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Let $\mathbb{R}^{n\times n}$ be the vector space of square matrices with real entries. For each $A\in \mathbb{R}^{n\times n}$ we consider the norms given by: $$ \displaystyle\|A\|_1=\max_{1\leq j\leq n}\sum_{i=1}^{n}|a_{ij}|; $$ $$ \displaystyle\|A\|_\infty=\max_{1\leq i\leq n}\sum_{j=1}^{n}|a_{ij}|; $$
$$ \displaystyle\|A\|_\text{max}=\max\{|a_{ij}|\}. $$ Matrix $A\in \mathbb{R}^{n\times n}$ is said to be positive definite iff $$ \langle Ax, x\rangle> 0 \quad \forall x\in\mathbb{R}^n\setminus\{0\}. $$ Let $S$ be the set of all positive definite matrices on $\mathbb{R}^{n\times n}$. Prove that $S$ is an open set in $(X,\|.\|_1)$, $(X,\|.\|_\infty$), $(X,\|.\|_\text{max})$.

I would like to thank all for their help and comments.

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So you're not assuming that a positive definite matrix is symmetric? –  littleO Nov 1 '12 at 2:32
@littleO: Thank Sir. –  blindman Nov 1 '12 at 2:34
All norms are equivalent on a finite dimensional vector space, so checking the openness in different norms is redundant. –  ronno Nov 1 '12 at 2:36
@ronno: Thank Sir for your comment. I understand what you mean. Actually, I would like to know which norm is easy to obtain the solution. –  blindman Nov 1 '12 at 2:41

1 Answer 1

up vote 2 down vote accepted

Restricting to the unit ball is always illustrating. Let $A$ be a given positive definite matrix, then there is $\delta>0$ such that \begin{equation} <Ax,x>\ge\delta \end{equation} for all $\|x\|=1$.

We use the 2-norm, defined by \begin{equation} \|A\|=\operatorname{sup}_{\|x\|=1}\|Ax\|, \end{equation} which is equivalent to any other norms.

If $B$ is very close to $A$, say, $\|B-A\|<\epsilon$, then \begin{equation} |<Bx,x>-<Ax,x>|=|<(B-A)x,x>|<\epsilon\|x\|^2, \end{equation} so if you restrict to the unit ball again then you can bound $<Bx,x>$ from below using positive definiteness of $A$ and controlling $\epsilon$, and this will lead to the positive definiteness of $B$.

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Thank you for your nice solution. –  blindman Nov 1 '12 at 4:46

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