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\title{Math 2135: Linear Algebra}
\author{Assignment 5}
\date{Solutions}

\begin{document}
\maketitle


\begin{enumerate}
  \item {\bf Let $V$ be a vector space and $T \in \cL(V)$, and suppose that $T^n = 0$ for some positive integer $n$. Prove that $I-T$ is invertible and that its inverse is
  $$ (I-T)^{-1} = I + T + T^2 + \dots + T^{n-1}.$$}
  
  We multiply
  \begin{multline*}
   (I-T)(I + T + T^2 + \dots + T^{n-1}) \\ = I - T + T - T^2 + T^2 - T^3 + \dots + T^{n-2} - T^{n-1} + T^{n-1} - T^n \\ = I + 0 + 0 + \dots + 0 - T^n = I \end{multline*}
   since $T^n = 0$.
  The same calculation shows that $(I + T + T^2 + \dots + T^{n-1})(I-T) = I$ (the two factors commute). So they are each other's inverses, and in particular both are invertible.
  
  \item {\bf Let $V$ be a vector space and $T \in \cL(V)$. Show that $9$ is an eigenvalue of $T^2$ if and only if at least one of $3$ and $-3$ is an eigenvalue of $T$.}
  
  If $9$ is an eigenvalue of $T^2$, then $T^2-9$ is not injective. But $T^2-9 = (T-3)(T+3)$, and the composition of injective operators is again injective. So at least one of $T-3$ and $T+3$ must not be injective, so at least one of $3$ and $-3$ must be an eigenvalue of $T$.
  
  If $3$ is an eigenvalue of $T$, let $v \neq 0$ be a corresponding eigenvector. Then $T^2 v = T T v = T 3 v = 3 T v = 3^2 v = 9 v$, and so $9$ is an eigenvalue of $T^2$. Similarly, if $-3$ is an eigenvalue of $T$, let $v \neq 0$ be a corresponding eigenvector, and compute $T^2 v= (-3)^2 v = 9v$.
  
  \item {\bf Let $V$ be a vector space and $T \in \cL(V)$, and suppose that $u,v\in V$ are eigenvectors of $T$ such that $u+v$ is also an eigenvector. Prove that the eigenvalues of $u$ and $v$ are equal.}
  
  Suppose $\lambda, \mu$ are the eigenvalues of $u,v$ respectively. If $\lambda \neq \mu$, then $u$ and $v$ are linearly independent. Compute
  $$ T(u+v) = Tu + Tv = \lambda u + \mu v.$$
  Since $u,v$ are linearly independent, if $\lambda u + \mu v = \alpha u + \beta v$, then $\lambda = \alpha$ and $\mu = \beta$. In particular, since $\lambda \neq \mu$, we cannot have an equality between $\lambda u + \mu v$ and $\alpha(u+v) = \alpha u + \alpha v$ for any $\alpha$.
  
  In other words, if $u,v$ have different eigenvalues, then $u+v$ cannot be an eigenvector.
  
  \item {\bf Let $V$ be a vector space and $S,T \in \cL(V)$ two operators on $V$ such that $ST = TS$. Show that $\ker(S)$ is invariant under $T$.}
  
  If $u \in \ker(S)$, then, using commutativity, we see that $STu = TSu = T0 = 0$, and so $Tu \in \ker(S)$.
  
  \item {\bf Let $V$ be an $n$-dimensional vector space and $S,T \in \cL(V)$ two operators on $V$ such that $ST = TS$. Suppose that $S$ has $n$ distinct eigenvalues. Show that $T$ is diagonalizable.
  
  \textbf{Hint:} Show that any eigenbasis of $S$ is also an eigenbasis of $T$.}
  
  Let $\lambda_1,\dots,\lambda_n$ be the eigenvalues of $S$, and $v_1,\dots,v_n$ some choice of corresponding eigenvectors (of $S$). We proved in class (and in the book) that $\{v_1,\dots,v_n\}$ are linearly independent and hence a basis (since we are in an $n$-dimensional vector space). We claim that this basis is an eigenbasis for $T$. For this, it suffices to show that each $v_i$ is an eigenvector for $T$.
  
  But, using commutativity, $STv_i = TSv_i = T\lambda_i v_i = \lambda_i Tv_i$. In other words, $\ker(S - \lambda_i)$, the $\lambda_i$th eigenspace  of $S$, is invariant under $T$. Since $S$ has $n$ distinct eigenvalues, its nontrivial eigenspaces are all one-dimensional. Thus $\ker(S - \lambda_i) = \operatorname{span}(v_i)$, and so $Tv_i = \mu_i v_i$ for some scalar $\mu_i$. This is what we wanted to prove.
  
  \item {\bf Suppose that $c_1,\dots,c_n \in \bR$ are distinct real numbers. Prove that the functions $e^{c_1 x},\dots,e^{c_n x}$ are linearly independent in the vector space $\bR^\bR$.
  
  \textbf{Hint:} Let $V := \operatorname{span}(e^{c_1 x},\dots,e^{c_n x})$ and define an operator $T \in \cL(V)$ by $T[f] = f'$, or in other words $T = \frac{\mathrm d}{\mathrm d x}$. What are its eigenvalues and eigenvectors?}
  
  Since sums and products of smooth functions are smooth, and since exponential functions are smooth, every function in $V$ is smooth. Moreover, $T[e^{c_ix}] = c_i e^{c_i x} \in V$. So $T$ is a well-defined linear operator $V \to V$. But $e^{c_i x}$ is an eigenvector under $T$ with eigenvalue $c_i$. Since the $c_i$'s are distinct, the $e^{c_ix}$'s must be linearly independent.
  
  \textbf{Remark:} One can also show the linear independence as follows. Order the numbers $c_1,\dots,c_n$ in increasing order $c_1 < c_2 < \dots < c_n$. Now suppose that we can find $\alpha_1,\dots,\alpha_n$ such that $0 = \alpha_1 e^{c_1 x} + \dots + \alpha_n e^{c_n x}$ for all $x$. Suppose that $\alpha_n \neq 0$. Let $x > \max_{i=1,\dots,n-1}\bigl(\frac1{c_n - c_i} \log(\frac{(n-1)|\alpha_i|}{|\alpha_n|})\bigr)$. Then $|\alpha_n e^{c_n x}| > |\alpha_1 e^{c_1 x}| + \dots + |\alpha_n e^{c_n x}| > |\alpha_1 e^{c_1 x} + \dots + \alpha_n e^{c_n x}|$, violating the functional equation $0 = \alpha_1 e^{c_1 x} + \dots + \alpha_n e^{c_n x}$. So $\alpha_n = 0$. Now induct: use the same argument to show that $\alpha_{n-1} = 0$, and so on.
  
  The advantage of the first proof is that it works with $\bR$ replaced by $\bC$.
 \end{enumerate}



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