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$K=\mathbb{R},\mathbb{C}$. 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.

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    $\begingroup$ 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
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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 $$ \Vert T(\hat{e}_1)\Vert< 0.8\Vert \hat{e}_1\Vert\qquad \Vert T(\hat{e}_2)\Vert< 0.8\Vert \hat{e}_2\Vert $$

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