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

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 $\alpha_2=\beta_2$.

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. On number 2.2.5, it is stated that (cA+B)^2 = AA + 2cAB + BB. It seems to me that this is incorrect since this assumes that AB = BA which is not always true with all matrices.

    1. You are correct. The solution is wrong. It is fixed.

Leave a Reply

Close Menu