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\title{Math 2135: Linear Algebra}
\author{Assignment 2}
\date{Solutions}

\begin{document}
\maketitle


\begin{enumerate}
  \item \begin{enumerate}
    \item \textbf{Suppose that $\{v_1,v_2,v_3,v_4\}$ spans a vector space $V$. Show that
    $ \{v_1 + v_2, v_2 + v_3, v_3 + v_4, v_4\}$
    also spans $V$.}
    
    Let $v \in V$ be a vector. Since $\{v_1,v_2,v_3,v_4\}$ spans, we can find numbers $\alpha_1,\alpha_2,\alpha_3,\alpha_4 \in \bF$ such that
    $$ v = \alpha_1 v_1 + \alpha_2 v_2 + \alpha_3 v_3 + \alpha_4 v_4.$$
    This can be rearranged to
    $$ v = \alpha_1 (v_1 + v_2) - \alpha_1 v_2 + \alpha_2 v_2 + \alpha_3 v_3 + \alpha_4 v_4 = \alpha_1(v_1 + v_2) + (\alpha_2 - \alpha_1) v_2 + \alpha_3 v_3 + \alpha_4 v_4.$$
    Repeating the trick gives:
    $$ v = \alpha_1(v_1 + v_2) + (\alpha_2 - \alpha_1) (v_2 + v_3) + (\alpha_3 - \alpha_2 + \alpha_1) (v_3 + v_4) + (\alpha_4 - \alpha_3 + \alpha_2 - \alpha_1) v_4.$$
    Thus we have found numbers $\alpha_1', \alpha_2',\alpha_3',\alpha_4'$ such that $v = \alpha_1'(v_1+v_2) + \alpha_2' (v_2+v_3) + \alpha_3' (v_3+v_4) + \alpha_4' v_4$, namely $\alpha_1' = \alpha_1$, $\alpha_2' = \alpha_2-\alpha_1$, $\alpha_3' = \alpha_3 - \alpha_2 + \alpha_1$, and $\alpha_4' = \alpha_4 - \alpha_3 + \alpha_2 - \alpha_1$.
    \item \textbf{Suppose that $\{v_1,v_2,v_3,v_4\}$ is linearly independent. Show that
    $ \{v_1 + v_2, v_2 + v_3, v_3 + v_4, v_4\}$
    is also linearly independent.}
    
    Suppose that we have found $\alpha_1,\alpha_2,\alpha_3,\alpha_4$ such that
    $$ 0 = \alpha_1 (v_1+v_2) + \alpha_2(v_2+v_3) + \alpha_3 (v_3+v_4) + \alpha_4 v_4.$$
    We wish to show that $\alpha_1 = \alpha_2 = \alpha_3 = \alpha_4 = 0$.
    
    Well, expanding and grouping like terms, we find:
    $$ 0 = \alpha_1 v_1 + (\alpha_1 + \alpha_2) v_2 + (\alpha_2 + \alpha_3) v_3 + (\alpha_3 + \alpha_4) v_4.$$
    Since $\{v_1,v_2,v_3,v_4\}$ is assumed linearly independent, we learn that $0 = \alpha_1 = \alpha_1+\alpha_2 = \alpha_2 + \alpha_3 = \alpha_3 + \alpha_4$. But if $0 = \alpha_1$ and $0 = \alpha_1 + \alpha_2$, then $0 = \alpha_2$. Repeating the argument with the other equations gives $0 = \alpha_3 = \alpha_4$.
  \end{enumerate}
  \item \textbf{For which numbers $t$ is the set 
  $ \{(3,1,4), (2,-3,5), (5,9,t)\}$
  a basis of $\bR^3$?}
  
  This set is a basis provided it is spanning and linearly independent. Suppose that linear independence fails. Then we would have numbers $a,b,c \in \bR$, not all zero, such that
  $$ (0,0,0) = a(3,1,4) + b(2,-3,5) + c(5,9,t) = (3a+2b+5c, a-3b+9c, 4a+5b+tc).$$
  Comparing the first two terms and solving for $c$ gives $c = -\frac35 a - \frac25 b$. Comparing the second terms and solving for $c$ gives $c = -\frac19 a + \frac13 b$. Let's set these equal and rearrange the equation:
  \begin{align*}
   -\frac35 a - \frac25 b & =  -\frac19 a + \frac13 b \\
   \left(-\frac35 + \frac19\right)a & = \left(\frac25 + \frac13\right)b \\
   \left(-3 + \frac59\right)a & = \left(2 + \frac53\right)b \\
   \left(-27 + 5\right)a & = \left(18 + 15\right)b \\
   -22 a & = 33 b \\
   -\frac23 a & b
  \end{align*}
  In particular, either both $a$ and $b$ are zero, in which case $c$ is, or neither $a$ nor $b$ is. In other words, if we are to find a nontrivial linear dependence, then we'd better have $a \neq 0$, and so we might as well devide through by $a$ and then redefine $b \leadsto \frac ba$ and $c \leadsto \frac ca$, and just decide that $a=1$. Then $b= -\frac23$, and looking again at $c = -\frac35 a - \frac25 b = -\frac19 a + \frac13 b$ gives $c = -\frac13$.
  
  This was just from looking at the first two terms. To have a linear dependence, we would also need
  $$ 0 = 4a+5b+tc = 4 - \frac{10}3 - \frac t 3$$
  or equivalently $t=2$.
  
  We have shown: \emph{The set $ \{(3,1,4), (2,-3,5), (5,9,t)\}$ is linearly independent if and only if $t\neq 2$.}
  
  We showed in class that a linearly independent set is a basis if and only if it has the correct size. This set does have the correct size to be a basis. \emph{The set  $ \{(3,1,4), (2,-3,5), (5,9,t)\}$ is a basis if and only if $t\neq 2$.}
  
  \item  \textbf{Let $U$ be the subspace of $\bC^5$ defined by
    $$ U := \{ (z_1,z_2,z_3,z_4,z_5) \in \bC^5 \st 6z_1 = z_2 \text{ and } z_3+2z_4+3z_5 = 0\}.$$}
    \begin{enumerate}
    \item \textbf{Find a basis for $U$.}
    
    Note that $\dim \bC^5 = 5$ and so $U$ is finite-dimensional. So we can build a basis from the ground up: start with the empty set $\emptyset$, and include vectors, while staying linearly independent the whole time, until we have to stop.
    
    $\Span(\emptyset) = \{0\} \neq U$. So we include some nonzero vector. For example, $(1,6,0,0,0) \in U$.
    
    $\Span((1,6,0,0,0))$ is not all of $U$. For example, $(0,0,2,-1,0) \in U \smallsetminus \Span((1,6,0,0,0))$.
    
    $\Span((1,6,0,0,0),(0,0,2,-1,0))$ is not all of $U$. For example, $(0,0,3,0,-1) \in U \smallsetminus \Span((1,6,0,0,0),(0,0,2,-1,0))$.
    
    The set $\{(1,6,0,0,0), (0,0,2,-1,0), (0,0,3,0,-1) \}$ is easily seen to be in $U$. Let's confirm that it is linearly independent. Well,
    $$ a (1,6,0,0,0) + b(0,0,2,-1,0) + c(0,0,3,0,-1) = (a,?,?,-b,-c)$$
    and I didn't feel like working out the middle two entries. The only way for this sum to equal $(0,0,0,0,0)$ is if $a=b=c=0$.
    
    We claim that $\{(1,6,0,0,0), (0,0,2,-1,0), (0,0,3,0,-1) \}$ spans $U$. Suppose that $v = (z_1,z_2,z_3,z_4,z_5) \in U$. Consider the vector
    $$ v' = z_1(1,6,0,0,0) - z_4(0,0,2,-1,0) - z_5(0,0,3,0,-1) = (z_1, 6z_1, -2z_4-3z_5, z_4,z_5).$$
    Then $v'$ and $v$ certainly agree in the first, fourth, and fifth entries. The second entry of $v'-v$ is $6z_1-z_2$, which must be zero if $v\in U$. Similarly, the third entry is $z_2 + 2z_4 + 3z_5$, which also vanishes by definition of $U$.
    
    So $v'=v$, and we have shown that $\{(1,6,0,0,0), (0,0,2,-1,0), (0,0,3,0,-1) \}$ spans. Since it was also linearly independent, it is a basis of $U$.
    
    \textbf{Remark:} This is not the only solution.
    
    \item \textbf{Extend the basis in part (a) to a basis for $\bC^5$.}
    
    We start with our basis $\{(1,6,0,0,0), (0,0,2,-1,0), (0,0,3,0,-1) \}$ and continue to include vectors. Note that $(0,1,0,0,0) \not\in U = \Span\{(1,6,0,0,0), (0,0,2,-1,0), (0,0,3,0,-1) \}$. So let's add it in. Another vector not in there: $(0,0,1,0,0)$.
    
    We claim that $\{(1,6,0,0,0), (0,0,2,-1,0), (0,0,3,0,-1),(0,1,0,0,0),(0,0,1,0,0) \}$ is a basis. Indeed:
    $$a(1,6,0,0,0)+b (0,0,2,-1,0)+c (0,0,3,0,-1)+d(0,1,0,0,0)+e(0,0,1,0,0) = (a, 6a+d, 2b+3c+e, -b, -c).$$
    For this sum to equal $(0,0,0,0,0)$ we must have $a=b=c=0$, in which case also $d=e=0$. So it is linearly independent. Conversely, the sum does equal $(z_1,z_2,z_3,z_4,z_5)$ provided $a=z_1$, $b=-z_4$, $c=-z_5$, $d = z_2-6a = z_2-6z_1$, and $e = z_3 - 2b-2c = z_3 + 2z_4 + 3z_5$.

    \textbf{Remark:} This is not the only solution.
  \end{enumerate}
  
%  \item \begin{enumerate} 
%  \item Suppose $\{v_1,v_2,v_3\}$ is a basis for a vector space $V$, and that $U \subset V$ is a vector subspace such that $v_1,v_2 \in U$ but $v_3 \not\in U$. Is $\{v_1,v_2\}$ necessarily a basis for $U$? Explain.
%  \item Suppose $\{v_1,v_2,v_3,v_4\}$ is a basis for a vector space $V$, and that $U \subset V$ is a vector subspace such that $v_1,v_2 \in U$ but $v_3,v_4 \not\in U$. Is $\{v_1,v_2\}$ necessarily a basis for $U$? Explain.
%  \end{enumerate} 
  
  \item \label{proof} {\bf We ended lecture on Monday having stated the following theorem, but we didn't supply a complete proof:
  
  \begin{quote}\textbf{Theorem:} Let $V$ be a finite-dimensional vector space, and $U_1,U_2 \subset V$ two vector subspaces. Then
  $$ \dim (U_1 + U_2) = \dim U_1 + \dim U_2 - \dim(U_1 \cap U_2).$$\end{quote}
  
  This exercise asks you to work through the remainder of the proof. A complete proof of this theorem can be found in \emph{Linear Algebra Done Right}, and you are welcome to read that discussion while thinking about this exercise. However, your answers must be in your own words. I recommend that you close the book before starting to write up your answers.
  
  To review, at the end of lecture we said the following. Since $U_1 \cap U_2 \subset V$ is a vector subspace, and since $V$ is finite-dimensional, we know that $U_1 \cap U_2$ is also finite-dimensional. Suppose that $A$ is any basis for $U_1 \cap U_2$. Then we argued that we can extend $A$ to a finite basis $A \cup B$ for $U_1$ and we can extend $A$ to a finite basis $A \cup C$ for $U_2$. We asserted, but did not prove, that $A \cup B \cup C$ is a basis for $U_1 + U_2$.
  
  Let's give names: $A = \{u_1,\dots,u_k\}$, $B = \{v_1,\dots,v_m\}$, and $C = \{w_1,\dots,w_n\}$.}
    
  \begin{enumerate}
    \item {\bf Implicit in the notation is that $A$ and $B$ are disjoint, and that $A$ and $C$ are disjoint. Explain why $B$ and $C$ are disjoint. (Two sets are \emph{disjoint} if their intersection is empty, i.e.\ if there are no elements in common.)}
    
    By definition, the elements of $C$ are in $U_2$ but not $U_1$, whereas the elements of $B$ are in $U_1$ (but not $U_2$).
    
    \item \textbf{Explain why the assertion ``$A \cup B \cup C$ is a basis for $U_1 + U_2$'' implies the theorem.}
    
    The dimension of a subspace is the size of any of its bases. Since they are bases, $\# A = \dim (U_1 \cap U_2)$, whereas $\dim U_1 = \#(A \cup B) = \# A + \# B$ and $\dim U_2 = \$(A\cup C) = \# A + \# C$. Thus
    $$ \#(A \cup B \cup C) = \# A + \# B + \# C = (\# A + \# B) + (\# A + \# C) - \# A,$$
    which is equal to the right-hand side of the equation we want to prove. On the other hand, if $A \cup B \cup C$ is a basis for $U_1 + U_2$, then also $\#(A \cup B \cup C)$ is equal to the left-hand side of the equation we want to prove.
        
    \item {\bf Explain why $\operatorname{Span}(A \cup B \cup C) = U_1 + U_2$. In other words, explain why if
    $$ v = \alpha_1 u_1 + \dots + \alpha_k u_k + \beta_1 v_1 + \dots + \beta_m v_m + \gamma_1 w_1 + \dots \gamma_1 w_k$$
    where all the $\alpha$'s, $\beta$'s, and $\gamma$'s are in $\bF$, then $v \in U_1 + U_2$, and conversely why any $v \in U_1 + U_2$ can be written as such a linear combination for some $\alpha$'s, $\beta$'s, and $\gamma$'s  in $\bF$.}
    
    If $v \in U_1 + U_2$, then we can find --- we've run out of letters --- let's say $x \in U_1$ and $y \in U_2$ such that $v = x+y$. Now, since $A \cup B$ is a basis for $U_1$, we can find numbers $\lambda_1,\dots,\lambda_k,\mu_1,\dots,\mu_m$ such that $$x = \lambda_1 u_1 + \dots + \lambda_k u_k + \mu_1 v_1 + \dots + \mu_m v_m.$$ Similarly, since $A \cup C$ is a basis for $U_2$, we can find numbers $\rho_1,\dots,\rho_k,\nu_1,\dots,\nu_n$ such that $$y = \rho_1 u_1 + \dots + \rho_k v_k + \nu_1 w_1 + \dots + \nu_n w_n.$$
    Adding the two expressions gives
    $$ v = x+y = (\lambda_1 + \rho_1)u_1 + \dots + (\lambda_k + \rho_k) u_k + \mu_1 v_1 + \dots + \mu_m v_m + \nu_1 w_1 + \dots + \nu_n w_n. $$
    In other words, we can win by setting $\alpha_i = \lambda_i + \rho_i$, $\beta_i = \mu_i$, and $\gamma_i = \nu_i$.
    
    \item {\bf (The most interesting part.) Explain why $A \cup B \cup C$ is linearly independent. In other words, we want to show that if
    $$ 0 = \alpha_1 u_1 + \dots + \alpha_k u_k + \beta_1 v_1 + \dots + \beta_m v_m + \gamma_1 w_1 + \dots \gamma_1 w_k$$
    for some $\alpha$'s, $\beta$'s, and $\gamma$'s in $\bF$, then all the $\alpha$'s, $\beta$'s, and $\gamma$'s are $0$. So assume that you have found some such solution, and let
    \begin{align*}
      u & := \alpha_1 u_1 + \dots + \alpha_k u_k \\
      v & := \beta_1 v_1 + \dots + \beta_m v_m \\
      w & := \gamma_1 w_1 + \dots \gamma_1 w_k
    \end{align*}
    Explain why $v \in U_1$. Explain why also $v \in U_2$. Conclude that $v \in U_1 \cap U_2$. Explain why this implies that there are numbers $\delta_1,\dots,\delta_k \in \bF$ such that
    $$ v = \delta_1 u_1 + \dots + \delta_k u_k.$$
    Explain why this implies that either all the $\beta$s are zero or that $A \cup B$ is linearly dependent. (Consider the difference of  two expressions for $v$.)
    
    But $A \cup B$ is linearly independent by assumption (which assumption?), so all the $\beta$s are zero. Explain why this, together with the assumption (which one?) that $A \cup C$ is linearly independent, implies that all the $\alpha$s and all the $\gamma$s are zero.}
    
    With these $u,v,w$, we have
    $$ 0 = u+v+w.$$
    Note that $v \in \Span(B) \subset \Span(A\cup B) = U_1$. Note also that $u+w \in \Span(A\cup C) = U_2$. But $v = -(u+w)$, and so $v \in U_2$. So $v \in U_1 \cap U_2$.
    
    But $A$ is a basis for $U_1 \cap U_2$, so there must exist $\delta$'s as above. So
    $$ 0 = v-v = (\delta_1 u_1 + \dots + \delta_k u_k) - ( \beta_1 v_1 + \dots + \beta_m v_m)  = \delta_1 u_1 + \dots + \delta_k u_k + (-\beta_1) v_1 + \dots + (-\beta_m) v_m.$$
    If any of the $\delta$'s or $\beta$ are nonzero, then we have found a linear dependency in $A \cup B$, which we assumed was a basis. So all the $\delta$s and all the $\beta$s are zero, and so $v=0$.
    
    But then our original assumed dependency becomes
    $$0 = u+w = \alpha_1 u_1 + \dots + \alpha_k u_k + \gamma_1 w_1 + \dots \gamma_1 w_k.$$
    On the other hand, since $A \cup C$ is linearly independent, it must happen that all the $\alpha$s and all the $\gamma$s vanish.
  \end{enumerate}
  \item {\bf Suppose that $V$ is finite-dimensional and contains three vector subspaces $U_1,U_2,U_3$. The theorem in Exercise~\ref{proof} might lead you to think that
  \begin{multline*}  \dim(U_1 + U_2 + U_3) \overset?= \dim(U_1) + \dim(U_2) + \dim(U_3) - \dim(U_1 \cap U_2)  \\ - \dim(U_1 \cap U_3) - \dim(U_2 \cap U_3) + \dim(U_1 \cap U_2 \cap U_3),\end{multline*}
  but this formula is \emph{not true} in general.}
  \begin{enumerate}
    \item {\bf Explain why $V = \bR^2$, with $U_1,U_2,U_3$ any three pairwise-distinct 1-dimensional subspaces, provides a counterexample.}
    
    If $U_1,U_2 \subset \bR^2$ are both one-dimensional, then either they are equal or they intersect only at the origin. In other words, if $U_1,U_2,U_3$ any three pairwise-distinct 1-dimensional subspaces, then we definitely have $U_1 \cap U_2 = U_1 \cap U_3 = U_2 \cap U_3 = U_1 \cap U_2 \cap U_3 = \{0\}$, and all of these spaces have dimension $0$. So the supposed equation becomes
    $$  \dim(U_1 + U_2 + U_3) \overset?= 1+1+1-0-0-0+0 = 3.$$
    But the sum $U_1 + U_2 + U_3$ is a subspace of $\bR^2$, and so cannot have dimension $>2$. So the supposed equation is false.
    
    In fact, if $U_1,U_2$ are distinct one-dimensional subspaces in $\bR^2$, then we already see from the theorem in Exercise~\ref{proof} that $\dim(U_1+U_2) = 1+1-0 = 2$, and so $U_1 + U_2 = \bR^2$. Thus $U_1 + U_2 + U_3 = \bR^2$ as well.
    \item \textbf{Suppose you tried to  repeat the proof from Exercise~\ref{proof} but with three subspaces. 
    Which step of the proof fails? Explain.}
    
    The step that fails is the ``interesting'' linear independence part. Let us explain:
    
    Some subscripts will help manage the alphabet soup. Let's choose a basis $A_{123}$ for $U_1 \cap U_2 \cap U_3$. We can extend it to a basis $A_{123} \cup A_{12}$ for $U_1 \cap U_2$, to a basis $A_{123} \cup A_{13}$ for $U_1 \cap U_3$, and to a basis $A_{123} \cup A_{23}$ for $U_2 \cap U_3$. Moreover, a version of the argument from Exercise~\ref{proof} shows that $A_{123} \cup A_{12} \cup A_{13}$ is a linearly independent set in $U$, and so extends to a basis $A_{123} \cup A_{12} \cup A_{13} \cup A_1$ for $U_1$. By the same token, there are bases $A_{123} \cup A_{12} \cup A_{23} \cup A_2$ for $U_2$ and $A_{123} \cup A_{13} \cup A_{23} \cup A_3$ for $U_3$.
    
    Then, to prove the supposed equation, it would suffice to prove that $A_{123} \cup A_{12} \cup A_{13} \cup A_{23} \cup A_1 \cup A_2 \cup A_3$ is a basis for  $U_1 + U_2 + U_3$.
    
    It is a spanning set, by a version of the argument from Exercise~\ref{proof}.
    
    But when we try to prove linear independence, we run into trouble. For example, we might get into a situation where some vector $v$ can be shown to be in $U_1$, and also in $U_2 + U_3$. If we could show it was in $U_1 \cup U_3$, then we could win: we'd be able to show it was zero through a version of the argument from Exercise~\ref{proof}. But all we can do is get into $U_2 + U_3$.
    
  \end{enumerate}
  \item {\bf Suppose, in Exercise~\ref{proof}, that $V$ is infinite-dimensional. Does this really matter for the theorem? Explain. \textbf{Hint:} What happens if, even though $V$ is infinite-dimensional, $U_1$ and $U_2$ are both finite-dimensional? What happens if one or both of them is infinite-dimensional?}
  
  Suppose that $U_1$ and $U_2$ are both finite-dimensional. Then so is $U_1 \cap U_2$, and we can simply proceed with the argument: choose bases; show that something is a basis for something.
  
  Suppose that $\dim U_1 = \infty$ but $\dim U_2 < \infty$. Then $\dim(U_1 + U_2)$ is also infinite, whereas $\dim (U_1 \cap U_2)$ is finite, and the equation we want to prove is the definitely-true statement
  $$ \infty = \infty + (\text{finite}) - (\text{finite}).$$
  Ditto with the roles of $U_1$ and $U_2$ switched.
  
  If $U_1$ and $U_2$ are both infinite-dimensional, then we have
  $$ \infty = \infty + \infty - \dim(U_1 \cap U_2).$$
  If $U_1 \cap U_2$ is finite-dimenisonal, then this is definitely a true statement. If $\dim(U_1 \cap U_2) = \infty$, then probably we might simply throw our hands in the air and decide that the statement has no content.
  
  Or we might decide: whatever these dimensions are, we definitely always have $\dim(U_2) \geq \dim(U_1 \cap U_2)$. So we should always interpret ``$\dim(U_2) - \dim(U_1 \cap U_2)$'' as being ``positive,'' perhaps infinite, even if it is $\infty-\infty$ and so not a specific number. Then we could decide that, if $\dim U_1 = \infty$, then the right-hand side is $\infty + (\text{positive})$, which is undeniably infinite. With this interpretation, the theorem remains true in all cases.
\end{enumerate}



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