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## Solution to Linear Algebra Hoffman & Kunze Chapter 2.2

#### Exercise 2.2.1

(a) This is not a subspace because for $(1,\dots,1)$ the additive inverse is $(-1,\dots,-1)$ which does not satisfy the condition.

(b) Suppose $(a_1,a_2,a_3,\dots,a_n)$ and $(b_1,b_2,b_3,\dots,b_n)$ satisfy the condition and let $c\in\Bbb R$. By Theorem 1 (page 35) we must show that $$c(a_1,a_2,a_3,\dots,a_n)+(b_1,b_2,b_3,\dots,b_n)=(ca_1+b_1,\dots,ca_n+b_n)$$ satisfies the condition. Now $$(ca_1+b_1)+3(ca_2+b_2)=c(a_1+3a_2)+(b_1+3b_2)=c(a_3)+(b_3)=ca_3+b_3.$$ Thus it does satisfy the condition so $V$ is a vector space.

(c) This is not a vector space because $(1,1)$ satisfies the condition since $1^2=1$, but $$(1,1,\dots)+(1,1,\dots)=(2,2,\dots)$$ and $(2,2,\dots)$ does not satisfy the condition because $2^2\not=2$.

(d) This is not a subspace. $(1,0,\dots)$ and $(0,1,\dots)$ both satisfy the condition, but their sum is $(1,1,\dots)$ which does not satisfy the condition.

(e) This is not a subspace. $(1,1,\dots,1)$ satisfies the condition, but $\pi(1,1,\dots,1)=(\pi,\pi,\dots,\pi)$ does not satisfy the condition.

#### Exercise 2.2.2

(a) Not a subspace. Let $f(x)=x$ and $g(x)=x^2$. Then both satisfy the condition: $f(x^2)=x^2=(f(x))^2$ and $g(x^2)=(x^2)^2=(g(x))^2$. But $(f+g)(x)=x+x^2$ and $(f+g)(x^2)=x^2+x^4$ while $$[(f+g)(x)]^2=(x+x^2)^2=x^4+2x^3+x^2.$$ These are not equal polynomials so the condition does not hold for $f+g$.

(b) Yes a subspace. Suppose $f$ and $g$ satisfy the property. Let $c\in\mathbb R$. Then $$(cf+g)(0)=cf(0)+g(0)=cf(1)+g(1)=(cf+g)(1).$$ Thus $(cf+g)(0)=(cf+g)(1)$. By Theorem 1 (page 35) the set of all such functions constitute a subspace.

(c) Not a subspae. Let $f(x)$ be the function defined by $f(3)=1$ and $f(x)=0$ for all $x\not=3$. Let $g(x)$ be the function defined by $g(-5)=0$ and $g(x)=1$ for all $x\not=-5$. Then both $f$ and $g$ satisfy the condition. But $$(f+g)(3)=f(3)+g(3)=1+1=2,$$ while $$1+(f+g)(-5)=1+f(-5)+g(-5)=1+0+0=1.$$ Since $1\not=2$, $f+g$ does not satisfy the condition.

(d) Yes a subspace. Suppose $f$ and $g$ satisfy the property. Let $c\in\mathbb R$. Then $$(cf+g)(-1)=cf(-1)+g(-1)=c\cdot0+0=0.$$ Thus $(cf+g)(-1)=0$. By Theorem 1 (page 35) the set of all such functions constitute a subspace.

(e) Yes a subspace. Let $f$ and $g$ be continuous functions from $\mathbb R$ to $\mathbb R$ and let $c\in\mathbb R$. Then we know from basic results of real analysis that the sum and product of continuous functions are continuous. Since the function $c\mapsto c$ is continuous as well as $f$ and $g$, it follows that $cf+g$ is continuous. By Theorem 1 (page 35) the set of all cotinuous functions constitute a subspace.

#### Exercise 2.2.3

I assume they meant $\mathbb R^4$. No, $(3,-1,0,-1)$ is not in the subspace. If we row reduce the augmented matrix
$$\left[\begin{array}{ccc|c}2&-1&1&3\\-1&1&1&-1\\3&1&9&0\\2&-3&-5&-1\end{array}\right]$$we obtain
$$\left[\begin{array}{ccc|c}1&0&2&2\\0&1&3&1\\0&0&0&-7\\0&0&0&-2\end{array}\right].$$The two bottom rows are zero rows to the left of the divider, but the values to the right of the divider in those two rows are non-zero. Thus the system does not have a solution (see comments bottom of page 24 and top of page 25).

#### Exercise 2.2.4

The matrix of the system is
$$\left[\begin{array}{ccccc}2&-1&4/3&-1&0\\1&0&2/3&0&-1\\9&-3&6&-3&-3\end{array}\right].$$Row reducing to reduced echelon form gives
$$\left[\begin{array}{ccccc}1&0&2/3&0&-1\\0&1&0&1&-2\\0&0&0&0&0\end{array}\right].$$Thus the system is equivalent to
$$x_1+2/3x_3-x_5=0$$$$x_2+x_4-2x_5=0.$$Thus the system is parametrized by $(x_3,x_4,x_5)$. Setting each equal to one and the other two to zero (as in Example 15, page 42), in turn, gives the three vectors $(-2/3,0,1,0,0)$, $(0,-1,0,1,0)$ and $(1,2,0,0,1)$. These three vectors therefore span $W$.

#### Exercise 2.2.5

(a) This is not a subspace. Let $A=\left[\begin{array}{cc}1&0\\0&1\end{array}\right]$ and let $B=\left[\begin{array}{cc}-1&0\\0&-1\end{array}\right]$. Then both $A$ and $B$ are invertible, but $A+B=\left[\begin{array}{cc}0&0\\0&0\end{array}\right]$ which is not invertible. Thus the subset is not closed with respect to matrix addition. Therefore it cannot be a subspace.

(b) This is not a subspace. Let $A=\left[\begin{array}{cc}1&0\\0&0\end{array}\right]$ and let $B=\left[\begin{array}{cc}0&0\\0&1\end{array}\right]$. Then neither $A$ nor $B$ is invertible, but $A+B=\left[\begin{array}{cc}1&0\\0&1\end{array}\right]$ which is invertible. Thus the subset is not closed with respect to matrix addition. Therefore it cannot be a subspace.

(c) This is a subspace. Suppose $A_1$ and $A_2$ satisfy $A_1B=BA_1$ and $A_2B=BA_2$. Let $c\in F$ be any constant. Then \begin{align*}(cA_1+A_2)B&=cA_1B+A_2B\\&=cBA_1+BA_2=B(cA_1)+BA_2=B(cA_1+A_2).\end{align*} Thus $cA_1+A_2$ satisfy the criteria. By Theorem 1 (page 35) the subset is a subspace.

(d) This is a subspace. Consider the case $A=\left[\begin{array}{cc}1&1\\0&0\end{array}\right]$ and $B=\left[\begin{array}{cc}0&0\\0&1\end{array}\right]$. It is clear that $A^2=A$ and $B^2=B$ but it is easy to check that $(A+B)^2\ne A+B$. So this subset is not closed under addition.

#### Exercise 2.2.6

(a) Let $V$ be a subspace of $\mathbb R^1$. Suppose $\exists$ $v\in V$ with $v\not=0$. Then $v$ is a vector but it is also simply an element of $\mathbb R$. Let $\alpha\in\mathbb R$. Then $\alpha=\frac{\alpha}{v}\cdot v$ where $\frac{\alpha}{v}$ is a scalar in the base field $\mathbb R$. Since $cv\in V$ $\forall$ $c\in\mathbb R$, it follows that $\alpha\in V$. Thus we have shown that if $V\not=\{0\}$ then $V=\mathbb R^1$.

(b) We know the subsests $\{(0,0)\}$ (example 6a, page 35) and $\mathbb R^2$ (example 1, page 29) are subspaces of $\mathbb R^2$. Also for any vector $v$ in any vector space over any field $F$, the set $\{cv\mid c\in F\}$ is a subspace (Theorem 3, page 37). Thus we will be done if we show that if $V$ is a subspace of $\mathbb R^2$ and there exists $v_1,v_2\in V$ such that $v_1$ and $v_2$ do not lie on the same line, then $V=\mathbb R^2$. Equivalently we must show that any vector $w\in\mathbb R^2$ can be written as a linear combination of $v_1$ and $v_2$ whenever $v_1$ and $v_2$ are not co-linear. Equivalently, by Theorem 13 (iii) (page 23), it suffices to show that if $v_1=(a,b)$ and $v_2=(c,d)$ are not colinear, then the matrix $A=[v_1^T\ \ v_2^T]$ is invertible. Suppose $a\not=0$ and let $x=c/a$. Then $xa=c$, and since $v_1$ and $v_2$ are not colinear, it follows that $xb\not=d$. Thus equivalently $ad-bc\not=0$. It follows now from Exercise 1.6.8 pae 27 that if $v_1$ and $v_2$ not colinear then the matrix $A^T$ is invertible. Finally $A^T$ is invertible implies $A$ is invertible, since clearly $(A^T)^{-1}=(A^{-1})^T$. Similarly if $a=0$ then it must be that $b\not=0$ so we can make the same argument. So in all cases $A$ is invertible.

(c) The subspaces are the zero subspace $\{0,0,0\}$, lines $\{cv\mid c\in\mathbb R\}$ for fixed $v\in\mathbb R^3$, planes $$\{c_1v_1+c_2v_2\mid c_1,c_2\in\mathbb R\}$$ for fixed $v_1,v_2\in\mathbb R^3$ and the whole space $\mathbb R^3$. By Theorem 3 we know these all are subspaces, we just must show they are the only subspaces. It suffices to show that if $v_1$, $v_2$ and $v_3$ are not co-planar then the space generated by $v_1, v_2,v_3$ is all of $\mathbb R^3$. Equivalently we must show if $v_1$ and $v_2$ are not co-linear, and $v_3$ is not in the plane generated by $v_1,v_2$ then any vector $w\in\mathbb R^3$ can be written as a linear combination of $v_1,v_2,v_3$. Equivalently, by Theorem 13 (iii) (page 23), it suffices to show the matrix $A=[v_1\ \ v_2\ \ v_3]$ is invertible. $A$ is invertible $\Leftrightarrow$ $A^T$ is invertible, since clearly $(A^T)^{-1}=(A^{-1})^T$. Now
$v_3$ is in the plane generated by $v_1,v_2$ $\Leftrightarrow$ $v_3$ can be written as a linear combination of $v_1$ and $v_2$ $\Leftrightarrow$ $A^T$ is row equivalent to a matrix with one of its rows equal to all zeros (this follows from Theorem 12, page 23) $\Leftrightarrow$ $A^T$ is {\it not}\/ invertible. Thus $v_3$ is not in the plane generated by $v_1,v_2$ $\Leftrightarrow$ $A$ is invertible.

#### Exercise 2.2.7

Assume the space generated by $W_1$ and $W_2$ is equal to their set-theoretic union $W_1\cup W_2$. Suppose $W_1\not\subseteq W_2$ and $W_2\not\subseteq W_1$. We wish to derive a contradiction. So suppose $\exists$ $w_1\in W_1\setminus W_2$ and $\exists$ $w_2\in W_2\setminus W_1$. Consider $w_1+w_2$. By assumption this is in $W_1\cup W_2$, so $\exists$ $w_1'\in W_1$ such that $w_1+w_2=w_1'$ or $\exists$ $w_2'\in W_2$ such that $w_1+w_2=w_2'$. If the former, then $w_2=w_1'-w_1\in W_1$ which contradicts the assumption that $w_2\not\in W_1$. Likewise the latter implies the contradiction $w_1\in W_2$. Thus we are done.

#### Exercise 2.2.8

(a) Let $f,g\in V_e$ and $c\in\mathbb R$. Let $h=cf+g$. Then $$h(-x)=cf(-x)+g(-x)=cf(x)+g(x)=h(x).$$ So $h\in V_e$. By Theorem 1 (page 35) $V_e$ is a subspace. Now let $f,g\in V_o$ and $c\in\mathbb R$. Let $h=cf+g$. Then $$h(-x)=cf(-x)+g(-x)=-cf(x)-g(x)=-h(x).$$ So $h\in V_o$. By Theorem 1 (page 35) $V_o$ is a subspace.

(b) Let $f\in V$. Let $f_e(x)=\frac{f(x)+f(-x)}{2}$ and $f_o=\frac{f(x)-f(-x)}{2}$. Then $f_e$ is even and $f_o$ is odd and $f=f_e+f_o$. Thus we have written $f$ as $g+h$ where $g\in V_e$ and $h\in V_o$.

(c) Let $f\in V_e\cap V_o$. Then $f(-x)=-f(x)$ and $f(-x)=f(x)$. Thus $f(x)=-f(x)$ which implies $2f(x)=0$ which implies $f=0$.

#### Exercise 2.2.9

Let $\alpha\in W$. Suppose $\alpha=\alpha_1+\alpha_2$ for $\alpha_i\in W_i$ and $\alpha=\beta_1+\beta_2$ for $\beta_i\in W_i$. Then $\alpha_1+\alpha_2=\beta_1+\beta_2$ which implies $\alpha_1-\beta_1=\beta_2-\alpha_2$. Thus $\alpha_1-\beta_1\in W_1$ and $\alpha_1-\beta_1\in W_2$. Since $W_1\cap W_2=\{0\}$ it follows that $\alpha_1-\beta_1=0$ and thus $\alpha_1=\beta_1$. Similarly, $\beta_2-\alpha_2\in W_1\cap W_2$ so also $\alpha_2=\beta_2$.