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\title{Math 2135: Linear Algebra}
\author{Assignment 1}
\date{Solutions}

\begin{document}
\maketitle

\begin{enumerate}
  \item \textbf{Show that $\frac12 + \frac{\sqrt{3}}2i \in \bC$ is a cube root of $-1$. Find two more cube roots of $-1$ in $\bC$.}
  
  We compute:
  \begin{multline*}
    \left(\frac12 + \frac{\sqrt{3}}2i\right)^3 = \left(\frac12 + \frac{\sqrt{3}}2i\right)^2\left(\frac12 + \frac{\sqrt{3}}2i\right) \\
    = \left( \left(\frac12\right)^2 + 2 \left(\frac12\right) \left(\frac{\sqrt{3}}2i\right) + \left(\frac{\sqrt{3}}2i\right)^2\right) \left(\frac12 + \frac{\sqrt{3}}2i\right) 
    = \left( \frac14 + \frac{\sqrt{3}}2i - \frac34 \right)\left(\frac12 + \frac{\sqrt{3}}2i\right) \\
    = \left(-\frac12 + \frac{\sqrt{3}}2i\right)\left(\frac12 + \frac{\sqrt{3}}2i\right) 
    = -\left(\frac12\right)^2 + \left(\frac{\sqrt{3}}2i\right)\left(\frac12\right) - + \left(\frac{\sqrt{3}}2i\right)\left(\frac12\right) + \left(\frac{\sqrt{3}}2i\right)^2 \\
    = -\frac14 + 0 - \frac34 = -1
  \end{multline*}
  
  Exactly the same calculations, with $-i$ in place of $i$, show that \fbox{$\frac12 - \frac{\sqrt{3}}2i$} is a cube root of $-1$. And of course \fbox{$-1$} is a (real) cube root of $-1$.
  
  \item \begin{enumerate}
    \item \textbf{Does there exist a $\lambda \in \bC$ such that $\lambda(2, 3i, 4+5i) = (6, 7i, 8-9i)$? Why or why not?}
    
    No. Note that the left-hand side is $(2\lambda, 3i\lambda, (4+5i)\lambda)$. For this to be equal to the right-hand side, we must have $2\lambda = 6$, which forces $\lambda = 3$. However, for this $\lambda$, we find that $3i\lambda = 9i \neq 7i$. So there is no equality.
    
    \item \textbf{Does there exist a $\lambda \in \bC$ such that $\lambda(2,1+i) = (1-i,1)$? Why or why not?}
    
    Yes. Indeed, $(1+i)(1-i) = 1^2 - i^2 = 1 - (-1) = 2$. So $\lambda = 1/(1+i)$ works. What is this $\lambda$? It is 
    $$ \frac1{1+i} = \frac{1-i}{(1+i)(1-i)} = \frac{1-i}{2} = \frac12 - \frac12 i.$$
  \end{enumerate}
  
  \item \textbf{Suppose that $\bF = \bR$ or $\bC$ and that $V$ is a vector space over $\bF$. Show that, if $\alpha \in \bF$ and $v \in V$, and if $\alpha v = 0$, then either $\alpha = 0$ or $v = 0$ (or both).}
  
  Equivalently, we are asked to show that if $\alpha \neq 0$, then $v = 0$. But if $\alpha \neq 0$, then there is a number $\alpha^{-1} \in \bF$ such that $\alpha\alpha^{-1} = 1$, in which case
  $$ 0 = \alpha^{-1} 0 = \alpha^{-1} \alpha v = 1 v = v.$$
   
  \item \textbf{The first three of the following six sets are subsets of the vector space $\bR^3$, and the last three are subsets of the vector space $\bR^\bR$. Three of these six subsets are actually vector subspaces, and the other three are not. For those that are, simply state that they are indeed vector spaces: you do not need to prove why. For those that are not, state that they are not vector spaces, and give a reason: explain one of the axioms for vector space that fails. (There might be more than one!)}
  \begin{enumerate}
    \item \textbf{The set of triples $(x,y,z) \in \bR^3$ such that $x+y=z$.}
    
    This \fbox{is} a vector subspace of $\bR^3$. Basically this is because the equation ``$x+y=z$'' is linear.
    \item \textbf{The set of triples $(x,y,z) \in \bR^3$ such that $xy = z$.}
    
    This \fbox{is not} a vector subspace of $\bR^3$. For example, the vector $v = (1,1,1)$ is in this subset, but $v+v = (2,2,2)$ is not, so it is not closed under $+$.
    \item \textbf{The set of triples $(x,y,z) \in \bR^3$ such that $x,y,z$ are nonnegative.}
    
    This \fbox{is not} a vector subspace of $\bR^3$. For example, the vector $v = (1,1,1)$ is in this subset, but $-v = (-1,-1,-1)$ is not, so it is not closed under scalar multiplication. (It is closed under $+$, on the other hand.)
    \item \textbf{The set of all smooth functions $f: \bR \to \bR$.}
    
    This \fbox{is} a vector subspace of $\bR^\bR$, basically because sums of products of smooth functions are again smooth. (A function is \emph{smooth} when it has a continuous derivative, which has a continuous derivative, which has a continuous derivative, ad infinitum.)
    \item \textbf{The set of all functions $f: \bR \to \bR$ which solve the differential equation $f''(x) = f(x)$, where $f''$ denotes the second derivative of $f$.}
    
    This \fbox{is} a vector subspace of $\bR^\bR$, basically because the equation ``$f'' = f$'' is linear. Indeed, $0$ solves $f''=f$ (because the derivative of the zero function is again zero); if $f_1,f_2$ both solve $f'' = f$, then so does $f_1+f_2$ (because derivative of a sum is sum of derivatives); and if $\alpha \in \bR$ then $(\alpha f)' = \alpha f'$, and so multiplying a solution to $f''=f$ by $\alpha$ produces another solution.
    
    \item \textbf{The set of functions $f : \bR \to \bR$ such that $f(0)$ is an integer.}
    
    This \fbox{is not} a vector subspace of $\bR^\bR$. It does pass some tests: it contains the $0$ function and is closed under $+$ and $-$. However, it is not closed under scalar multiplication. Indeed: the constant function $f(x)\equiv 1$ is in this set, but multiplying this by the number $\alpha=\frac12$ produces the constant function $f(x)\equiv \frac12$, which is not in this set.
  \end{enumerate}
  \item \textbf{Is $\bR$ naturally a vector space over $\bC$? Is $\bC$ naturally a vector space over $\bR$?}
  
  $V=\bC$ \fbox{is} naturally a vector space over $\bF=\bR$. First, $\bC$ comes equipped with an addition law which satisfies all the requirements for addition in a vector space. To be a vector space over $\bR$, the other datum $\bC$ needs is a way of taking a ``scalar'' $\alpha\in \bR$ and a ``vector'' $v \in \bC$ and multiply them to produce a new ``vector'' $\alpha v \in \bC$. Since the real numbers are naturally a subset of the complex numbers, we can use the complex multiplication for this: just read ``$\alpha v$'' as that product of complex numbers. This multiplication law does indeed satisfy the necessary associativity and distributivity rules.
  
  $V=\bR$ \fbox{is not} naturally a vector space over $\bF\bC$. It does come with an addition law satisfying all requirements. However, in order to be a vector space, we would need a way to multiply a ``scalar'' $\alpha \in \bC$ and a ``vector'' $v \in\bR$ and produce a ``vector'' $\alpha v \in \bR$. Let's try to find such a multiplication law; by trying and failing, we will amass evidence that it might be impossible. (We will not produce a proof that it is impossible, however.) Well, we could remember that $\bR$ is a subset of $\bC$ and so evaluate $\alpha v$ as a complex multiplication. But the result will typically be a complex number which is not real. We could take just its real part. But that will not be suitably associative.
  
  Actually proving that $V=\bR$ is not naturally a vector space over $\bF=\bC$ is hard because it requires saying more carefully what the word ``naturally'' means. Let's agree that in order to be natural, the vector addition should be the addition in $\bR$. Let's also agree that ``naturality'' requires that the multiplication law $\alpha v$, where $\alpha \in \bC$ and $v \in \bR$, have the following property: if $\alpha$ happens to be real, then $\alpha v$ is means the usual multiplication of real numbers. With these agreements, we can prove that $\bR$ is not naturally a vector space over $\bC$. Indeed, suppose that it were, and take $\alpha = \sqrt{-1}$ and $v = 1$. Then, by supposition, there would be some real number $r = \alpha v$. But this is also equal to $\beta v$ where $\beta = r \in \bR \subset \bC$ and still $v=1$. So $(\alpha-\beta)v = 0$. By exercise 2, this forces $\alpha-\beta= 0$ or $v=0$ or both. Well, $v \neq 0$, so it forces $\alpha-\beta = 0$. But $\alpha = \sqrt{-1}$ whereas $\beta$ is real, and so we have a contradiction.
  
  \item \textbf{Let $V := \bR_{>0}$ denote the set of positive real numbers. Let's define, for $v,w \in V$, their ``$V$-sum'' to be $v\, +_V \,w := vw$, where the subscript on the left-hand side the subscript reminds that it is a new notion of ``sum'' in the set $V$, and on the right-hand side we mean the product of positive numbers. Also, for $\lambda \in \bR$ and $v \in V$, let's define their ``product'' $\lambda \cdot_V v := v^\lambda$, where the right-hand side means the exponential of real numbers. Is this $V$, with these notions of addition and scalar multiplication, a vector space over $\bR$? Explain.}
  
  This $V$, with these notions of addition and scalar multiplication, \fbox{is} a vector space over $\bR$! 
  
  There is a sneaky reason for this. Given $v \in V$, consider the real number $r = \log v \in \bR^1$; conversely, given $r \in \bR^1$, consider the positive real number $v = \exp v \in V = \bR_{>0}$. (Mathematicians typically write ``$\exp$'' for the natural, i.e.\ base $e$, exponential function, and ``$\log$'' for the natural, i.e.\ base $e$, logarithm. But any base will work so long as they are the same: if you want to think of $\log$ as the base-10 logarithm, then just decide that $\exp(r) := 10^r$.) These two functions $\exp: r \leftrightarrow v:\log$ provide a perfect matching between $\bR^1$ and $V$. Moreover, they relate the usual addition and scalar multiplication in $\bR^1$  to the funny versions for $V$.
  
  There is also a non-sneaky reason: we can just check the axioms directly. First, we need to check whether $+_V$ is commutative and associative: If $u,v,w\in V$, do we have $v+_V w \overset?= w +_V v$ and $(u +_V v) +_V w \overset?= u +_V (v+_V w)$? Yes, because these questions are equivalent to asking, for the usual multiplication of real numbers, whether $vw \overset?= wv$ and $(uv)w \overset?= u(vw)$, and the answers to both questions are yes. Second, we need to check whether there is an element $0_V \in V$ such that, for every $v \in V$, $0_V +_V v = v$? Yes: take $0_V := 1$, and use that for usual multiplication, $1v= v$. And is there negation? I.e.\ given $v \in V$, is there $-_V v$ such that $v +_V (-_V v) = 0_V$? Yes: take $-_V v := v^{-1}$. Third, we need to check some properties of multiplication. We need to check whether, given $\alpha,\beta \in \bR$ and $v \in V$, is $\alpha \cdot_V (\beta \cdot_V v) \overset?= (\alpha\beta) \cdot_V v$? Yes, because this unpacks to the question of whether $(v^\alpha)^\beta \overset?= v^{\alpha\beta}$. We need to check whether, given $v \in V$, is $1 \cdot_V v \overset?= v$? Yes, because this unpacks to the question of whether $v^1 \overset?= v$. And last, we need to check some distributivity laws, which unpack to whether $(vw)^\alpha \overset?= v^\alpha w^\alpha$ and whether $v^{\alpha+\beta} \overset? v^\alpha v^\beta$.
    
  \item \textbf{Let $V = \bR \sqcup \{+\infty,-\infty\}$. In other words, we take the set of real numbers, and add two new elements to the set, named ``$+\infty$'' and ``$-\infty$.'' Equip $V$ with an ``addition'' law which is the usual addition in $\bR$, extended by
  \begin{gather*} v + (+\infty) = (+\infty) + v = \infty, \qquad v + (-\infty) = (-\infty) + v = -\infty, \\
  (+\infty) + (+\infty) = (+\infty), \qquad (+\infty) + (-\infty) = (-\infty) + (+\infty) = 0, \qquad (-\infty) + (-\infty) = -\infty \end{gather*}
  Also, define a ``scalar multiplication'' law which is the usual multiplication in $\bR$, and, if $\alpha \in \bR$, then
  $$ \alpha(+\infty) = \begin{cases} -\infty, & \alpha < 0, \\ 0, & \alpha = 0, \\ +\infty, & \alpha > 0,\end{cases} \qquad \alpha(-\infty) = \begin{cases} +\infty, & \alpha < 0, \\ 0, & \alpha = 0, \\ -\infty, & \alpha > 0.\end{cases} $$
  Is this $V$, with these notions of addition and scalar multiplication, a vector space over $\bR$? Explain.}
  
  This $V$, with these notions of addition and scalar multiplication,  \fbox{is not} a vector space over $\bR$. Almost every axiom fails. For example, associativity of $+$ fails: $((+\infty) + (+\infty)) + (-\infty) = (+\infty) + (-\infty) = 0$, but $(+\infty) + ((+\infty) + (-\infty)) = (+\infty) + 0 = +\infty$.
  
  \item \textbf{Let $V$ be a vector space.}
  \begin{enumerate}
  \item \textbf{Prove that the intersection of two vector subspaces of $V$ is always another vector subspace of $V$.}
  
  Let $U_1, U_2 \subset V$ be vector subspaces. To show that $U := U_1 \cap U_2$ is a vector subspace of $V$, it suffices to show: (i) $0 \in U$; (ii) if $u,v \in U$ then $u+v \in U$; (iii) if $u \in U$ and $\alpha \in \bR$, then $\alpha u \in U$.
  
  To show (i), note that $0 \in U_1$ and $0 \in U_2$ and hence $0 \in U_1 \cap U_2 = U$.
  
  To show (ii), note that, if $u,v \in U$, then $u,v \in U_1$ and so $u+v \in U_1$, but also $u,v \in U_2$ so $u+v \in U_2$, and hence $u+v \in U_1 \cap U_2 = U$.
  
  To show (iii), note that, if $u \in U$, then $u \in U_1$, and so $\alpha u \in U_1$ for every $\alpha \in \bR$. But also $u \in U_2$, and so $\alpha u \in U_2$. Since $\alpha u$ is in both $U_1$ and $U_2$, it is in $U = U_1 \cap U_2$.
  
  
  \item \textbf{Prove that the union of two vector subspaces of $V$ is another vector subspace of $V$ only if one of the two original subspaces contains the other one.}
  
  Let $U_1, U_2 \subset V$ be vector subspaces. We are asked to show: the only way for $U_1 \cup U_2$ to be a vector subspace is if $U_1 \subset U_2$ or $U_2 \subset U_1$ (or both). Equivalently, we are asked to show that if $U_1 \not\subset U_2$ and $U_2 \not\subset U_1$, then $U_1 \cup U_2$ is not a vector subspace.
  
  Well, if $U_1 \not\subset U_2$, then we can choose an element $u_1 \in U_1$ which is not also in $U_2$. Similarly, if $U_2 \not\subset U_1$, then we can choose an element $u_2 \in U_2$ which is not also in $U_1$. So let's make such choices. Then these $u_1$ and $u_2$ are both in $U_1 \cup U_2$. We will show, however, that their sum $u_1 + u_2$ is not in $U_1 \cup U_2$. This will prove that $U_1 \cup U_2$ is not closed under $+$, and so is not a vector subspace.
  
  First, we claim that $u_1 + u_2$ is not in $U_1$. If it were, then since also $u_1 \in U_1$ and since $U_1$ is a vector subspace, we would have $u_2 = (u_1 + u_2) - u_1 \in U_1$. But when we selected $u_2$, we explicitly selected it to not be in $U_1$. Second, we claim that $u_1 + u_2$ is not in $U_2$. If it were, then since also $u_2 \in U_2$ and since $U_2$ is a vector subspace, we would have $u_1 = (u_1 + u_2) - u_2 \in U_2$. But when we selected $u_1$, we explicitly selected it to not be in $U_2$. So both statements ``$u_1 + u_2 \in U_1$'' and ``$u_1 + u_2 \in U_2$'' are false. This means that $u_1 + u_2$ is also not in the union $U_1 \cup U_2$, which is what we wanted to show.
  
  \end{enumerate}
  \item \textbf{Let $V$ be a vector space.}
  \begin{enumerate}
    \item \textbf{Is the addition of vector subspaces associative? In other words, given three vector subspaces $U_1,U_2,U_3 \subset V$, is}
    $$ U_1 + (U_2 + U_3) \overset?= (U_1 + U_2) + U_3 $$
    
    Yes. Indeed, the left-hand side unpacks to the set of all vectors in $V$ which can be written as $u_1 + (u_2 + u_3)$ for some elements $u_1 \in U_1$, $u_2 \in U_2$, and $u_3 \in U_3$. The right-hand side unpacks to the set of all vectors in $V$  which can be written as $(u_1 + u_2) + u_3$ for some elements $u_1 \in U_1$, $u_2 \in U_2$, and $u_3 \in U_3$. The associativity of vector addition implies that these two sets are the same
    \item \textbf{Is the addition of vector subspaces commutative? In other words, given two vector subspaces $U_1,U_2 \subset V$, is}
    $$ U_1 + U_2 \overset?= U_2 + U_1$$
    Yes. After unpacking, the commutativity of vector addition implies that these two sets are the same.
    \item \textbf{Is there a vector subspace ``$O$'' $\subset V$ such that $U + O = U$ for every vector subspace $U \subset V$?}
    
    Yes. The subspace $O := \{0\}$ works. Indeed, for this $U+O$ is the set of vectors of the form $u+o$ where $u \in U$ and $o \in O$. But the only choice for $o$ is $o=0$, and so $U+O$ is the set of vectors of the form $u+0$ where $u\in U$. But this is precisely the set $U$.
  \end{enumerate}
  \item \textbf{Let $U_1 \subset \bR^\bR$ be the set of functions $f : \bR \to \bR$ such that $f(x) = 0$ if $x < 0$. Let $U_2 \subset \bR^\bR$ be the set of functions $f : \bR \to \bR$ such that $f(x) = 0$ if $x \geq 0$.}
  \begin{enumerate}
    \item \textbf{Show that $U_1$ and $U_2$ are vector subspaces of $\bR^\bR$.}
    
    For each $U = U_1$ or $U_2$, it suffices to show: (i) $0 \in U$; (ii) if $f,g \in U$ then $f+g \in U$; (iii) if $f\in U$ and $\alpha \in \bR$, then $\alpha f \in U$.
    
    Well, the zero function $0$ certainly vanishes on all negative numbers, and so it is in $U_1$. It also vanishes on all nonnegative numbers, so it is in $U_2$. This confirms (i) for both $U_1$ and $U_2$.
    
    Suppose that $f,g \in U_1$, i.e.\ if $x <0$, then $f(x) = g(x) = 0$. Then, continuing to assume that $x<0$, we find that $(f+g)(x) = f(x)+g(x) = 0+0 = 0$. This shows that $f+g \in U_1$, establishing (ii) for $U_1$. Essentially the same proof works for $U_2$: suppose that $f,g \in U_2$, i.e.\ that if $x\geq 0$ then $f(x)=g(x)=0$; thus if $x \geq 0$ then $(f+g)(x) = f(x)+g(x) = 0+0=0$; thus $f+g \in U_2$.
    
    Property (iii) is proved similarly. Suppose that $f\in U_1$ and that $\alpha \in \bR$ and that $x < 0$. Then $(\alpha f)(x) = \alpha\, f(x) = \alpha\, 0 = 0$. Since this was true for all $x<0$, we find that $\alpha f \in U_1$. Suppose that $f \in U_2$ and that $\alpha \in \bR$ and that $x \geq 0$. Then $(\alpha f)(x) = \alpha\, f(x) = \alpha\, 0 = 0$.  Since this was true for all $x\geq 0$, we find that  $\alpha f \in U_1$
    \item \textbf{Show that $U_1 + U_2 = \bR^\bR$.}
    
    We must show that, for any function $f : \bR \to \bR$, there exists functions $f_1 \in U_1$ and $f_2 \in U_2$ such that $f = f_1 + f_2$. Set
    $$ f_1(x) := \begin{cases} f(x), & x \geq 0, \\ 0, & x < 0,\end{cases} \qquad  f_2(x) := \begin{cases}  0, & x \geq 0, \\ f(x), & x < 0. \end{cases}$$
    Then we find that indeed $f_1$ does satisfy the requirements to be in $U_1$ (i.e.\ it is a function which vanishes on negative numbers), and $f_2$ does satisfy the requirements to be in $U_2$ (i.e.\ it is a function which vanishes on nonnegative numbers). We can add these functions by considering the two cases:
    $$ (f_1+f_2)(x) = f_1(x) + f_2(x) = \begin{cases} f(x) + 0, & x \geq 0, \\ 0 + f(x), & x < 0. \end{cases}$$
    But this is just $f(x)$ in all cases, and so $f_1 + f_2 = f$.
    
    \item \textbf{Show that the sum is direct.}
    
    It suffices to show that $U_1 \cap U_2 = \{0\}$. In other words, we claim that the only function which is in both $U_1$ and $U_2$ is the zero function. Well, if $f \in U_1 \cap U_2$, then certainly $f \in U_1$ and so $f(x)=0$ for all negative $x$. On the other hand, if $f \in U_1 \cap U_2$, then certainly $f \in U_2$ and so $f(x) = 0$ for all nonnegative $x$. But every real $x$ is either negative or nonnegative, and so we see that if $f \in U_1 \cap U_2$, then $f(x) = 0$ for all $x$. And so $f=0$, which is what we wanted to prove.
  \end{enumerate}
\end{enumerate}



\end{document}  