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\title{Math 2135: Linear Algebra}
\author{Assignment 3}
\date{Solutions}

\begin{document}
\maketitle


\begin{enumerate}
  \item {\bf There is an interesting function $\bC \to \bC$ called \emph{complex conjugation} and denoted $z \mapsto \bar z$. It takes a complex number $z = a+b\sqrt{-1}$, and $a,b \in\bR$, to the \emph{conjugate} number $\bar z = a - b\sqrt{-1}$.}
  \begin{enumerate}
    \item {\bf Think of $V= \bC$ as a vector space over $\bF = \bR$. Is complex conjugation a linear map?}
    
    Yes. Given complex numbers $z_1,z_2 \in V = \bC$, we do have $\overline{z_1+z_2} = \overline{z_1} + \overline{z_2}$, so complex conjugation respects vector addition. Furthermore, if $\lambda \in \bF = \bR$, and if $z \in V = \bC$, then $\overline{\lambda z} = \overline\lambda \overline z = \lambda \overline z$, so complex conjugation respects scalar multiplication.
    
    \item {\bf Think of $V = \bC$ as a vector space over $\bF =\bC$. Is complex conjugation a linear map?}
    
    No. Let $z \in V = \bC$ be a nonzero complex number, for example $z=1$. Let $\lambda \in \bF = \bC$ be a complex number which is not purely real, for example $\lambda = \sqrt{-1}$. Then $\overline{\lambda z} = \overline\lambda \overline z \neq \lambda \overline z$. So complex conjugation does not respect scalar multiplication.
  \end{enumerate}
  \item {\bf Given any $b,c \in \bR$, define a function $T : \bR^3 \to \bR^2$ by 
  $$ T(x,y,z) = (2x-4y+3z+b, 6x+cxyz).$$
  Show that $T$ is linear if and only if $b=c=0$.}
  
  Let us check whether $T$ respects vector addition. In other words, given two vectors $v_1 = (x_1,y_1,z_1)$ and $v_2 = (x_2,y_2,z_2) \in \bR^3$, we wish to check whether $T(v_1+v_2) \overset?= T(v_1) + T(v_2)$. Spelled out, this is asking whether
  \begin{multline*} (2x_1-4y_1+3z_1+b, 6x_1+cx_1y_1z_1) + (2x_2-4y_2+3z_2+b, 6x_2+cx_2y_2z_2) \\ \overset?= (2(x_1+x_2)-4(y_1+y_2)+3(z_1+z_2)+b, 6(x_1+x_2)+c(x_1+x_2)(y_1+y_2)(z_1+z_2))\end{multline*}
  is always an equality of vectors in $\bR^2$. Matching the coordinates, we are asking:
  \begin{gather*}
    2x_1-4y_1+3z_1+b + 2x_2-4y_2+3z_2+b \overset?= 2(x_1+x_2)-4(y_1+y_2)+3(z_1+z_2)+b \\
    6x_1+cx_1y_1z_1 + 6x_2+cx_2y_2z_2 \overset?= 6(x_1+x_2)+c(x_1+x_2)(y_1+y_2)(z_1+z_2)
  \end{gather*}
  
  After some light rearranging, we see that the first equation holds if and only if
  $$ 2b \overset?= b$$
  and the second holds if and only of, for every $x_1,y_1,z_1,x_2,y_2,z_2 \in \bR$, we have
  $$ c(x_1y_1z_1 + x_2y_2z_2) \overset?= c(x_1+x_2)(y_1+y_2)(z_1+z_2).$$
  The first equation holds if and only if $b=0$. The second holds if and only if $c=0$: the if direction is obvious, and the only if direction is readily seen by setting $x_1=\dots=z_2=1$.
  
  We have thus shown that $T$ respects vector addition if and only if $b=c=0$. It remains to show that if $b=c=0$, then $T$ respects scalar multiplication. But, letting $\lambda \in\bR$, and taking $b=c=0$, we compute:
  $$ T\lambda(x,y,z) = T(\lambda x,\lambda y, \lambda z) = (2\lambda x-4\lambda y + 3\lambda z, 6\lambda x) = \lambda(2x-4y+3z,6x) = \lambda T(x,y,z).$$
  
  \item {\bf Let $V\subset \bR^\bR$ be the vector space of all differentiable functions. Given $f \in V$, denote its derivative by $f'$.}
  \begin{enumerate}
    \item {\bf Is the function $V \to \bR^2$ that sends $f \mapsto \bigl(f(1) + f'(2), \int_0^3 f(x)\,\mathrm{d}x\bigr)$ linear? Why or why not?}
    
    \fbox{Yes}, this function is linear. To see this, it is helpful to inspect the pieces of the function. 
    
    First, for any real number $r\in \bR$, the map $V \to \bR$ sending $f \mapsto f(r)$ is linear. Indeed, the function $f+g$ is defined so that  $(f+g)(r) = f(r)+g(r)$ for every $r$, and for $\lambda \in \bR$ the function $\lambda f$ is defined so that $(\lambda f)(r) = \lambda f(r)$ for every $r$. So for example $f \mapsto f(1)$ is a linear transformation.
    
    Second, taking derivatives is linear. Indeed, $(f+g)' = f'+g'$ and $(\lambda f)' = \lambda' f + \lambda f' = \lambda f'$ since $\lambda \in \bR$ is constant (so $\lambda' = 0$). Since evaluating is also linear, we see that $f \mapsto f'(2)$ is linear.
    
    Third, definite integrals are linear: for fixed real numbers $a,b \in \bR$, and for any integrable functions $f,g$, we know that $\int_a^b(f(x)+g(x))\,\mathrm{d}x = \int_a^b f(x)\,\mathrm{d}x + \int_a^b g(x)\,\mathrm{d}x$, and for any integrable function $f$ and any real number $\lambda$, we know that $\int_a^b \lambda f(x)\, \mathrm{d}x = \lambda \int_a^b f(x)\,\mathrm{d}x$. (All differential functions are continuous, and all continuous functions are integrable.)
    
    Sums of linear functions are linear. And the addition and scalar multiplication in $\bR^2$ are defined component wise: this implies that if you have a transformation valued in $\bR^2$ such that each component is linear, then the whole transformation is linear.
    
    More directly, what we are saying is: For any differentiable functions $f,g$ and any constant $\lambda$,
    \begin{gather*}
      \hspace{-1in} \left((f+g)(1) + (f+g)'(2), \int_0^3 (f+g)(x)\,\mathrm{d}x\right) = \left(f(1) + f'(2), \int_0^3 f(x)\,\mathrm{d}x\right) + \left(g(1) + g'(2), \int_0^3 g(x)\,\mathrm{d}x\right), \hspace{-1in} \\
      \left(\lambda f(1) + \lambda f'(2), \int_0^3 \lambda f(x)\,\mathrm{d}x\right) = \lambda \left(f(1) + f'(2), \int_0^3 f(x)\,\mathrm{d}x\right).
    \end{gather*}
    
    \item {\bf Is the function $V \to \bR$ that sends $f \mapsto \int_0^3 x^2 f(x)\,\mathrm{d}x$ linear? Why or why not?}
    
    \fbox{Yes}, this function is linear. We have already remarked that the assignment $f\mapsto \int_0^3 f(x)\,\mathrm{d}x$ is linear. The question asks about what you get if you precompose this operation with the assignment $f(x) \mapsto x^2 f(x)$. In general, fix any differentiable function $h(x)$. Then the function $V \to V$ which sends $f \mapsto hf$ is linear:
    $$\hspace{-1in} h(x)\,(f(x)+g(x)) = h(x)\,f(x) + h(x)\,g(x), \quad h(x)\,\lambda f(x) = \lambda h(x)\,f(x), \quad \forall f,g \in V, \, \lambda \in \bR\hspace{-1in}$$
    
    More directly, what we are saying is: For any differentiable functions $f,g$ and any constant $\lambda$,
    \begin{gather*} \int_0^3 x^2 (f(x)+g(x))\,\mathrm{d}x = \int_0^3\bigl( x^2 f(x) + x^2 g(x)\bigr),\mathrm{d}x = \int_0^3 x^2 f(x)\,\mathrm{d}x + \int_0^3 x^2 g(x)\,\mathrm{d}x, \\
    \int_0^3 x^2 \lambda f(x)\,\mathrm{d}x = \int_0^3 \lambda x^2 f(x)\,\mathrm{d}x = \lambda \int_0^3 x^2 f(x)\,\mathrm{d}x.
    \end{gather*}
    \item {\bf Is the function $V \to \bR$ that sends $f \mapsto f'(2)^2$ linear? Why or why not?}
    
    \fbox{No}, this function is not linear. Let us write $T$ for this operation, i.e.\, $T(f) = f'(2)^2$. Now consider the case $f(x)=g(x) = x$. Then $f'(2) = g'(2) = 1$, and so
    $$ T(f) + T(g) = 1+1 = 2.$$
    On the other hand, $(f+g)(x) = 2x$, and so
    $$ T(f+g) = T(2x) = 2^2 = 4 \neq 2.$$
    Thus $T(f+g) \neq T(f) + T(g)$ for some $f,g \in V$, and so $T$ does not respect vector addition.
    
    $T$ also does not respect scalar multiplication, which can be seen for example by taking $\lambda = 2$ and $f(x) = x$.
    
    \textbf{Remark:} There do exist functions $f,g$ such that $T(f+g)=T(f)+T(g)$, for example $f=g=0$. There also exist numbers $\lambda$ such that $T(\lambda f) = \lambda T(f)$, for example $\lambda = 0$ or $\lambda = 1$. However, for \emph{most} functions $f,g$ and for most numbers $\lambda$, you will find that $T(f+g) \neq T(f)+T(g)$ and $T(\lambda g) \neq \lambda T(f)$.
  \end{enumerate}
  \item {\bf Suppose $T : \bF^4 \to \bF^2$ is a linear map such that $$\ker(T) = \{(x_1,x_2,x_3,x_4) \st x_1=5x_2 \text{ and } x_3=x_1+x_4\}.$$ Show that $T$ is surjective.}
  
  We know that $\dim \bF^4 = 4$ and $\dim \bF^2 = 2$. We also know that
  $$ \dim \bF^4 = \dim \ker T + \dim \im T. $$
  Thus, if we can show that $\dim \ker T = 2$, then we will know that $\dim \im T = 2$. But $\im T \subseteq \bF^2$, and so if $\dim \im T = 2$, then we must have equality $\im T = \bF^2$.
  
  It thus remains to understand $\ker T$. We claim that $v_1 := (5,1,5,0)$ and $v_2 := (0,0,1,1)$ are together a basis for $\ker T$. They are obviously linearly independent, and obviously both in $\ker T$. So the only thing to check is whether they span $\ker T$. Well, suppose that $v = (x_1,x_2,x_3,x_4) \in \ker T$. Then $x_1 = 5x_2$ and $x_3 = 5x_2 + x_4$. Set $\alpha_1 := x_2$ and $\alpha_2 := x_4$. Then
  \begin{multline*}
   \alpha_1 v_1 + \alpha_2 v_2 = x_2(5,1,5,0) + x_4(0,0,1,1) = (5x_2,x_2,5x_2,0) + (0,0,x_4,x_4) \\ = (5x_2,x_2,5x_2+x_4,x_4) = (x_1,x_2,x_3,x_4) = v.
   \end{multline*}
  So these two vectors span, and hence are a basis for, $\ker T$. So $\dim \ker T = 2$, which is what we wanted to prove.
  \item \begin{enumerate}
   \item {\bf Find a linear map $T : \bF^2\to \bF^2$ such that $\ker(T) = \im(T)$ or show that one does not exist.}
   
   For example,
   $$ T(x,y) := (y,0) $$
   has $\ker T = \{(x,y) : y=0\} = \{(x,0)\}$, where $x$ is arbitrary, whereas $\im T = \{(y,0)\}$ where $y$ is arbitrary. So $\ker T = \im T$.
   
   \item {\bf Find a linear map $T : \bF^3\to \bF^3$ such that $\ker(T) = \im(T)$ or show that one does not exist.}
   
   There does not exist such a map. Indeed, for any map $T : \bF^3 \to \bF^3$, we definitely have
   $$ 3 = \dim \bF^3 = \dim \ker T + \dim \im T.$$
   Suppose that $\ker T = \im T$ and that $n = \dim \ker T = \dim \im T$. Then we would have $3=2n$, i.e.\ $n=\frac32$. But dimensions are always integers.
   
   \end{enumerate}
  \item {\bf Let $S : U \to V$ and $T : V \to W$ be linear maps. Show that if any two of the three maps $S$, $T$, and their composition $TS : U \to W$, are invertible, then so is the third one.}
  
  Suppose first that $S$ and $T$ are invertible, with inverses $S^{-1}$ and $T^{-1}$ respectively. In other words, $SS^{-1} = I_V$, $S^{-1}S = I_U$, $TT^{-1} = I_W$, and $T^{-1}T = I_V$. Set $X := S^{-1}T^{-1}$. We claim that $X$ is an inverse to $TS$. In other words, we claim that $XTS = I_U$ and $TSX = I_W$. We now check these:
  $$ S^{-1}T^{-1}TS = S^{-1}I_V S = S^{-1}S = I_U, \qquad TSS^{-1}T^{-1} = TI_V T^{-1} = I_W.$$
  These equations confirm that $X= S^{-1}T^{-1}$ is indeed an inverse to $TS$
  
  Next, suppose that $S$ and $TS$ are invertible, with inverses $S^{-1}$ and $(TS)^{-1}$ respectively. In other words, $SS^{-1} = I_V$, $S^{-1}S = I_U$, 
  $(TS)^{-1}TS = I_U$, and $TS(TS)^{-1} = I_W$.
   Set $X := S(TS)^{-1}$. We wish to claim that $X$ is an inverse to $T$, i.e.\ that $TX = I_W$ and $XT = I_V$. One of these is easy:
  $$ TX = TS(TS)^{-1} = I_W.$$
  To show that $X$ is an inverse to $T$, we must also show that $XT = I_V$. This is harder because $XT = S(TS)^{-1}T$. If we could move the $S$ over next to the $T$, then we'd be able to use the equation $(TS)^{-1}TS = I_U$.
  Well, let's remember that we have access to $S^{-1}$ such that $SS^{-1} = I_V$. Then:
  $$ XT = S(TS)^{-1}T = S(TS)^{-1}TI_V = S(TS)^{-1}TSS^{-1} = SI_US^{-1} SS^{-1} = I_V.$$
  So indeed $X$ is an inverse to $T$.
  
  The third case is almost the same as the second case. Suppose that $T$ and $TS$ are invertible, with $T^{-1}$ and $(TS)^{-1}$ respectively. We claim that $X := (TS)^{-1}T$ is an inverse to $S$. To check this, we compute:
  \begin{gather*}
   XS = (TS)^{-1}TS = I_U, \\
   SX = S(TS)^{-1}T = I_VS(TS)^{-1}T=  T^{-1}T S(TS)^{-1}T = T^{-1}I_WT  = T^{-1}T = I_V.
  \end{gather*}
 \end{enumerate}



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