MAT 3770 Homework 2/19/2018
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HolyAvocados TV
Last Updated:
il y a 7 ans
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Résumé:
hw
\begin
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\title{MAT 3770 Homework 2/19/2018}
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\begin{document}
\noindent \textbf{\Large MAT 2550 Hw due 2/26 \hfill Myles Miller}
\begin{enumerate}
\item
\begin{proposition}
Prove that if $A:V\rightarrow W$ is an isomorphism and $\textbf{v}_1,\textbf{v}_2,\textbf{v}_3,...,\textbf{v}_n$ is a basis for $V$, then $A\textbf{v}_1,A\textbf{v}_2,A\textbf{v}_3,...,A\textbf{v}_n$ is a basis in $W$
\end{proposition}
\begin{proof}
Proving by contradiction.
Assuming that the basis of $W$ is contains some addition vector $Av_{n+1}$, we know that, by performing the reverse of the isomorphism, we will get $v_{n+1}$, a vector unreachable through the basis of $V$, thus, since it is not in $V$'s basis, is not in $V$.
Therefore, $W$'s basis contains no extra vectors.
Assuming that the basis of $W$ is smaller than $V$'s, such that some $Av_i$ is reachable through the other vectors in $W$'s basis.
Performing the inverse isomorphism, we result in $v_i$, which is in $v$'s basis.
This is a contradiction, so we know that $W$'s basis is not smaller than $V$'s.
Given that we assumed the two to be isomorphic, and their basis' are of the same size, we know that their basis' are isomorphic, so $A\textbf{v}_1,A\textbf{v}_2,A\textbf{v}_3,...,A\textbf{v}_n$ forms $W$'s basis.
\end{proof}
\item
\begin{proposition}
Find all right inverses to the $1 \times 2$ matrix $A = (1,1)$. Show that there is no left inverse.
\end{proposition}
\begin{proof}
We know that the matrix we are looking for results the following:
$$A * A^{-1}_r = 1$$
Given that we know that $A = (1,1)$, we know that $\Ain{r}$ looks like $ \begin{bmatrix}
x\\y
\end{bmatrix}$
Thus:
$$\begin{bmatrix}
1&1
\end{bmatrix} \times \begin{bmatrix}
x\\y
\end{bmatrix} = \left[x*1 + y*1\right] = \left[1\right]$$
Therefore, we know that $x+y=1$. In efforts to eliminate variables, we recognize that $y = 1-x$, so $\Ain{r} = \begin{bmatrix}
x\\1-x
\end{bmatrix}$
Assuming there to be some $\Ain{L}$ such that $\AL \times A = 1$, we know that $\AL$ is no taller than $1$ and no wider than $1$, so that we may multiply the two.
Thus, $\AL = \begin{bmatrix}
x
\end{bmatrix}$.
However, $\begin{bmatrix}
x
\end{bmatrix}\times \begin{bmatrix}
1&1
\end{bmatrix} = \begin{bmatrix}
x&x
\end{bmatrix}\neq\begin{bmatrix}
1
\end{bmatrix}$
\end{proof}
\item
\begin{proposition}
Find all left inverses of $\bm{1&2&3}^T$
\end{proposition}
\begin{proof}
Letting $A = \bm{1\\2\\3}$
We know that $\AL \times A = 1$, which means that $\AL$ looks like $\bm{x&y&z}$
That means:$\AL \times A = \bm{x&y&z} \times \bm{1\\2\\3} = \bm{x*1+2*y + 3*z} = 1$
Which means that $x+2y+3z = 1$, so $x =1 -2y-3z$. thus $\AL = \bm{1-2y-3z&y&z}$
Proving this:
$$\bm{1-2y-3z&y&z} \times \bm{1\\2\\3} = \bm{1*\left(1-2y-3z\right)+\left(2*y\right)+\left(3*z\right)}=$$
$$\bm{1-2y-3z+2y+3z} =\bm{1}$$
\end{proof}
\item
\begin{proposition}
Is the column $(1,2.3)^T$ invertible?
\end{proposition}
\begin{proof}
Letting $\AR = \bm{x}$, the only properly sized matrix which can be multiplied with $A$ in that order.
This results $A\times \AR = \bm{1\\2\\3} \times \bm{x} = \bm{x\\2x\\3x} \neq \bm{1}$
\end{proof}
\setcounter{enumi}{5}
\item
\begin{proposition}
Suppose the product AB is invertible. Show that A is right invertible and B is left invertible.
\end{proposition}
\begin{proof}
\end{proof}
\item
\begin{proposition}
Prove that i
\end{proposition}
\begin{proof}
\end{proof}
\setcounter{enumi}{12}
\item
\begin{proposition}
Prove that i
\end{proposition}
\begin{proof}
\end{proof}
\setcounter{enumi}{0}
\item
\begin{proposition}
Prove that i
\end{proposition}
\begin{proof}
\end{proof}
\setcounter{enumi}{2}
\item
\begin{proposition}
Prove that i
\end{proposition}
\begin{proof}
\end{proof}
\end{enumerate}
$$\subset ,\subseteq ,\supset , \supseteq,\cup, \cap,\in, \not\in$$
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