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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 $(cx_1+cy_1,\dots,cx_n+cy_n)$. 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

Linearity

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.

This Post Has 2 Comments

  1. Hi!

    Your blog has helped me a lot. Thanks!

    In Exercise 2.1.6, I was wondering, what happens if x=0? Isn´t f(x) real valued?

    I´ld really aprecciate your point of view

  2. The first question Exercise 2.1.1 and Example 1.
    Part 7 of the argument when ‘c’ is multiplied to the Nth term of the added tuple (third line), it should’ve been CXn+CYn instead of CXn+XYn as it is written (probable misprint). You may correct that.

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