Matrix norm question. When do we know that $|Ax| \geq \lambda|x|$ for all $x$ where $A$ is some matrix and $\lambda$ is some constant > 0. Is it enough if all of the eigenvalues are positive? If so, can you please prove this. 
Thank you for your help.
 A: This is equivalent to say that $A$ is invertible. assume that $|Ax|\geq \lambda |x|$. This shows that $A$, as a linear map is injective, so invertible.
Now assume that $A$ is invertible. then you can put $\lambda= \parallel A^{-1} \parallel$. The later norm is the operator-norm which has the following definition:
For  a  matrice $B$, $\parallel B \parallel=\sup \frac{|Bx|}{|x|}, x\neq 0$
A: As Ali indicates, this is equivalent to injectivity (that is, to $A$ being invertible).
One direction should be clear: If $x\ne y$, then $0<\lambda|x-y|\le |A(x-y)|=|Ax-Ay|$, so $A$ is injective.
For the other direction, let me expand on their suggestion: Suppose $A$ is injective. The function $x\mapsto Ax$ is continuous. When restricted to the boundary of the unit sphere, it attains its minimum, since this boundary is compact. Since $A$ is injective, $Ax\ne0$ for all $x$ in this boundary, so the minimum is strictly positive. That is, there is a $\lambda>0$ such that $|Ax|\ge \lambda$ for all $x$ with $|x|=1$. Now, given an arbitrary $x\ne0$, $Ax=|x|A(x/|x|)\ge\lambda|x|$, using that $x/|x|$ has norm $1$. Finally, if $x=0$, $|Ax|=0=\lambda |x|$, so the inequality holds in all cases.
(What I did was to work explicitly the relevant details coded in the fact that $\|B\|$, as defined in Ali's post, is a norm.) 
Note in particular that if all eigenvalues of $A$ are positive, then $A$ is invertible, so the result holds.  
