$K=\mathbb{R},\mathbb{C}$. Let $V$ be a finite dimensional normed $K$-vector space, $T:V\rightarrow V$ a linear map and $\{b_1,\dots,b_n\}\subset V$ a basis of $V$ s.t. there exist constants $0<c_i<1$, $i=1,\dots,n$ with $\|T b_i\|\le c_i\| b_i\|$.

Is $T$ a contraction?

I suppose the answer is yes.

  • 1
    $\begingroup$ What is the norm like? $\endgroup$
    – Hui Yu
    Nov 13 '12 at 12:58
  • $\begingroup$ If you suppose that the answer is yes, try to prove it. Start with the most simple but still non-trivial example. $V = \mathbb{R}^2$, with the usual inner product. Identify linear maps $T : V \rightarrow V$ with matrices. Play with diagonal matrices, and then with non-diagonal ones. $\endgroup$
    – levap
    Nov 13 '12 at 13:06
  • $\begingroup$ I think that the linear map $T$ must be a continous map. $\endgroup$
    – user48941
    Nov 13 '12 at 13:28
  • $\begingroup$ @Alisad: $V$ is finite-dimensional, so every linear map is continuous. $\endgroup$ Nov 13 '12 at 13:52
  • $\begingroup$ @MartinArgerami thanks for anamnesis. $\endgroup$
    – user48941
    Nov 13 '12 at 13:57

The answer is no. Consider map $T$ given by matrix $$ [T]= \begin{pmatrix} 2 & 0\\ 0 & 0.5 \end{pmatrix} $$ in the standard basis $\{e_1,e_2\}$ of $\mathbb{R}^2$ with euqlidean norm. The map $T$ is not a contraction, since $\Vert T(e_1)\Vert=2\Vert e_1\Vert$, so $\Vert T\Vert\geq 2$. Consider new basis $$ \hat{e}_1=e_1+0.1e_2\qquad\hat{e}_2=e_1-0.1 e_2 $$ It is straight forward to check $$ \Vert T(\hat{e}_1)\Vert< 0.8\Vert \hat{e}_1\Vert\qquad \Vert T(\hat{e}_2)\Vert< 0.8\Vert \hat{e}_2\Vert $$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.