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\title{Math 2135: Linear Algebra}
\author{Assignment 6}
\date{Solutions}

\begin{document}
\maketitle

\subsection*{Primary questions}
\begin{enumerate}
  \item {\bf Give an example of $T \in \cL(\bR^2)$ such that $T^4 + 1 = 0$.}
  
  For example, rotation by $45^\circ$.
  
  \item {\bf Is the matrix
  $$ \begin{pmatrix} 1 & 2 & 3 \\ 0 & 1 & 4 \\ 0 & 0 & 5 \end{pmatrix} $$
  diagonalizable? Justify your answer.}
  
  This operator is not diagonalizable. To see this, note that its only eigenvalues are $1$ and $5$, and the corresponding eigenspaces are both $1$-dimensional.
  
  \item \begin{enumerate}
  \item {\bf Suppose that $R,T \in \cL(\bF^3)$ each have $2,6,7$ as eigenvalues. Prove that there exists an invertible operator $S \in \cL(\bF^3)$ such that $R = STS^{-1}$.}
  
  We can find eigenbases for both of these matrices, since they have all-distinct eigenvalues. So both are conjugate to the diagonal matrix with diagonal entries $2,6,7$, and hence they are conjugate to each other.
  
  \item {\bf Find a pair of operators $R,T \in \cL(\bF^4)$ with the following properties: each of them have $2,6,7$ as eigenvalues and neither has any other eigenvalues; there does not exist an an invertible operator $S \in \cL(\bF^4)$ such that $R = STS^{-1}$.}
  
  For example, the diagonal matrices
  $$ \begin{pmatrix} 2 \\ & 2 \\ && 6 \\ &&& 7\end{pmatrix}, \begin{pmatrix} 2 \\ & 6 \\ && 6 \\ &&&7 \end{pmatrix}.$$
  They cannot be conjugate, since their eigenspaces are of different dimensions.
  \end{enumerate}
  \item {\bf Suppose that $T \in \cL(\bF^5)$ and that the eigenspace $E(8,T)$ is 4-dimensional. Show that at least one of $T-2I$ and $T-6I$ is invertible.}
  
  The sum of eigenspaces is always direct. Thus the eigenspaces $E(2,T)$ and $E(6,T)$ cannot both be at-least-one-dimensional since $4+1+1 > 5$.
    
  \item {\bf Suppose $T \in \cL(V)$. Prove that $T/\im(T) = 0$.} \label{Tmodim}
  
  By definition, $T/\im(T)$ is the operator in $V/\im(T) = \{v + \im(T): v\in V\}$ that takes $v + \im(T)$ to $Tv + \im(T)$. But $Tv + \im(T) = \im(T)$, since $Tv \in \im(T)$, and $\im(T)  = 0+\im(T) \in V/\im(T)$ is the zero vector. So $T/\im(T)$ is the operator on $V/\im(T)$ that takes everything to $0$, and so it is the zero operator.
  
  \item {\bf Suppose that $V$ is a finite-dimensional vector space, $T \in \cL(V)$, and $U \subset V$ is a $T$-invariant vector subspace. Show that each eigenvalue of $T/U$ is also an eigenvalue of $T$.} \label{lowerblock} 
  \setcounter{rolling}{\value{enumi}}
  
  Let $\lambda$ be an eigenvalue of $T/U$. Note that $(T/U)-\lambda = (T-\lambda)/U$; $\lambda$ being an eigenvalue of $T/U$ is the statement that this operator is not injective.  
  Since $V/U$ is finite-dimensional (its dimension is at most $\dim V$), non-injectivity of an operator on $V/U$ is equivalent to non-surjectivity. So let $v + U \in V/U$ be an element which is not in the image of $(T-\lambda)/U$. In other words, there does not exist a $w \in V$ such that $(T-\lambda)w + U = v+U$. Then certainly there is no $w$ solving $(T-\lambda)w = v$. So $v$ is not in the image of $T-\lambda$, and so $T-\lambda$ is not surjective. Since $V$ is finite-dimensional, non-surjectivity of an operator on $V$ is equivalent to non-injectivity. So $T-\lambda$ is not injective, and $\lambda$ is an eigenvalue of $T$.  
\end{enumerate}

\subsection*{Bonus questions}
\begin{enumerate}\setcounter{enumi}{\value{rolling}}
  \item {\bf Show that the finite-dimensionality is necessary in exercise \ref{lowerblock}. Specifically, find a (necessarily infinite-dimensional) vector space $V$ and an operator $T \in \cL(V)$ such that $T$ has no eigenvalues at all, but such that there is an invariant subspace $U$ for which $T/U$ has an eigenvalue. \label{bonus}}
  
  For example, let $V$ be the vector space $\bF[x]$ of polynomials, and $T = \hat{x}$ the operator which multiplies a polynomial by $x$. Then $T$ cannot have any eigenvalues, because if $f(x) \in \bF[x]$ is nonzero, then the degree of $Tf$ is one more than the degree of $\lambda f$ (and so they cannot be equal). 
  
  Let $U = \im(T)$. Then $1 \not\in U$, and so $V/U$ is not the zero vector space. By exercise \ref{Tmodim}, $T/U = 0$, which has $0$ as an eigenvalue.
    
  \item {\bf In exercises \ref{lowerblock} and \ref{bonus}, what would happen if you used generalized eigenvalues rather than eigenvalues?}
  
  The statement still fails. Let $V = \bR^\bR$ be the space of all functions, and $T$ the operator that shifts $f(x) \mapsto f(x-1)$. Then $T$ is invertible, and so $0$ is not a generalized eigenvalue of $T$. 
  
  Let $U$ be the subspace of functions which vanish on negative values of $x$. In other words
  $$ U = \{f \in \bR^\bR \st f(x) = 0 \forall x < 0\}.$$
  Then we claim that $U$ is $T$-invariant. Indeed, suppose $f \in U$ and $x < 0$. Then $x-1 < 0$ so $f(x-1) = 0$. So the function $Tf$, defined by $(Tf)(x) = f(x-1)$, is in $U$.
  
  Note that $T|_U$ is not invertible. Indeed, its image consists of the functions which vanish on $x<1$.
  
  What is the quotient space $V/U$? We can choose a splitting $V = U \oplus W$ where $W = \{f \in \bR^\bR \st f(x) = 0 \forall x \geq 0\}$. This splitting selects an isomorphism $V/U \cong W$. Under this isomorphism, the quotient operator $T/U$ sends a function $f$ to the function $Tf$ defined by
  $$ (Tf)(x) = \begin{cases} f(x-1), & x < 0 \\ 0, & x \geq 0 \end{cases} $$
  This operator is not invertible: its kernel consists of the functions supported in the interval $(0,1]$. 
  
  Thus $0$ is a generalized eigenvalue of $T/U$ (indeed, an actual eigenvalue). But $0$ was not an eigenvalue of $T$.  
\end{enumerate}



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