If you find any mistakes, please make a comment! Thank you.

Solution to Linear Algebra Hoffman & Kunze Chapter 2.1

Exercise 2.1.1

Example 1 starts with any field and defines the objects, the addition rule and the scalar multiplication rule. We must show the set of $n$-tuples satisfies the eight properties required in the definition.

(1) Addition is commutative. Let $\alpha=(x_1,\dots,x_n)$ and $\beta=(y_1,\dots,y_n)$ be two $n$-tuples. Then $$\alpha+\beta = (x_1+y_1,\dots,x_n+y_n).$$ And since $F$ is commutative this equals $(y_1+x_1,\dots,y_n+x_n)$, which equals $\beta+\alpha$. Thus $\alpha+\beta=\beta+\alpha$.

(2) Addition is associative. Let $\alpha=(x_1,\dots,x_n)$, $\beta=(y_1,\dots,y_n)$ and $\gamma=(z_1,\dots,z_n)$ be three $n$-tuples. Then $$(\alpha+\beta)+\gamma=((x_1+y_1)+z_1,\dots,(x_n+y_n)+z_n).$$ And since $F$ is associative this equals
$$(x_1+(y_1+z_1),\dots,x_n+(y_n+z_n)),$$ which equals $\alpha+(\beta+\gamma)$.

(3) We must show there is a unique vector $0$ in $V$ such that $\alpha+0=\alpha$ $\forall$ $\alpha\in V$. Consider $(0_F,\dots,0_F)$ the vector of all $0$’s of length $n$, where $0_F$ is the zero element of $F$. Then this vector satisfies the property that $$(0_F,\dots,0_F)+(x_1,\dots,x_n)=(0_F+x_1,\dots,0_F+x_n)=(x_1,\dots,x_n)$$ since $0_F+x=x$ for all $x\in F$. Thus $(0_F,\dots,0_F)+\alpha=\alpha$ $\forall \alpha\in V$. We must just show this vector is unique with respect to this property. Suppose $\beta=(x_1,\dots,x_n)$ also satisfies the property that $\beta+\alpha=\alpha$ for all $\alpha\in V$. Let $\alpha=(0_F,\dots,0_F)$. Then $$(x_1,\dots,x_n)=(x_1+0_F,\dots,x_n+0_F)=(x_1,\dots,x_n)+(0_F,\dots,0_F)$$ and by definition of $\beta$ this equals $(0_F,\dots,0_F)$. Thus $(x_1,\dots,x_n)=(0_F,\dots,0_F)$. Thus $\beta=\alpha$ and the zero element is unique.

(4) We must show for each vector $\alpha$ there is a unique vector $\beta$ such that $\alpha+\beta=0$. Suppose $\alpha=(x_1,\dots,x_n)$. Let $\beta=(-x_1,\dots,-x_n)$. Then $\beta$ has the required property $\alpha+\beta=0$. We must show $\beta$ is unique with respect to this property. Suppose also $\beta’=(x’_1,\dots,x’_n)$ also has this property. Then $\alpha+\beta=0$ and $\alpha+\beta’=0$. So $$\beta=\beta+0=\beta+(\alpha+\beta’)=(\beta+\alpha)+\beta’=0+\beta’=\beta’.$$

(5) Let $1_F$ be the multiplicative identity in $F$. Then $1_F \cdot(x_1,\dots,x_n)=(1\cdot x_1,\dots,1\cdot x_n)=(x_1,\dots,x_n)$ since $1_F\cdot x=x$ $\forall$ $x\in F$. Thus $1_F\alpha=\alpha$, for all $\alpha\in V$.

(6) Let $\alpha=(x_1,\dots,x_n)$. Then $(c_1c_2)\alpha=((c_1c_2)x_1,\dots, (c_1c_2)x_n)$ and since multiplication in $F$ is associative this equals $$(c_1(c_2x_1),\dots,c_1(c_2x_n))=c_1(c_2x_1,\dots c_2x_n)=c_1\cdot(c_2\alpha).$$

(7) Let $\alpha=(x_1,\dots,x_n)$ and $\beta=(y_1,\dots,y_n)$. Then $$c(\alpha+\beta)=c(x_1+ y_1,\dots,x_n+y_n)=(c(x_1+y_1),\dots,c(x_n+y_n))$$ and since multiplication is distributive over addition in $F$ this equals $(cx_1+cy_1,\dots,cx_n+xy_n)$. This then equals $$(cx_1,\dots,cx_n)+(cy_1,\dots,cy_n)=c(x_1,\dots,x_n)+c(y_1,\dots,y_n)=c\alpha+c\beta.$$ Thus $c(\alpha+\beta)=c\alpha+c\beta$.

(8) Let $\alpha=(x_1,\dots,x_n)$. Then $(c_1+c_2)\alpha=((c_1+c_2)x_1,\dots,(c_1+c_2)x_n)$ and since multiplication distributes over addition in $F$ this equals \begin{align*}&(c_1x_1+c_2x_1,\dots, c_1x_n+c_2x_n)\\=&(c_1x_1,\dots,c_1x_n)+(c_2x_1,\dots,c_2x_n)\\=&c_1(x_1,\dots,x_n)+c_2(x_1,\dots,x_n)\\=&c_1\alpha+c_2\alpha.\end{align*} Thus $(c_1+c_2)\alpha=c_1\alpha+c_2\alpha$.

Exercise 2.1.2

This follows associativity and commutativity properties of $V$:$$(\alpha_1+\alpha_2)+(\alpha_3+\alpha_4)$$$$=(\alpha_2+\alpha_1)+(\alpha_3+\alpha_4)$$$$=\alpha_2+[\alpha_1+(\alpha_3+\alpha_4)]$$$$=\alpha_2+[(\alpha_1+\alpha_3)+\alpha_4]$$$$=[\alpha_2+(\alpha_1+\alpha_3)]+\alpha_4$$$$=[\alpha_2+(\alpha_3+\alpha_1)]+\alpha_4.$$

Exercise 2.1.3

If we make a matrix out of these three vectors
$$A=\left[\begin{array}{ccc}1&0&1\\0&1&1\\-1&1&1\end{array}\right]$$then if we row-reduce the augmented matrix
$$\left[\begin{array}{ccc|ccc}1&0&1&1&0&0\\0&1&1&0&1&0\\-1&1&1&0&0&1\end{array}\right]$$we get
$$\left[\begin{array}{ccc|ccc}1&0&0&0&1&-1\\0&1&0&-1&2&-1\\0&0&1&1&-1&1\end{array}\right].$$Therefore the matrix is invertible and $AX=Y$ has a solution $X=A^{-1}Y$ for any $Y$. Thus any vector $Y\in\mathbb C^3$ can be written as a linear combination of the three vectors. Not sure what the point was of making the base field $\mathbb C$.

Exercise 2.1.4

No, it is not a vector space because $(0,2)=(0,1)+(0,1)=2(0,1)=(2\cdot0,1)=(0,1)$. Thus we must have $(0,2)=(0,1)$ which implies $1=2$ which is a contradiction in the field of real numbers.

Exercise 2.1.5

(1) $\oplus$ is not commutative since $(0,\dots,0)\oplus(1,\dots,1)=(-1,\dots,-1)$ while $(1,\dots,1)\oplus(0,\dots,0)=(1,\dots,1)$. And $(1,\dots,1)\not=(-1,\dots,-1)$.

(2) $\oplus$ is not associative since $$((1,\dots,1)\oplus(1,\dots,1))\oplus(2,\dots,2)=(0,\dots,0)\oplus(2,\dots,2)=(-2,\dots,-2)$$ while $$(1,\dots,1)\oplus((1,\dots,1)\oplus(2,\dots,2))=(1,\dots,1)\oplus(-1,\dots,-1)=(2,\dots,2).$$

(3) There does exist a right additive identity, i.e. a vector $0$ that satisfies $\alpha+0=\alpha$ for all $\alpha$. The vector $\beta=(0,\dots,0)$ satisfies $\alpha+\beta=\alpha$ for all $\alpha$. And if $\beta’=(b_1,\dots,b_n)$ also satisfies $$(x_1,\dots,x_n)+\beta’=(x_1,\dots,x_n)$$ then $x_i-b_i=x_i$ for all $i$ and thus $b_i=0$ for all $i$. Thus $\beta=(0,\dots,0)$ is unique with respect to the property $\alpha+\beta=\alpha$ for all $\alpha$.

(4) There do exist right additive inverses. For the vector $\alpha=(x_1,\dots,x_n)$ clearly only $\alpha$ itself satisfies $\alpha\oplus\alpha=(0,\dots,0)$.

(5) The element $1$ does not satisfy $1\cdot\alpha=\alpha$ for any non-zero $\alpha$ since otherwise we would have $$1\cdot(x_1,\dots,x_n)=(-x_1,\dots,-x_n)=(x_1,\dots,x_n)$$ only if $x_i=0$ for all $i$.

(6) The property $(c_1c_2)\cdot\alpha=c_1\cdot(c_2\cdot\alpha)$ does not hold since $(c_1c_2)\alpha=(-c_1c_2)\alpha$ while $$c_1(c_2\alpha)=c_1(-c_2\alpha)=(-c_1(-c2\alpha))=+c_1c_2\alpha.$$ Since $c_1c_2\not=-c_1c_2$ for all $c_1,c_2$ they are not always equal.

(7) It does hold that $c\cdot(\alpha\oplus\beta)=c\cdot\alpha\oplus c\cdot\beta$. Firstly, $$c\cdot(\alpha\oplus\beta)=c\cdot(\alpha-\beta)=-c(\alpha-\beta)=-c\alpha+c\beta.$$ And secondly $$c\cdot\alpha\oplus c\cdot\beta=(-c\alpha)\oplus(-c\beta)=-c\alpha-(-c\beta)=-c\alpha+c\beta.$$ Thus they are equal.

(8) It does not hold that $(c_1+c_2)\cdot\alpha=(c_1\cdot\alpha)\oplus(c_2\cdot\alpha)$. Firstly, $$(c_1+c_2)\cdot\alpha=-(c_1+c_2)\alpha=-c_1\alpha-c_2\alpha.$$ Secondly, $$c_1\cdot\alpha\oplus c_2\cdot\alpha=(-c_1\cdot\alpha)\oplus(-c_2\cdot\alpha)=-c_1\alpha+c_2\alpha.$$ Since $-c_1\alpha-c_2\alpha\not=-c_1\alpha-c_2\alpha$ for all $c_1,c_2$ they are not equal.

Exercise 2.1.6

We will use the basic fact that $\overline{a+b}=\overline{a}+\overline{b}$ and $\overline{ab}=\overline{a}\cdot\overline{b}$.

Before we show $V$ satisfies the eight properties we must first show vector addition and scalar multiplication as defined are actually well-defined in the sense that they are indeed operations on $V$. In other words if $f$ and $g$ are two functions in $V$ then we must show that $f+g$ is in $V$. In other words if $f(-t)=\overline{f(t)}$ and $g(-t)=\overline{g(t)}$ then we must show that $(f+g)(-t)=\overline{(f+g)(t)}$. This is true because $$(f+g)(-t)=f(-t)+g(-t)=\overline{f(t)}+\overline{g(t)}=\overline{(f(t)+g(t)}=\overline{(f+g)(t)}.$$

Similary, if $c\in\mathbb R$, $(cf)(-t)=cf(-t)=c\overline{f(t)}=\overline{cf(t)}$ since $\overline{c}=c$ when $c\in\mathbb R$.

Thus the operations are well defined. We now show the eight properties hold:

(1) Addition on functions in $V$ is defined by adding in $\mathbb C$ to the values of the functions in $\mathbb C$. Thus since $\mathbb C$ is commutative, addition in $V$ inherits this commutativity.

(2) Similar to 1, since $\mathbb C$ is associative, addition in $V$ inherits this associativity.

(3) The zero function $g(t)=0$ is in $V$ since $-0=\overline{0}$. And $g$ satisfies $f+g=f$ for all $f\in V$. Thus $V$ has a right additive identity.

(4) Let $g$ be the function $g(t)=-f(t)$. Then $$g(-t)=-f(-t)=-\overline{f(t)}=\overline{-f(t)}=\overline{g(t)}.$$ Thus $g\in V$. And $$(f+g)(t)=f(t)+g(t)=f(t)-f(t)=0.$$ Thus $g$ is a right additive inverse for $f$.

(5) Clearly $1\cdot f=f$ since $1$ is the multiplicative identity in $\mathbb R$.

(6) As before, associativity in $\mathbb C$ implies $(c_1c_2)f=c_1(c_2f)$.

(7) Similarly, the distributive property in $\mathbb C$ implies $c(f+g)=cf+cg$.

(8) Similarly, the distributive property in $\mathbb C$ implies $(c_1+c_2)f=c_1f+c_2f$.

An example of a function in $V$ which is not real valued is $f(x)=ix$. Since $f(1)=i$, $f$ is not real-valued. And $f(-x)=-ix=\overline{ix}$ since $x\in\mathbb R$, so $f\in V$.

Exercise 2.1.7

This is not a vector space because there would have to be an additive identity element $(a,b)$ which has the property that $(a,b)+(x,y)=(x,y)$ for all $(x,y)\in V$. But this is impossible, because $(a,b)+(0,1)=(a,0)\not=(0,1)$ no matter what $(a,b)$ is. Thus $V$ does not satisfy the third requirement of having an additive identity element.

From http://greggrant.org


This website is supposed to help you study Linear Algebras. Please only read these solutions after thinking about the problems carefully. Do not just copy these solutions.

Leave a Reply

Close Menu