7.2: Vector Space - Examples

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 at most \(2\) 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.

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.

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.

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.