Translates of a subspace, Quotient spaces

Andrew Thangaraj

Aug-Nov 2020


  • Vector space \(V\) over a scalar field \(F\)
    • \(F\): real field \(\mathbb{R}\) or complex field \(\mathbb{C}\) in this course
  • \(m\times n\) matrix A represents a linear map \(T:F^n\to F^m\)
    • null\((A)=\) null \(T=\{v\in V:Tv=0\}\), colspace\((A)\) = range \(T=\{Tv:v\in V\}\)
    • dim null \(T\) + dim range \(T\) = dim \(V\)
  • Linear equation: \(Ax=b\)
    • Solved using elementary row operations
    • Solution (if it exists): \(u+\) null\((A)\)

Translates of a subspace

\(U\subseteq V\), 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

Quotient space, denoted \(V/U\), is the set of all translates of \(U\).

  • How to find all translates?

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 \(V=U\oplus W\).

  • Translates of \(U\) partition \(V\)

Quotient space of null of a linear map

\(T:V\to W\), 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 \(w\in W\)

  • All vectors in \(w+\text{null }T\) mapped to same vector \(Tw\) by \(T\)

\(Tw_1\ne Tw_2\) for \(w_1,w_2\in W\), \(w_1\ne w_2\)

  • Two different vectors in \(W\) mapped to distinct vectors by \(T\)

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\(\{(1,0)\}\)

\(\begin{bmatrix} 1&3\\ 2&6 \end{bmatrix}\left(a\begin{bmatrix}1\\0\end{bmatrix}+b\begin{bmatrix}-3\\1\end{bmatrix}\right)=a\begin{bmatrix}1\\2\end{bmatrix}\)

Solving linear equations

Linear map \(T\) maps the quotient space \(V/\text{null }T\) to range \(T\) in a one-to-one manner.


  • if \(b\) is not in range \(A\)

    • no solution
  • if \(b\) is in range \(A\)

    • find one \(u\) such that \(Au=b\)

    • solution: \(u+\) null \(A\)