Linear Assignment 1
Autor
Arielle Knittel
Letzte Aktualisierung
vor 4 Jahren
Lizenz
Creative Commons CC BY 4.0
Abstrakt
First assignment dealing with nullspace and rowspace
First assignment dealing with nullspace and rowspace
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\begin{document}
\firstpageheader{Mathematics 284}{ }{\makebox[3 in]{Name: ~Arielle Knittel}}
\noindent Latex Assignment $\#1$\\
October 7, 2019
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\runningheader{Latex Assignment $\#1$}{}{Page \thepage\ of \numpages}
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Consider the matrix given below. \[A=\begin{bmatrix}~1 & 2 & 4 & 0\\ -3~ & 1 & 5 & 2\\ -2~ & 3 & 9 & 2\end{bmatrix}.\]
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\begin{questions}
%Question #1
\question The transformation associated with $A$ maps $\RR^4 \to \RR^3$.
%Question #2
\question Describe the row space of $A$. \\ \\\textbf{\textit{{\color{coral} Solution:}}} The row space of $A$ is the subspace of $\RR^n$ spanned by its rows, or the collection of all the linear combinations of the rows of $A$. When we reduce $A$, we find that \[R=\begin{bmatrix}~1 & 0 & \frac{-6}{7} & \frac{-4}{7} \\[6pt] 0 & 1 & \frac{17}{7} & \frac{2}{7}\\ 0 & 0 & 0 & 0\end{bmatrix} \] \\Here, we can see that the last row of $R$ is all zeros, meaning the last row of $A$ is a linear combination of the first two rows of $A$. The linear combination of the first two rows of $A$ is therefore our row space, and can be expressed as \[\text{rowspace(A)}=\left \{a\begin{bmatrix} 1 & 2 & 4 & 0 \end{bmatrix}+b\begin{bmatrix} -3 & 1 & 5 & 2 \end{bmatrix} \quad \bigg| \quad a,b,c,d \in \RR \right\}\]
%Question #3
\question Describe the column space of $A$.\\ \\\textbf{\textit{{\color{coral} Solution:}}} The column space of $A$ is the subspace of the columns of $A$, or the collection of all linear combinations of the columns of $A$. We can see from $R$ that the first and second comlumns are linearly independent, therefore the first two columns of $A$ are independent and constitute the column space of $A$, or \[\text{col(A)}=\left \{a\begin{bmatrix} 1 \\ -3 \\ -2 \end{bmatrix} + b\begin{bmatrix} 2 \\ 1 \\3 \end{bmatrix} \quad \bigg| \quad a, b \in \RR \right\}\]
%Question #4
\question What is the Rank of $A$?\\ \\\textbf{\textit{{\color{coral} Solution:}}} The rank of $A$ is the dimension of the row space and column space; that is, the maximum number of independent rows or columns. As we can see from both the row and column spaces, that number is 2. Therefore, Rank(A) = 2
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%Question #5
\question What is the Nullity of $A$?\\ \\\textbf{\textit{{\color{coral} Solution:}}}The nullity of $A$ is the dimension of the null space of $A$. The number of
columns n equals the rank r plus the nullity. Thus
nullity = n - r = 4 - 2 = 2
%Question #6
\question Describe the null space of $A$\\ \\\textbf{\textit{{\color{coral} Solution:}}} The null space space of $A$ is the collection of vectors $x$ for which $Ax=0$. We can use the reduced row echelon form $R$ of the matrix $A$ to find the basis vectors for the nullspace of $A$. Call these basis vectors $n_1$ and $n_2$ and let the matrix $N$ have $n_1$ and $n_2$ as columns.The identity matrix fills in the remaining rows associated with the pivot variables. \[R=\begin{bmatrix} {\color{purple}1} & {\color{purple}0} & {\color{blue}\frac{-6}{7}} & {\color{blue}\frac{-4}{7}}\\[6pt] {\color{purple}0} & {\color{purple}1} & {\color{blue}\frac{17}{7}} & {\color{blue}\frac{2}{7}} \\ 0 & 0 & 0 & 0 \end{bmatrix}\implies N=\begin{bmatrix} {\color{purple}1} & {\color{purple}0}\\ ~{\color{purple}0} & ~{\color{purple}1} \\ ~{\color{blue}\frac{6}{7}} & {\color{blue}\frac{4}{7}} \\[6pt] ~{\color{blue}\frac{-17}{7}} & ~{\color{blue}\frac{-2}{7}} \end{bmatrix}\]
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