ \vdots \\ bx_n \end{array} \right] \right) \\ &= \left[ \begin{array}{c} a(bx_1) \\ a(bx_2) \\ \vdots \\ a(bx_n) \end{array} \right] \\ &= \left[ \begin{array}{c} (ab)x_1 \\ (ab)x_2 \\ \vdots \\ (ab)x_n \end{array} \right] \\ &= (ab) \left[ \begin{array}{c} x_1 \\ x_2 \\ \vdots \\ x_n \end{array} \right] \\ &= (ab)\vec{x}\end{aligned}\]
• Finally, we need to show that \(1\vec{x} = \vec{x}\). \[\begin{aligned} 1\vec{x} &= 1 \left[ \begin{array}{c} x_1 \\ x_2 \\ \vdots \\ x_n \end{array} \right] \\ &= \left[ \begin{array}{c} 1x_1 \\ 1x_2 \\ \vdots \\ 1x_n \end{array} \right] \\ &= \left[ \begin{array}{c} x_1 \\ x_2 \\ \vdots \\ x_n \end{array} \right] \\ &= \vec{x}\end{aligned}\]

By the above proofs, it is clear that \(\mathbb{R}^n\) satisfies the vector space axioms. Hence, \(\mathbb{R}^n\) is a vector space under the usual operations of vector addition and scalar multiplication.

We now consider some examples of vector spaces.

Example \(\PageIndex{2}\): Vector Space of Polynomials

Let \(\mathbb{P}_2\) be the set of all polynomials of at most degree \(2\) as well as the zero polynomial. Define addition to be the standard addition of polynomials, and scalar multiplication the usual multiplication of a polynomial by a number. Then \(\mathbb{P}_2\) is a vector space.

Solution

We can write \(\mathbb{P}_2\) explicitly as \[\mathbb{P}_2 = \left\{ a_2x^2 + a_1x + a_0 | a_i \in \mathbb{R} \; \mbox{for all} \; i \right\}\nonumber \] To show that \(\mathbb{P}_2\) is a vector space, we verify the axioms. Let \(p(x), q(x), r(x)\) be polynomials in \(\mathbb{P}_2\) and let \(a,b,c\) be real numbers. Write \(p(x)=p_2x^2 + p_1x + p_0\), \(q(x)=q_2x^2 + q_1x + q_0\), and \(r(x)=r_2x^2 + r_1x + r_0\).

• We first prove that addition of polynomials in \(\mathbb{P}_2\) is closed. For two polynomials in \(\mathbb{P}_2\) we need to show that their sum is also a polynomial in \(\mathbb{P}_2\). From the definition of \(\mathbb{P}_2\), a polynomial is contained in \(\mathbb{P}_2\) if it is of degree at most \(2\) or the zero polynomial. \[\begin{aligned} p(x) + q(x) &= p_2x^2 + p_1x + p_0 + q_2x^2+ q_1x + q_0 \\ &= (p_2+q_2)x^2 + (p_1+q_1)x + (p_0+q_0) \end{aligned}\] The sum is a polynomial of degree \(2\) and therefore is in \(\mathbb{P}_2\). It follows that \(\mathbb{P}_2\) is closed under addition.
• We need to show that addition is commutative, that is \(p(x)+q(x) = q(x) + p(x)\). \[\begin{aligned} p(x) + q(x) &= p_2x^2 + p_1x + p_0 + q_2x^2 + q_1x + q_0\\ &= (p_2+q_2)x^2 + (p_1+q_1)x + (p_0+q_0) \\ &= (q_2+p_2)x^2 + (q_1+p_1)x + (q_0+p_0) \\ &= q_2x^2 + q_1x + q_0 + p_2x^2 + p_1x + p_0\\ &= q(x) + p(x)\end{aligned}\]
• Next, we need to show that addition is associative. That is, that \((p(x) + q(x)) + r(x) = p(x) + (q(x)+r(x))\). \[\begin{aligned} (p(x) + q(x)) + r(x) &= \left( p_2x^2 +p_1x + p_0 + q_2x^2 + q_1x + q_0 \right) + r_2x^2 +r_1x + r_0 \\ &= (p_2+q_2)x^2 + (p_1+q_1)x + (p_0 +q_0) + r_2x^2 + r_1x + r_0\\ &= (p_2+q_2+r_2)x^2 + (p_1+q_1+r_1)x + (p_0+q_0+r_0) \\ &= p_2x^2 + p_1x + p_0 + (q_2+r_2)x^2 + (q_1+r_1)x + (q_0+r_0) \\ &= p_2x^2 + p_1x + p_0 + \left( q_2x^2 +q_1x + q_0 + r_2x^2 + r_1x + r_0 \right)\\ &= p(x) + \left( q(x) + r(x) \right)\end{aligned}\]
• Next, we must prove that there exists an additive identity. Let \(0(x)=0x^2+0x+0\). \[\begin{aligned} p(x) + 0(x) &= p_2x^2 + p_1x + p_0 + 0x^2 + 0x + 0 \\ &= (p_2 + 0)x^2 + (p_1 + 0)x + (p_0 + 0)\\ &= p_2x^2 + p_1x + p_0 \\ &= p(x)\end{aligned}\] Hence an additive identity exists, specifically the zero polynomial.
• Next we must prove that there exists an additive inverse. Let \(-p(x) = -p_2x^2 - p_1x - p_0\) and consider the following: \[\begin{aligned} p(x) + (-p(x)) &= p_2x^2 + p_1x + p_0 + \left( - p_2x^2 - p_1x - p_0\right) \\ &= (p_2 - p_2)x^2 + (p_1 - p_1)x + (p_0 - p_0) \\ &= 0x^2 + 0x + 0 \\ &= 0(x)\end{aligned}\] Hence an additive inverse \(-p(x)\) exists such that \(p(x) + (-p(x)) = 0(x)\).

We now need to verify the axioms related to scalar multiplication.

• First we prove that \(\mathbb{P}_2\) is closed under scalar multiplication. That is, we show that \(ap(x)\) is also a polynomial of degree at most \(2\). \[ap(x) = a\left( p_2x^2 + p_1x + p_0 \right) = ap_2x^2 +ap_1x+ ap_0\nonumber \] Therefore \(\mathbb{P}_2\) is closed under scalar multiplication.
• We need to show that \(a(p(x) + q(x)) = ap(x) + aq(x)\). \[\begin{aligned} a(p(x) + q(x)) &= a \left( p_2x^2 + p_1x + p_0 + q_2x^2 + q_1x + q_0 \right)\\ &= a \left( (p_2+q_2)x^2 + (p_1+q_1)x + (p_0+q_0) \right)\\ &= a(p_2+q_2)x^2 + a(p_1+q_1)x + a(p_0 + q_0) \\ &= (ap_2 + aq_2)x^2 + (ap_1+aq_1)x + (ap_0 + aq_0) \\ &= ap_2x^2 + ap_1x + ap_0 + aq_2x^2 +aq_1x + aq_0\\ &= ap(x) + aq(x) \end{aligned}\]
• Next we show that \((a+b) p(x) = ap(x) + bp(x)\). \[\begin{aligned} (a+b) p(x) &= (a+b) ( p_2x^2 + p_1x + p_0)\\ &= (a+b)p_2x^2 + (a+b)p_1x + (a+b)p_0 \\ &= ap_2x^2 + ap_1x + ap_0 + bp_2x^2 +bp_1x + bp_0\\ &= ap(x) + bp(x)\end{aligned}\]
• The next axiom which needs to be verified is \(a(bp(x)) = (ab)p(x)\). \[\begin{aligned} a(bp(x)) &= a \left( b \left(p_2x^2 + p_1x +p_0\right)\right) \\ &= a \left( bp_2x^2 +bp_1x + bp_0 \right)\\ &=abp_2x^2+abp_1x+abp_0 \\ &=(ab)(p_2x^2+p_1x+p_0) \\ &=(ab)p(x)\end{aligned}\]
• Finally, we show that \(1p(x) = p(x)\). \[\begin{aligned} 1p(x) &= 1 \left( p_2x^2 + p_1x + p_0\right)\\ &= 1p_2x^2 + 1p_1x + 1p_0\\ &= p_2x^2 + p_1x + p_0\\ &= p(x)\end{aligned}\] Since the above axioms hold, we know that \(\mathbb{P}_2\) as described above is a vector space.

Another important example of a vector space is the set of all matrices of the same size.

Example \(\PageIndex{3}\): Vector Space of Matrices

Let \(\mathbb{M}_{2,3}\) be the set of all \(2 \times 3\) matrices. Using the usual operations of matrix addition and scalar multiplication, show that \(\mathbb{M}_{2,3}\) is a vector space.

Solution

Let \(A, B\) be \(2 \times 3\) matrices in \(\mathbb{M}_{2,3}\). We first prove the axioms for addition.

• In order to prove that \(\mathbb{M}_{2,3}\) is closed under matrix addition, we show that the sum \(A+B\) is in \(\mathbb{M}_{2,3}\). This means showing that \(A+B\) is a \(2 \times 3\) matrix. \[\begin{aligned} A+B &= \left[ \begin{array}{rrr} a_{11} & a_{12} & a_{13}\\ a_{21} & a_{22} & a_{23} \end{array} \right] + \left[ \begin{array}{rrr} b_{11} & b_{12} & b_{13}\\ b_{21} & b_{22} & b_{23} \end{array} \right] \\ &= \left[ \begin{array}{rrr} a_{11} + b_{11} & a_{12}+b_{12} & a_{13}+b_{13}\\ a_{21} +b_{21}& a_{22}+b_{22} & a_{23}+b_{23} \end{array} \right]\end{aligned}\] You can see that the sum is a \(2\times 3\) matrix, so it is in \(\mathbb{M}_{2,3}\). It follows that \(\mathbb{M}_{2,3}\) is closed under matrix addition.
• The remaining axioms regarding matrix addition follow from properties of matrix addition. Therefore \(\mathbb{M}_{2,3}\) satisfies the axioms of matrix addition.

We now turn our attention to the axioms regarding scalar multiplication. Let \(A, B\) be matrices in \(\mathbb{M}_{2,3}\) and let \(c\) be a real number.

• We first show that \(\mathbb{M}_{2,3}\) is closed under scalar multiplication. That is, we show that \(cA\) a \(2 \times 3\) matrix. \[\begin{aligned} cA &= c\left[ \begin{array}{rrr} a_{11} & a_{12} & a_{13}\\ a_{21} & a_{22} & a_{23} \end{array} \right] \\ &= \left[ \begin{array}{rrr} ca_{11} & ca_{12} & ca_{13}\\ ca_{21} & ca_{22} & ca_{23} \end{array} \right]\end{aligned}\] This is a \(2 \times 3\) matrix in \(\mathbb{M}_{2,3}\) which proves that the set is closed under scalar multiplication.
• The remaining axioms of scalar multiplication follow from properties of scalar multiplication of matrices. Therefore \(\mathbb{M}_{2,3}\) satisfies the axioms of scalar multiplication.

In conclusion, \(\mathbb{M}_{2,3}\) satisfies the required axioms and is a vector space.

While here we proved that the set of all \(2 \times 3\) matrices is a vector space, there is nothing special about this choice of matrix size. In fact if we instead consider \(\mathbb{M}_{m,n}\), the set of all \(m \times n\) matrices, then \(\mathbb{M}_{m,n}\) is a vector space under the operations of matrix addition and scalar multiplication.

We now examine an example of a set that does not satisfy all of the above axioms, and is therefore not a vector space.

Example \(\PageIndex{4}\): Not a Vector Space

Let \(V\) denote the set of \(2 \times 3\) matrices. Let addition in \(V\) be defined by \(A + B = A\) for matrices \(A,B\) in \(V\). Let scalar multiplication in \(V\) be the usual scalar multiplication of matrices. Show that \(V\) is not a vector space.

Solution

In order to show that \(V\) is not a vector space, it suffices to find only one axiom which is not satisfied. We will begin by examining the axioms for addition until one is found which does not hold. Let \(A,B\) be matrices in \(V\).

• We first want to check if addition is closed. Consider \(A+B\). By the definition of addition in the example, we have that \(A+B = A\). Since \(A\) is a \(2 \times 3\) matrix, it follows that the sum \(A+B\) is in \(V\), and \(V\) is closed under addition.
• We now wish to check if addition is commutative. That is, we want to check if \(A + B = B + A\) for all choices of \(A\) and \(B\) in \(V\). From the definition of addition, we have that \(A + B = A\) and \(B + A = B\). Therefore, we can find \(A\), \(B\) in \(V\) such that these sums are not equal. One example is \[A = \left[ \begin{array}{rrr} 1 & 0 & 0 \\ 0 & 0 & 0 \end{array}\right], B = \left[ \begin{array}{rrr} 0 & 0 & 0 \\ 1 & 0 & 0 \end{array}\right]\nonumber \] Using the operation defined by \(A+B=A\), we have \[\begin{aligned}A+B&=A \\ &=\left[\begin{array}{ccc}1&0&0\\0&0&0\end{array}\right] \\ B+A&=B \\ &=\left[\begin{array}{ccc}0&0&0\\1&0&0\end{array}\right]\end{aligned}\] It follows that \(A+B\neq B+A\). Therefore addition as defined for \(V\) is not commutative and \(V\) fails this axiom. Hence \(V\) is not a vector space.

Consider another example of a vector space.

Example \(\PageIndex{5}\): Vector Space of Functions

Let \(S\) be a nonempty set and define \(\mathbb{F}_S\) to be the set of real functions defined on \(S\). In other words, we write \(\mathbb{F}_S: S \mapsto \mathbb{R}\). Letting \(a,b,c\) be scalars and \(f,g,h\) functions, the vector operations are defined as \[\begin{aligned} \left( f+g\right) \left( x\right) &=f\left( x\right) +g\left( x\right) \\ \left( af\right) \left( x\right) &=a\left( f\left( x\right) \right)\end{aligned}\] Show that \(\mathbb{F}_S\) is a vector space.

Solution

To verify that \(\mathbb{F}_S\) is a vector space, we must prove the axioms beginning with those for addition. Let \(f, g, h\) be functions in \(\mathbb{F}_S\).

• First we check that addition is closed. For functions \(f, g\) defined on the set \(S\), their sum given by \[(f+g)(x) = f(x)+g(x)\nonumber \] is again a function defined on \(S\). Hence this sum is in \(\mathbb{F}_S\) and \(\mathbb{F}_S\) is closed under addition.
• Secondly, we check the commutative law of addition: \[\left( f+g\right) \left( x\right) =f\left( x\right) +g\left( x\right) =g\left( x\right) +f\left( x\right) =\left( g+f\right) \left( x\right)\nonumber \] Since \(x\) is arbitrary, \(f+g=g+f\).
• Next we check the associative law of addition: \[\left( \left( f+g\right) +h\right) \left( x\right) = \left( f+g\right) \left( x\right) +h\left( x\right) =\left( f\left( x\right) +g\left( x\right) \right) +h\left( x\right)\nonumber \] \[=f\left( x\right) +\left( g\left( x\right) +h\left( x\right) \right) =\left( f\left( x\right) +\left( g+h\right) \left( x\right) \right) =\left( f+\left( g+h\right) \right) \left( x\right)\nonumber \] and so \(\left( f+g\right) +h=f+\left( g+h\right) .\)
• Next we check for an additive identity. Let \(0\) denote the function which is given by \(0\left( x\right) =0.\) Then this is an additive identity because \[\left( f+0\right) \left( x\right) =f\left( x\right) +0\left( x\right) =f\left( x\right)\nonumber \] and so \(f+0=f\).
• Finally, check for an additive inverse. Let \(-f\) be the function which satisfies \(\left( -f\right) \left( x\right) = -f\left( x\right) .\) Then \[\left( f+\left( -f\right) \right) \left( x\right) = f\left( x\right) +\left( -f\right) \left( x\right) = f\left( x\right) +-f\left( x\right) =0\nonumber \] Hence \(f+\left( -f\right) =0\).

Now, check the axioms for scalar multiplication.

• We first need to check that \(\mathbb{F}_S\) is closed under scalar multiplication. For a function \(f(x)\) in \(\mathbb{F}_S\) and real number \(a\), the function \((af)(x) = a(f(x))\) is again a function defined on the set \(S\). Hence \(a(f(x))\) is in \(\mathbb{F}_S\) and \(\mathbb{F}_S\) is closed under scalar multiplication.
• \[\left( \left( a+b\right) f\right) \left( x\right) = \left( a+b\right) f\left( x\right) =af\left( x\right) +bf\left( x\right) = \left( af+bf\right) \left( x\right)\nonumber \] and so \(\left( a+b\right) f=af+bf\).
• \[\left( a\left( f+g\right) \right) \left( x\right) = a\left( f+g\right) \left( x\right) = a\left( f\left( x\right) +g\left( x\right) \right)\nonumber \] \[=af\left( x\right) +bg\left( x\right) = \left( af+bg\right) \left( x\right)\nonumber \] and so \(a\left( f+g\right) =af+bg\).
• \[\left( \left( ab\right) f\right) \left( x\right) = \left( ab\right) f\left( x\right) =a\left( bf\left( x\right) \right) = \left( a\left( bf\right) \right) \left( x\right)\nonumber \] so \(\left( abf\right) =a\left( bf\right)\).
• Finally \(\left( 1f\right) \left( x\right) = 1f\left( x\right) =f\left( x\right)\) so \(1f=f\).

It follows that \(V\) satisfies all the required axioms and is a vector space.

One interseting example is to find the zero elment and the addivtie invse for the first set in the tabvle 
 

Example \(\PageIndex{1}\)

Finding the Zero Element and Additive Inverse

Let the zero element be \(\vec{0} = (a, b)\). To satisfy the zero property, we must have: \[ (x, y) + (a, b) = (x, y) \] Using the given operation: \[ (x, y) + (a, b) = (x + a + 1,\, y + b) \] Set the result equal to \((x, y)\): \[ (x + a + 1,\, y + b) = (x, y) \] Comparing components gives: \[ a + 1 = 0 \quad \text{and} \quad b = 0 \] Therefore, the zero element is: \[ \vec{0} = (-1,\, 0) \] --- Now, let the additive inverse of \((x, y)\) be \((u, v)\) such that: \[ (x, y) + (u, v) = \vec{0} \] Substitute the operation: \[ (x + u + 1,\, y + v) = (-1,\, 0) \] From this we get: \[ u = -x - 2 \quad \text{and} \quad v = -y \] Hence, the additive inverse of \((x, y)\) is: \[ -(x, y) = (-x - 2,\, -y) \]

 

 

Consider the following important theorem.

Theorem \(\PageIndex{1}\): Uniqueness

In any vector space, the following are true:

1. \(\vec{0}\), the additive identity, is unique
2. \(-\vec{x}\), the additive inverse, is unique
3. \(0\vec{x}=\vec{0}\) for all vectors \(\vec{x}\)
4. \(\left( -1\right) \vec{x}=-\vec{x}\) for all vectors \(\vec{x}\)

Proof

1. When we say that the additive identity, \(\vec{0}\), is unique, we mean that if a vector acts like the additive identity, then it is the additive identity. To prove this uniqueness, we want to show that another vector which acts like the additive identity is actually equal to \(\vec{0}\). Suppose \(\vec{0}^{\prime }\) is also an additive identity. Then, \[\vec{0} + \vec{0}^{\prime} = \vec{0}\nonumber \] Now, for \(\vec{0}\) the additive identity given above in the axioms, we have that \[\vec{0}^{\prime} + \vec{0} = \vec{0}^{\prime}\nonumber \] So by the commutative property: \[0 = 0 + 0^{\prime} = 0^{\prime} + 0 = 0^{\prime}\nonumber \] This says that if a vector acts like an additive identity (such as \(\vec{0}^{\prime}\)), it in fact equals \(\vec{0}\). This proves the uniqueness of \(\vec{0}\).
2. When we say that the additive inverse, \(-\vec{x}\), is unique, we mean that if a vector acts like the additive inverse, then it is the additive inverse. Suppose that \(\vec{y}\) acts like an additive inverse: \[\vec{x}+\vec{y}=\vec{0}\nonumber \] Then the following holds: \[\vec{y} = \vec{0} + \vec{y} = (-\vec{x} + \vec{x}) + \vec{y} = -\vec{x} + (\vec{x} + \vec{y}) = -\vec{x} + \vec{0} = -\vec{x}\nonumber \] Thus if \(\vec{y}\) acts like the additive inverse, it is equal to the additive inverse \(-\vec{x}\). This proves the uniqueness of \(-\vec{x}\).
3. This statement claims that for all vectors \(\vec{x}\), scalar multiplication by \(0\) equals the zero vector \(\vec{0}\). Consider the following, using the fact that we can write \(0=0+0\): \[0\vec{x}=\left( 0+0\right) \vec{x}=0\vec{x}+0\vec{x}\nonumber \] We use a small trick here: add \(-0\vec{x}\) to both sides. This gives \[\begin{aligned} 0\vec{x} + (-0\vec{x})&=0\vec{x}+0\vec{x}+(-\vec{x})\\ \vec{0} + 0 &=0\vec{x} + 0 \\ \vec{0} &= 0\vec{x}\end{aligned}\] This proves that scalar multiplication of any vector by \(0\) results in the zero vector \(\vec{0}\).
4. Finally, we wish to show that scalar multiplication of \(-1\) and any vector \(\vec{x}\) results in the additive inverse of that vector, \(-\vec{x}\). Recall from \(2.\) above that the additive inverse is unique. Consider the following: \[\begin{aligned} \left( -1\right) \vec{x}+\vec{x} & =\left( -1\right) \vec{x}+1\vec{x}\\ &=\left(-1+1\right) \vec{x} \\ &=0\vec{x} \\ &=\vec{0}\end{aligned}\] By the uniqueness of the additive inverse shown earlier, any vector which acts like the additive inverse must be equal to the additive inverse. It follows that \(\left( -1\right) \vec{x}=-\vec{x}\).

An important use of the additive inverse is the following theorem.

Theorem \(\PageIndex{2}\)

Let \(V\) be a vector space. Then \(\vec{v} + \vec{w} = \vec{v} + \vec{z}\) implies that \(\vec{w} = \vec{z}\) for all \(\vec{v}, \vec{w}, \vec{z} \in V\)

Proof

The proof follows from the vector space axioms, in particular the existence of an additive inverse (\(-\vec{u}\)). The proof is left as an exercise to the reader.

Exercise

Exercise \(\PageIndex{1}\)

Suppose you have \(\mathbb{R}^2\) and the \(+\) operation is as follows: \[(a,b) + (c,d) = (a+d,b+c).\nonumber\] Scalar multiplication is defined in the usual way. Is this a vector space? Explain why or why not.

Exercise \(\PageIndex{2}\)

Suppose you have \(\mathbb{R}^2\) and the \(+\) operation is as follows: \[(a,b) + (c,d) = (0,b+d)\nonumber\] Scalar multiplication is defined in the usual way. Is this a vector space? Explain why or why not.

Exercise \(\PageIndex{3}\)

Suppose you have \(\mathbb{R}^2\) and scalar multiplication is defined as \(c(a,b) = (a, cb)\) while vector addition is defined as usual. Is this a vector space? Explain why or why not.

Exercise \(\PageIndex{4}\)

Suppose you have \(\mathbb{R}^2\) and the \(+\) operation is defined as follows. \[(a,b) + (c,d) = (a−c,b−d)\nonumber\] Scalar multiplication is same as usual. Is this a vector space? Explain why or why not.

Exercise \(\PageIndex{5}\)

Consider all the functions defined on a non empty set which have values in \(\mathbb{R}\). Is this a vector space? Explain. The operations are defined as follows. Here \(f ,g\) signify functions and \(a\) is a scalar \[\begin{aligned} (f+g)(x)&=f(x)+g(x) \\ (af)(x)&=a(f(x))\end{aligned}\]

Exercise \(\PageIndex{6}\)

Denote by \(\mathbb{R}^{\mathbb{N}}\) the set of real valued sequences. For \(\vec{a} ≡ \{a_n\}_{n=1}^∞\), \(\vec{b} ≡ \{b_n\}_{n=1}^\infty\) two of these, define their sum to be given by \[\vec{a}+\vec{b}=\{a_n+b_n\}_{n=1}^\infty\nonumber\] and define scalar multiplication by \[c\vec{a}=\{ca_n\}_{n=1}^\infty\text{ where }\vec{a}=\{a+n\}_{n=1}^\infty\nonumber\] Is this a special case of Exercise \(\PageIndex{5}\)? Is this a vector space?

Exercise \(\PageIndex{7}\)

Let \(\mathbb{C}^2\) be the set of ordered pairs of complex numbers. Define addition and scalar multiplication in the usual way. \[(z,w) + (\hat{z},\hat{w}) = (z+\hat{z},w+\hat{w}), u(z,w) ≡ (uz,uw)\nonumber\] Here the scalars are from \(\mathbb{C}\). Show this is a vector space.

Exercise \(\PageIndex{8}\)

Let \(V\) be the set of functions defined on a nonempty set which have values in a vector space \(W\). Is this a vector space? Explain.

Exercise \(\PageIndex{9}\)

Consider the space of \(m\times n\) matrices with operation of addition and scalar multiplication defined the usual way. That is, if \(A,B\) are two \(m\times n\) matrices and \(c\) a scalar, \[(A+B)_{ i j} = A_{i j} +B_{i j}, \:(cA)_{ i j} ≡ c (A_{ij})\nonumber\]

Exercise \(\PageIndex{10}\)

Consider the set of \(n\times n\) symmetric matrices. That is, \(A = A^T\). In other words, \(A_{i j} = A_{ji}\). Show that this set of symmetric matrices is a vector space and a subspace of the vector space of \(n\times n\) matrices.