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\noindent {\sc Linear Algebra II
			\hfill Midterm 1  \vspace{0.3cm} \\
			Name:\underline{\hspace*{1.1in}} Id No.:\underline{\hspace*{1.9in}}  Class: 	\underline{\hspace*{0.9in}}	} \\


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{\bf Problem 1:}
Let $A=\begin{pmatrix} 4 & 3 \\ 5 & 6 \end{pmatrix} \in M_{2 \times 2}(\mathbb{R})$,
\begin{enumerate}
\item (4 points) Determine all the eigenvalues and the corresponding eigenvectors of $A$.
\begin{proof}[solution]
$\text{Eigenvalue}: \lambda_1=1, \text{Eigenvector}: v_1=t\begin{pmatrix} 1 \\ -1 \end{pmatrix}, t\not=0, t \in \mathbb{R}.$\\
$\text{Eigenvalue}: \lambda_2=9, \text{Eigenvector}:v_2=t\begin{pmatrix} 3 \\ 5 \end{pmatrix}, t\not=0, t \in \mathbb{R}.$
\end{proof}
\item (2 points) Determine an invertible matrix $Q$ and a diagonal matrix $D$ such that $Q^{-1}AQ=D.$
\begin{proof}[Solution]
$Q=\begin{pmatrix} 1 & 3 \\ -1 & 5 \end{pmatrix}.$
\end{proof}
\item (4 points) Use (1) and (2) to compute $e^A.$
\begin{proof}
$\begin{pmatrix} \frac{5e+3e^9}{8} &  \frac{-3e+3e^9}{8} \\  \frac{-5e+5e^9}{8} &  \frac{3e+5e^9}{8}\end{pmatrix}.$
\end{proof}
\end{enumerate}

{\bf Problem 2:}  
\begin{enumerate}
\item (2 points) Let $T \in \mathcal{L}(V)$ and $\dim (V) < \infty$. Let $W$ be the $T$-cyclic subspace of $V$ generated by a vector $v \in V\backslash \left\{ 0 \right\}$, and $\dim (W) = 3.$ Suppose that $-4I(v)+3T(v)-2T^2(v)+T^3(v)$ is a zero vector of $V$. Find the characteristic polynomial $P_{T_W}(t)$ of $T_W$.
\begin{proof}[Solution]
$-t^3+2t^2-3t+4.$
\end{proof}
\item (4 points) Let $T$ be the linear operator on $M_{2 \times 2}(\mathbb{R})$ such that $T(A)=A^t$. Find the characteristic polynomial $P_T(t)$ of $T$. 
\begin{proof}[Solution]
$(t-1)^3(t+1).$
\end{proof}
\end{enumerate}

{\bf Problem 3:} Let $T:P_2(\mathbb{R}) \to P_2(\mathbb{R})$ be the linear operator defined by $T(f(x))=f(x)+(1+x)f'(x)$, where $P_2(\mathbb{R})$ is the set of all polynomials with real coefficients with degree at most $2$ and $f'(x)$ is the derivative of $f(x)$. 
\begin{enumerate}
\item (3 points) Find all the eigenvalues of the operator $T$.
\begin{proof}[solution]
$1,2,3.$
\end{proof}
\item (2 points) Find all the eigenvalues of the operator $T^5+2T^3+5T.$
\begin{proof}[Solution]
$1^5+2\cdot1^3+5\cdot 1, 2^5+2\cdot 2^3+5\cdot 2, 3^5+2\cdot 3^3+5\cdot 3.$
\end{proof}
\item (3 points) Find a basis $\beta$ for $P_2(\mathbb{R})$ such that $[T]_\beta$ is a diagonal matrix.
\begin{proof}[Solution]
$\beta=\left\{ 1,1+x,1+2x+x^2 \right\}.$
\end{proof}
\end{enumerate}

{\bf Problem 4:}
(5 points) Let $T:P_3(\mathbb{R}) \to P_3(\mathbb{R})$ be the linear operator defined by $T(f(x))=f'(x)+f''(x).$ Test $T$ for diagonalizability.
\begin{proof}[Solution]
Not diagonalizable.
\end{proof}
{\bf Problem 5:} 
(4 points) Let $T$ be a linear operator on an inner product space $V$, and suppose that $\Vert T(x)\Vert = \Vert x \Vert$ for all $x$. Prove that $T$ is one-to-one.
\begin{proof}
Let $x \in V$ be an arbitrary vector. Then $T(x)=0 \Rightarrow \Vert T(x) \Vert = \Vert x \Vert =0 \Rightarrow x=0.$ Hence $T$ is one-to-one.
\end{proof}

{\bf Problem 6:}
(4 points) Let $V=C([0,1])$, and define
\[
<f,g>=\int_0^{3/4}f(t)g(t)dt.
\]
Is this an inner product on $V.$
\begin{proof}[Solution]
Let $f(x)=0, \, \text{if} \, x\le 3/4$ and $f(x)=x-3/4 \, \text{if} \, x>3/4$. Then $<f,f>=0$, but $f\not=0.$ Hence it is not an inner product on $V$.
\end{proof}

{\bf Problem 7:}
%\begin{enumerate}
%\item (5 points) Suppose that $A \in M_{n \times n}(F)$ has two distinct eigenvalues, $\lambda_1$ and $\lambda_2$, and that $\dim(E_{\lambda_1})=n-1$. Prove that $A$ is diagonalizable.
%\item 
(5 points) Prove that similar matrices have the same characteristic polynomial.
\begin{proof}
Assume that the $n \times n$ matrix $A$ is similar to the $n \times n$ matrix $B$, then there exists an invertible $n \times n$ matrix $Q$ such that $B=Q^{-1}AQ$. 
Now 
\begin{eqnarray}
\det(B-\lambda I) & = & \det(Q^{-1}AQ-\lambda I) \nonumber \\ & = & \det(Q^{-1}(A-\lambda I)Q) \nonumber \\ & = & \det(Q^{-1})\det(A-\lambda I)\det(Q) \nonumber \\ & = & \det(A-\lambda I). \nonumber
\end{eqnarray}
Hence the similar matrices $A$ and $B$ have the same characteristic polynomial.
\end{proof}
% \end{enumerate}
 
 {\bf Problem 8:}
Show that if a matrix $A$ is diagonalizable, then 
 \begin{enumerate}
 \item (4 points) the determinant of $A$, $\det (A)$, is the product of its eigenvalues (counting with multiplicities).
 \begin{proof}
 Since $A$ is diagonalizable, $A$ is similar to a diagonal matrix whose diagonal entries are eigenvalues. Since similar matrices have the same determinant. The determinant of $A$, $\det (A)$, is equal to the product of its eigenvalues (counting with multiplicities).
 \end{proof}
 \item (4 points) the trace of $A$, $\operatorname{tr}(A)$, is the sum of its eigenvalues (counting with multiplicities).
 \begin{proof}
  Since $A$ is diagonalizable, $A$ is similar to a diagonal matrix whose diagonal entries are eigenvalues. Since similar matrices have the same trace. The trace of $A$, $\operatorname{tr} (A)$, is equal to the sum of its eigenvalues (counting with multiplicities).
 \end{proof}
 \end{enumerate}
\end{document}