Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

share|improve this question
1  
What is the norm like? –  Hui Yu Nov 13 '12 at 12:58
    
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. –  levap Nov 13 '12 at 13:06
    
I think that the linear map $T$ must be a continous map. –  Alisad Nov 13 '12 at 13:28
    
@Alisad: $V$ is finite-dimensional, so every linear map is continuous. –  Martin Argerami Nov 13 '12 at 13:52
    
@MartinArgerami thanks for anamnesis. –  Alisad Nov 13 '12 at 13:57

1 Answer 1

up vote 2 down vote accepted

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

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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