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TEX program = pdflatex + +\documentclass[solution]{seu-ml-assign} + +\title{Assignment} +\author{Teddy van Jerry} +\studentID{61520522} +\instructor{TeX - LaTeX Stack Exchange} +\date{\today} +\duedate{20:00 March 21, 2022} +\assignno{1} +\semester{SEU --- 2022 Spring} + +\begin{document} + +\maketitle + +% \startsolution[print] + +\problem{Basic Vector Operations} +\subproblem{} +$\|\mathbf{a}\|_2=\sqrt{1^2+2^2+3^2}=\sqrt{14},\quad \|\mathbf{b}\|_2=\sqrt{(-8)^2+1^2+2^2}=\sqrt{69}$. + +\subproblem{} +$\|\mathbf{a}-\mathbf{b}\|_2=\sqrt{9^2+1^2+1^2}=\sqrt{83}$. + +\subproblem{} +$\mathbf{a}$ and $\mathbf{b}$ are orthogonal. +\begin{proof} +The inner product of vectors $\mathbf{a}$ and $\mathbf{b}$ is +\begin{equation} +\langle\mathbf{a},\mathbf{b}\rangle=\mathbf{a}^T\mathbf{b}=1\times(-8)+2\times 1+3\times 2=0, +\end{equation} +therefore $\mathbf{a}$ and $\mathbf{b}$ are orthogonal. +\end{proof} + +\problem{Basic Matrix Operations} +According to the consensus, the matrix notation should be the bold upper-case letter like $\mathbf{A}$ or $\bm{A}$, not $A$. + +\subproblem{} +\begin{equation} + \begin{aligned} + [\mathbf{A}, \mathbf{I}_3]&= + \begin{bNiceArray}{rrr:rrr} + 1&-3&3&1&0&0\\3&-5&3&0&1&0\\6&-6&4&0&0&1 + \end{bNiceArray}\sim + \begin{bNiceArray}{rrr:rrr} + 1&-3&3&1&0&0\\0&4&-6&-3&1&0\\0&12&-14&-6&0&1 + \end{bNiceArray}\sim + \begin{bNiceArray}{rrr:rrr} + 1&-3&3&1&0&0\\0&4&-6&-3&1&0\\0&0&4&3&-3&1 + \end{bNiceArray}\\ + &\sim + \begin{bNiceArray}{rrr:rrr} + 1&-3&0&-\frac{5}{4}&\frac{9}{4}&\frac{3}{4}\\[0.3em]0&4&0&\frac{3}{2}&-\frac{7}{2}&-\frac{3}{2}\\[0.3em]0&0&1&\frac{3}{4}&-\frac{3}{4}&\frac{1}{4} + \end{bNiceArray}\sim + \begin{bNiceArray}{rrr:rrr} + 1&0&0&-\frac{1}{8}&-\frac{3}{8}&\frac{3}{8}\\[0.3em]0&1&0&\frac{3}{8}&-\frac{7}{8}&\frac{3}{8}\\[0.3em]0&0&1&\frac{3}{4}&-\frac{3}{4}&\frac{1}{4} + \end{bNiceArray}, + \end{aligned} +\end{equation} +where $\mathbf{I}_3$ is the $3\times 3$ identity matrix. +Therefore we have +\begin{equation}\label{eq:2-1-inv} + \mathbf{A}^{-1}= + \begin{bNiceArray}{rrr} + -\frac{1}{8}&-\frac{3}{8}&\frac{3}{8}\\[0.3em]\frac{3}{8}&-\frac{7}{8}&\frac{3}{8}\\[0.3em]\frac{3}{4}&-\frac{3}{4}&\frac{1}{4} + \end{bNiceArray}. +\end{equation} +The determinant of matrix $\mathbf{A}$ can be calculated as +\begin{equation}\label{eq:2-1-det} + \mathrm{det}(\mathbf{A})=1\times\begin{vNiceArray}{rr} + -5&3\\-6&4 + \end{vNiceArray}-(-3)\times\begin{vNiceArray}{rr} + 3&3\\6&4 + \end{vNiceArray}+3\times\begin{vNiceArray}{rr} + 3&-5\\6&-6 + \end{vNiceArray}=1\times(-2)+3\times(-6)+3\times 12=16, +\end{equation} +where $|\cdot|$ denotes the determinant. + +\subproblem{} +The rank of matrix $\mathbf{A}$ is $3$ because as is shown in Eq.~\eqref{eq:2-1-inv} the matrix $\mathbf{A}$ is invertible. + +\subproblem{} +The trace of matrix $\mathbf{A}$ is +\begin{equation} + \mathrm{tr}(\mathbf{A})=\sum_{i=1}^{3}a_{ii}=1+(-5)+4=0. +\end{equation} + +\begin{equation} + \mathbf{A}+\mathbf{A}^{T}=\begin{bNiceArray}{rrr} + 1&-3&3\\3&-5&3\\6&-6&4 + \end{bNiceArray}+ + \begin{bNiceArray}{rrr} + 1&3&6\\-3&-5&-6\\3&3&4 + \end{bNiceArray}= + \begin{bNiceArray}{rrr} + 2&0&9\\0&-10&-3\\9&-3&8 + \end{bNiceArray}. +\end{equation} + +\subproblem{} +\begin{equation} + \mathbf{A}+\mathbf{A}^{T}=\begin{bNiceArray}{rrr} + 1&-3&3\\3&-5&3\\6&-6&4 + \end{bNiceArray}+ + \begin{bNiceArray}{rrr} + 1&3&6\\-3&-5&-6\\3&3&4 + \end{bNiceArray}= + \begin{bNiceArray}{rrr} + 2&0&9\\0&-10&-3\\9&-3&8 + \end{bNiceArray}. +\end{equation} + +\subproblem{} +$\mathbf{A}$ is not an orthogonal matrix. +\begin{proof} + Assume $\mathbf{A}$ is an orthogonal matrix, + therefore + \begin{equation} + \mathbf{AA}^{T}=\mathbf{I}_3, + \end{equation} + Take the determinant at both side, it can be derived that + \begin{equation} + |\mathrm{det}(\mathbf{A})|=\sqrt{|\mathbf{A}||\mathbf{A}^T|}=|\mathrm{det}(\mathbf{I}_3)|=1, + \end{equation} + which contradicts with Eq.~\eqref{eq:2-1-det}. + Therefore, the assumption is false. +\end{proof} + +\subproblem{} +Let $f(\lambda)$ be the characteristic function of matrix $\mathbf{A}$ and +\begin{equation}\label{eq:2-6-f} + f(\lambda)=\begin{vNiceArray}{ccc} + \lambda-1&3&-3\\-3&\lambda+5&-3\\-6&6&\lambda-4 + \end{vNiceArray}=(\lambda-4)(\lambda+2)^2, +\end{equation} +therefore the eigenvalues are $\lambda_1=4, \lambda_2=\lambda_3=-2$. +Let the corresponding eigenvectors be $\bm{\alpha}_i$, $i=1,2,3$. +\begin{equation} + (\mathbf{A}-\lambda_i\mathbf{I}_3)\bm{\alpha}_i=\mathbf{0},\quad i=1,2,3, +\end{equation} +and the corresponding eigenvectors are +\begin{equation} + \bm{\alpha}_1=\begin{bNiceArray}{ccc}1&1&2\end{bNiceArray}^T,\quad + \bm{\alpha}_{2,3}=\begin{bNiceArray}{ccc}1&1+c_{2,3}&c_{2,3}\end{bNiceArray}^T, +\end{equation} +where $c_{2,3}\in\mathbb{R}$. +Without loss of generality, we take $c_2=0$ and $c_3=-1$, and we have $\bm{\alpha}_2=\begin{bNiceArray}{ccc}1&1&0\end{bNiceArray}^T$ and $\bm{\alpha}_2=\begin{bNiceArray}{ccc}1&0&-1\end{bNiceArray}^T$. + +\subproblem{} +Use the result from Eq.~\eqref{eq:2-6-f}, the matrix $\mathbf{A}$ can be diagonalized as +\begin{equation} + \bm{\Lambda}=\begin{bNiceArray}{rrr} + 4&0&0\\0&-2&0\\0&0&-2 + \end{bNiceArray}. +\end{equation} + +\subproblem{} +The $\ell_{2,1}$ norm of $\mathbf{A}$ is +\begin{equation} + \|\mathbf{A}\|_{2,1}=\sum_{i=1}^3\sqrt{\sum_{j=1}^3a_{ij}^2}=\sqrt{46}+\sqrt{70}+\sqrt{34}\approx 20.98, +\end{equation} +and the Frobenius norm of $\mathbf{A}$ is +\begin{equation} + \|\mathbf{A}\|_F=\sqrt{\sum_{i,j=1,\mathrlap{2,3}}a_{ij}^2}=\sqrt{150}=5\sqrt{6}\approx 12.247. +\end{equation} + +\subproblem{} +The nuclear norm of $\mathbf{A}$ is +\begin{equation} + \|\mathbf{A}\|_*=\mathrm{tr}(\sqrt{\mathbf{A}\mathbf{A^*}})=\sum_{i=1}^3\sigma_i(\mathbf{A})\approx 14.728, +\end{equation} +and the spectral norm of $\mathbf{A}$ is +\begin{equation} + \|\mathbf{A}\|_2=\max\sigma_i(\mathbf{A})\approx 12.065. +\end{equation} + +\vspace{2mm} +\begin{lstlisting}[language=Matlab, title={MATLAB Code for Check}] +A = [1, -3, 3; 3, -5, 3; 6, -6, 4]; % define the matrix A +inv(A) % calculate and print the inverse of A +det(A) % the determinant of A +rank(A) % the rank of A +trace(A) % the trace of A +A + A.' % the sum of A and the transpose of A +sum(sum(A * A.' ~= eye(3))) % check if A is orthogonal +[X, D] = eig(A) % the eigenvectors and the corresponding eigenvalues of A +sum(sqrt(sum(A .^ 2))) % l-2,1 norm of A +norm(A, 'fro') % Frobenius norm of A +sum(svd(A)) % nuclear norm of A +max(svd(A)) % spectral norm of A +\end{lstlisting} + +\problem{Linear Equations} +\subproblem{} +It is evident to solve the linear equation +\begin{equation}\label{eq:3-1} + \left\{ + \begin{aligned} + x_1&=-1, \\ + x_2&=0, \\ + x_3&=1. + \end{aligned} + \right. +\end{equation} + +\subproblem{} +Let +\begin{equation} + \mathbf{A}=\begin{bNiceArray}{rrr} + 2&2&3\\1&-1&0\\-1&2&1 + \end{bNiceArray},\quad + \mathbf{b}= + \begin{bNiceArray}{r} + 1\\-1\\2 + \end{bNiceArray}, +\end{equation} +and we have $\mathbf{Ax}=\mathbf{b}$ as +\begin{equation} + \begin{bNiceArray}{rrr} + 2&2&3\\1&-1&0\\-1&2&1 + \end{bNiceArray} + \begin{bNiceArray}{r} + x_1\\x_2\\x_3 + \end{bNiceArray}= + \begin{bNiceArray}{r} + 1\\-1\\2 + \end{bNiceArray}. +\end{equation} + +\subproblem{} +Since there is a unique solution shown in Eq.~\eqref{eq:3-1}, we know +\begin{equation}\label{eq:3-3} + \mathrm{rank}(\mathbf{A})=3. +\end{equation} + +\subproblem{} +\begin{equation} + \begin{aligned} + [\mathbf{A}, \mathbf{I}_3]&= + \begin{bNiceArray}{rrr:rrr} + 2&2&3&1&0&0\\1&-1&0&0&1&0\\-1&2&1&0&0&1 + \end{bNiceArray}\sim + \begin{bNiceArray}{rrr:rrr} + 2&2&3&1&0&0\\1&-1&0&0&1&0\\0&1&1&0&1&1 + \end{bNiceArray}\sim + \begin{bNiceArray}{rrr:rrr} + 1&1&\frac{3}{2}&\frac{1}{2}&0&0\\1&-1&0&0&1&0\\0&1&1&0&1&1 + \end{bNiceArray}\\ + &\sim + \begin{bNiceArray}{rrr:rrr} + 1&1&\frac{3}{2}&\frac{1}{2}&0&0\\[0.3em]0&-2&-\frac{3}{2}&-\frac{1}{2}&1&0\\[0.3em]0&1&1&0&1&1 + \end{bNiceArray}\sim + \begin{bNiceArray}{rrr:rrr} + 1&1&\frac{3}{2}&\frac{1}{2}&0&0\\[0.3em]0&-1&-\frac{3}{4}&-\frac{1}{4}&\frac{1}{2}&0\\[0.3em]0&0&\frac{1}{4}&-\frac{1}{4}&\frac{3}{2}&1 + \end{bNiceArray}\sim + \begin{bNiceArray}{rrr:rrr} + 1&1&0&2&-9&-6\\[0.3em]0&-1&0&-1&5&3\\[0.3em]0&0&1&-1&6&4 + \end{bNiceArray}\\ + &\sim + \begin{bNiceArray}{rrr:rrr} + 1&0&0&1&-4&-3\\0&1&0&1&-5&-3\\0&0&1&-1&6&4 + \end{bNiceArray}, + \end{aligned} +\end{equation} +therefore the inverse of $\mathbf{A}$ is +\begin{equation}\label{eq:3-4-inv} + \mathbf{A}^{-1}=\begin{bNiceArray}{rrr} + 1&-4&-3\\1&-5&-3\\-1&6&4 + \end{bNiceArray}. +\end{equation} +The determinant of $\mathbf{A}$ can be calculated as +\begin{equation} + \mathrm{det}(\mathbf{A})=2\times\begin{vNiceArray}{rr} + -1&0\\2&1 + \end{vNiceArray}-2\times\begin{vNiceArray}{rr} + 1&0\\-1&1 + \end{vNiceArray}+3\times\begin{vNiceArray}{rr} + 1&-1\\-1&2 + \end{vNiceArray}=2\times(-1)-2\times 1+3\times 1=-1. +\end{equation} + +\subproblem{} +As is shown in Eq.~\eqref{eq:3-3}, $\mathbf{A}$ is invertible and with the result in Eq.~\eqref{eq:3-4-inv} +\begin{equation} + \mathbf{x}=\mathbf{A}^{-1}\mathbf{b}= + \begin{bNiceArray}{rrr} + 1&-4&-3\\1&-5&-3\\-1&6&4 + \end{bNiceArray} + \begin{bNiceArray}{r} + 1\\-1\\2 + \end{bNiceArray}= + \begin{bNiceArray}{r} + -1\\0\\1 + \end{bNiceArray}, +\end{equation} +and it is exactly the same result with Eq.~\eqref{eq:3-1}. + +\subproblem{} +The inner product +\begin{equation} + \langle\mathbf{x},\mathbf{b}\rangle=\mathbf{x}^T\mathbf{b}=1\times 1+0\times(-1)+1\times 2=1, +\end{equation} +and the outer product is +\begin{equation} + \mathbf{x}\otimes\mathbf{b}=\mathbf{x}\mathbf{b}^T= + \begin{bNiceArray}{r} + -1\\0\\1 + \end{bNiceArray} + \begin{bNiceArray}{rrr} + 1&-1&2 + \end{bNiceArray}= + \begin{bNiceArray}{rrr} + -1&1&-2\\0&0&0\\1&-1&2 + \end{bNiceArray}. +\end{equation} + +\subproblem{} +$\|\mathbf{b}\|_1=|1|+|-1|+|2|=4,\quad\|\mathbf{b}\|_2=\sqrt{1^2+(-1)^2+2^2}=\sqrt{6},\quad\|\mathbf{b}\|_{\infty}=\max\{|1|,|-1|,|2|\}=2.$ + +\subproblem{} +Let $\mathbf{y}=\begin{bNiceArray}{rrr} + y_1&y_2&y_3 +\end{bNiceArray}^T$, +we have +\begin{equation} + \mathbf{y}^T\mathbf{Ay}= + \begin{bNiceArray}{rrr} + y_1&y_2&y_3 + \end{bNiceArray} + \begin{bNiceArray}{rrr} + 2&2&3\\1&-1&0\\-1&2&1 + \end{bNiceArray} + \begin{bNiceArray}{r} + y_1\\y_2\\y_3 + \end{bNiceArray}=2y_1^2-y_2^2+y_3^2+3y_1y_2+2y_2y_3+2y_1y_3, +\end{equation} +and +\begin{equation} + \bigtriangledown_{\mathbf{y}}\mathbf{y}^T\mathbf{Ay}= + \begin{bNiceArray}{r} + \frac{\partial}{\partial y_1}\mathbf{y}^T\mathbf{Ay} \\[.3em] + \frac{\partial}{\partial y_2}\mathbf{y}^T\mathbf{Ay} \\[.3em] + \frac{\partial}{\partial y_3}\mathbf{y}^T\mathbf{Ay} + \end{bNiceArray}= + \begin{bNiceArray}{r} + 4y_1+3y_2+2y_3 \\ + 3y_1-2y_2+2y_3 \\ + 2y_1+2y_2+2y_3 + \end{bNiceArray}. +\end{equation} + +\subproblem{} +The equation $\mathbf{A}_1\mathbf{x}=\mathbf{b}_1$ can be represented as +\begin{equation} + \begin{bNiceArray}{rrr} + 2&2&3\\1&-1&0\\-1&2&1\\-1&2&1 + \end{bNiceArray} + \begin{bNiceArray}{r} + x_1\\x_2\\x_3 + \end{bNiceArray}= + \begin{bNiceArray}{r} + 1\\-1\\2\\2 + \end{bNiceArray}. +\end{equation} + +\subproblem{} +$\mathrm{rank}(\mathbf{A}_1)=3$. +\begin{proof} + On one hand, $\mathrm{rank}(\mathbf{A}_1)\geq\mathrm{rank}(\mathbf{A})=3$ which is shown in Eq.~\eqref{eq:3-3}. + On the other hand, $\mathrm{rank}(\mathbf{A}_1)\leq\min\{3,4\}=3$. + Therefore, $\mathrm{rank}(\mathbf{A}_1)=3$. + We can also find the first three equations are linearly independent while the last equation is actually the same with the third equation which makes it meaningless. +\end{proof} + +\subproblem{} +Yes. +\begin{proof} + Since $\mathrm{rank}(\mathbf{A}_1)=\|\mathbf{x}\|_0$, i.e. rank of $\mathbf{A}_1$ is equal to the dimension of $\mathbf{x}$, the formula can be solved with a unique solution the same as Eq.~\eqref{eq:3-1}. +\end{proof} + +\end{document} |