# Contraction of a finite dimensional space

$K=\mathbb{R},\mathbb{C}$. 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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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. – user48941 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. – 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 $$\Vert T(\hat{e}_1)\Vert< 0.8\Vert \hat{e}_1\Vert\qquad \Vert T(\hat{e}_2)\Vert< 0.8\Vert \hat{e}_2\Vert$$