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Let $A$ be a positive definite matrix such that $\|A\|>1$. Can we say that $A-I$ also is a positive matrix?

We can generalize this question for a unital $C^*$-algebra: Let ${\cal A}$ be a unital $C^*$-algebra with unit $1_{\cal A}$. If for some $a\in {\cal A}^+$ such taht $\|a\|>1$, can we say that $a -1_{\cal A}$ is positive?

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good point, also $A-I$ wants to pos def if all eigenvalues of A are bigger than 1, so all of $A-I > 0$. – mike Apr 16 '12 at 11:27
What is the norm on the set of matrices? If the considered dimension $d$ is $\geq 2$, then taking the Forbenius norm, which is sub-multiplicative, we get, taking $A=aI$, with $\sqrt d a>1$, (for example $a=2/\sqrt d$) then $A-I=(a-1)I=(\frac 2{\sqrt d}-1)I$ which is not positive definite. So we need at least a norm such that $\lVert I\rVert=1$. – Davide Giraudo Apr 16 '12 at 11:38
up vote 1 down vote accepted

Don't think so... Let A be $\begin{pmatrix} 1 & 0 \\\\ 2 & 2 \end{pmatrix}$, Its determinant is 2 and for any $z = \begin{pmatrix} x \\\\ y \end{pmatrix}$, $z^TAz = (x+y)^2 + y^2$. So it seems to satisfy the conditions. However, $A-I$ is $\begin{pmatrix} 0 & 0 \\\\ 2 & 1 \end{pmatrix}$ and so fails the condition with $z = \begin{pmatrix} -1 \\\\ 1 \end{pmatrix}$

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