Andrew Thangaraj
Aug-Nov 2020
\(U\subseteq V\), subspace
For \(v\in V\), the translate of \(U\) by \(v\) is defined as \(v+U\).
If \(v\in U\), \(v+U=U\)
If \(v\notin U\), \((v+U) \cap U=\emptyset\)
For \(v,w\in V\), the translates \(v+U\) and \(w+U\) are either equal or disjoint. Partial overlap of two translates is not possible.
if \(v+U\) and \(w+U\) are disjoint, we are done.
if not, let \(x\in v+U\) and \(x\in w+U\).
\(x=v+u_1=w+u_2\), where \(u_1,u_2\in U\).
\(w=v+u_3\), where \(u_3=u_1-u_2\in U\).
so, \(w+U=v+(u_3+U)=v+U\) and the two translates are equal.
Quotient space, denoted \(V/U\), is the set of all translates of \(U\).
Let \(W\) be a subspace s.t. \(V=U\oplus W\).
For any translate \(v+U\), there exists \(w\in W\) such that \(v+U=w+U\).
For \(w_1,w_2\in W\), \(w_1\ne w_2\), the translates \(w_1+U\) and \(w_2+U\) are disjoint.
\(V/U=\{w+U:w\in W\}\), where \(W\) is a subspace satisfying \(V=U\oplus W\).
\(T:V\to W\), linear map. Let us consider \(V/\)null \(T\).
\(W\): subspace such that \(V=\text{null }T\oplus W\)
\[V/\text{null }T=\{w+\text{null }T:w\in W\}\]
\(T(w+\text{null }T)=Tw\) for any \(w\in W\)
\(Tw_1\ne Tw_2\) for \(w_1,w_2\in W\), \(w_1\ne w_2\)
Linear map \(T\) maps the quotient space \(V/\text{null }T\) to range \(T\) in a one-to-one manner.
\(A=\begin{bmatrix} 1&3\\ 2&6 \end{bmatrix}\)
range \(T=\) span\(\{(1,2),(3,6)\}=\) span\(\{(1,2)\}\), dim null \(T=1\)
Elementary row operations \[Ax\to \begin{bmatrix} 1&3\\ 0&0 \end{bmatrix}\]
null \(T=\) span\(\{(-3,1)\}\)
\(W=\) span\(\{(1,0)\}\)
\(\begin{bmatrix} 1&3\\ 2&6 \end{bmatrix}\left(a\begin{bmatrix}1\\0\end{bmatrix}+\begin{bmatrix}-3\\1\end{bmatrix}\right)=a\begin{bmatrix}1\\2\end{bmatrix}\)
Linear map \(T\) maps the quotient space \(V/\text{null }T\) to range \(T\) in a one-to-one manner.
\(Ax=b\)
if \(x\) is not in range \(A\)
if \(x\) is in range \(A\)
find one \(u\) such that \(Au=b\)
solution: \(u+\) null \(A\)