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Linear Algebra Final Exam TeX

先生が最後のテスト、返してくれなかったので自分で作り直しました。TeXです。

PDFは画像として最後に置いときます。

問題(TeX), 答え(解説なし), 問題と答えのpdfを画像化したもの、の順です。

答えは極めて雑。

 

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%Yuma Nakai
%Kyoto University, Civil Engineering ICP
%January 30, 2018
%Because prpfessor returned back our final exams during feedback class, I remade it by myself.
%You should use this TeX file and pdf file only in private.

\documentclass{article}
\usepackage{amssymb,amsmath}
\title{\textbf{Linear Algebra B Final Exam \\with Solutions Fall 2017 }}
\author{\textit{Professor ********** ***} \\ \textit{Yuma Nakai (editor of this pdf)}}
\date{January 29,2017}

\begin{document}

\maketitle

\section*{Problem I (20 points)}
Determine whether the following statements are True or False. No explanation needed. Let \textbf{A},\textbf{B} $\in$ \textit{$M_{n\times n}$} ($\mathbb{R}$) be a real number $n \times n$ square matrices, and \emph{V} a vector space over $\mathbb{R}$.
\\ \\ \\
a) If a finite set \emph{S} spans \emph{V}, then \emph{S} is a basis of \emph{V}. \\
b) If $\beta$,$\beta^{'}$ are two different orthogonal basis for \emph{V}, then the change-of-basis matrix
\emph{$P^{\beta}_{\beta^{'}}$} is an orthogonal matrix. \\
c) If \textbf{A} is a projection matrix, then \begin{center} $\bf A^{777} = A$. \end{center}
d) If \textbf{AB} is similar to \textbf{BA}, then either \textbf{A} or \textbf{B} is invertible.\\
e) If the eigenvectors of \textbf{A} span $\mathbb{R}^{n}$, then \textbf{A} is diagonalizable.\\
f) Any nonzero vector is an eigenvector of the zero matrix \textbf{0}.\\
g) If \textbf{p}($\lambda$) is a nonzero polynomial, and \begin{center} ${\bf p}({\bf A}) = {\bf 0}$ \end{center}
then \textbf{p}($\lambda$) is the characteristic polynomial of \textbf{A}.\\
h) If \textbf{A},\textbf{B} are orthogonal, then \begin{center} ${\bf B}^{T}{\bf A}{\bf B}^{-1}$ \end{center} is also orthogonal.\\
i) If \[ W_1=\{ {\bf x}\in \mathbb{R}^{n}: {\bf Ax}={\bf 0} \},\] \[ W_2=\{ {\bf x}\in \mathbb{R}^{n}: {\bf A^{T}x}={\bf 0}\} \] then $\mathbb{R}^{n} =W_1\oplus W_2$.\\
j) If  $\{ {\bf v_1},{\bf v_2},{\bf v_3} \}$ is an orthogonal basis for \emph{V}, then multiplying ${\bf v_3}$ by a scalar $c \in \mathbb{R}$ gives a new orthogonal basis $\{ {\bf v_1},{\bf v_2},c{\bf v_3} \}$ for \emph{V}.

\newpage

\section*{Problem I\hspace{-.1em}I (20 points)}
Let ${\bf A} = \begin{pmatrix} 0 & 2 & -1\\ 2 & 3 & -2 \\ -1 & -2 & 0 \end{pmatrix}$ \\ \\ \\
a) (2 points) Explains \textbf{without calculation} why \textbf{A} can be diagnolized.\\
b) (14 points) Find an orthogonal matrix \textbf{P} and diagnol matrix \textbf{D} such that
\begin{center} ${\bf A}= {\bf PDP}^{T}$. \end{center}
c) (4 points) What is the characteristic polynomial of ${\bf A}^{2018}$. \\ \\

\section*{Problem I\hspace{-.1em}I\hspace{-.1em}I (20 points) }
a) (12 points) Construct a 3 $\times$ 3 matrix \textbf{A} such that \\
$ Nul{\bf A}=span \begin{pmatrix} 1 \\ 1\\ 1\\ \end{pmatrix},
{\bf Ae_3} = \begin{pmatrix} 1 \\ 2 \\ 3 \\ \end{pmatrix},
 Tr({\bf A}) =4,
\begin{pmatrix} 1 \\ 0\\ 1\\ \end{pmatrix}$ is eigenvector of {\bf A}. \\ \\ \\
b) (8 points) Professor says he can diagnolize \textbf{A} you found above. \\ Do you believe in him? Explain why.\\ \\

\section*{Problem I\hspace{-.1em}V (20 points)}
Let ${\bf A} = \begin{pmatrix} 0&1&1&0\\1&1&1&-1\\1&1&1&-1\\0&1&1&0 \end{pmatrix}$.\\ \\ \\
a) (4 points) Show that the characteristic polynomial of {\bf A} is $p(\lambda)=\lambda^{4}-2\lambda^{3}$. \\
b) (6 points) Find the eigenvalues and eigenspaces of \textbf{A}.\\
c) (4 points) What is the algebraic and  geometrix multiplicities of the eigenvalues? \\
d) (4 points) Using (c), find the Jordan Canonical Form of \textbf{A}.\\
e) (2 points) Using (d), find the minimal polynomial $m(\lambda)$ of \textbf{A}

\newpage
\section*{Problem V (20 points)}
Let \textbf{A} be an invertible n $\times$ n matrix. Show that
\[ det {\bf A}^{T} = det {\bf A}\]
using the \emph{QR} decomposition as follows (fill in the blanks, 1 mark each).\\ \\ \\
a) ${\bf A} =\underline{\hspace{3em}}$. \\ \\
b) ${\bf A}^{T} =\underline{\hspace{1.5 em}} \cdot \underline{\hspace{1.5 em}}$.\\ \\
c) Because \textbf{Q} is \underline{\hspace{10 em}}, det${\bf Q}^{T}$=det$\underline{\hspace{1.5 em}}$.\\ \\
d) By properties of determinant, $\underline{\hspace{3 em}} = \underline{\hspace{3 em}}$. \\ \\
e) Because det\textbf{Q} =\underline{\hspace{1.5 em}} or \underline{\hspace{1.5 em}}, \underline{\hspace{1.5 em}} = det\textbf{Q}. \\ \\
f) Therefore $\mbox{det}{\bf Q}^{T}=\mbox{det}{\bf Q}$ \\ \\
g) On the other hand, since \textbf{R} and ${\bf R}^{T}$ are both \underline{\hspace{10 em}}, and they have the same \underline{\hspace{7 em}}, we have det${\bf R}^{T}=\mbox{det}{\bf R}$. \\ \\
h) Combining, we have
\[
\begin{split}
\mbox{det}{\bf A}^{T}&=\mbox{det}\underline{\hspace{4 em}} \hspace{6 em} \mbox{by(b)}\\
&=\mbox{det}\underline{\hspace{1.5 em}}\mbox{det}\underline{\hspace{1.5 em}} \\
&=\mbox{det}\underline{\hspace{1.5 em}}\mbox{det}\underline{\hspace{1.5 em}} \\
&=\mbox{det}\underline{\hspace{1.5 em}}\mbox{det}\underline{\hspace{1.5 em}} \hspace{6 em} \mbox{by(f) and (g)}\\
&=\mbox{det}\underline{\hspace{4 em}} \\
&=\mbox{det}{\bf A}
\end{split}
\]
as required.

\end{document}

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Answer

<Problem 1>

F F T F T T F T F F

<Problem 2>

a) Because A is symmetric

b)  P=1/√2 & -1/√3 & -1/√6 \\ 0 & 1/√3 & -2/√6 \\ 1/√2 & 1/√3 & 1/√6

     D = -1&0&&0 \\ 0&-1&0 \\ 0&0&5

c) p(λ)=(λ-1)^2 *(λ-5^2018)

<Problem 3>

a) A=1&-2&1 \\ -2&0&2 \\ -1&-2&3

b) No because minimal polynomial m(λ)=λ(λ-2)^2 (←A(A-2I)≠0 so m(λ)≠λ(λ-2)  )

<Problem 4>

a) (Skip)

b) λ=0,2

λ=0...(1&0&0&1),(0&1&-1&0) ; λ=2...(1&1&1&1)

c) λ=0 : alge=3 geom=2

λ=2:alge=1 geom=1

<Problem 5>

abc) QR, R^T・Q^T,  orthogonal matrix, detQ^(-1)

def) detQ^(-1)=(detQ)^(-1), 1 or -1, (detQ)^(-1)

g) triangular matrix, diagnol

h) R^T・Q^T, R^T, Q^T, R, Q, Q, R, QR, (A)

 

 

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