Linear Assignment 1
Autor:
Arielle Knittel
Last Updated:
hace 5 años
License:
Creative Commons CC BY 4.0
Resumen:
First assignment dealing with nullspace and rowspace
\begin
Discover why 18 million people worldwide trust Overleaf with their work.
First assignment dealing with nullspace and rowspace
\begin
Discover why 18 million people worldwide trust Overleaf with their work.
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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. \[A=\begin{bmatrix}~1 & 2 & 4 & 0\\ -3~ & 1 & 5 & 2\\ -2~ & 3 & 9 & 2\end{bmatrix}.\]
\vspace{2mm}
\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. \[R=\begin{bmatrix} {\color{purple}1} & {\color{purple}0} & {\color{blue}\frac{-6}{7}} & {\color{blue}\frac{-4}{7}}\\[6pt] {\color{purple}0} & {\color{purple}1} & {\color{blue}\frac{17}{7}} & {\color{blue}\frac{2}{7}} \\ 0 & 0 & 0 & 0 \end{bmatrix}\implies N=\begin{bmatrix} {\color{purple}1} & {\color{purple}0}\\ ~{\color{purple}0} & ~{\color{purple}1} \\ ~{\color{blue}\frac{6}{7}} & {\color{blue}\frac{4}{7}} \\[6pt] ~{\color{blue}\frac{-17}{7}} & ~{\color{blue}\frac{-2}{7}} \end{bmatrix}\]
\end{questions}
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