Determining the extremal values of a linear transformation I have a real, orthogonal matrix $\mathbf{Q}$ (of dimension $n \times n$) which I am using as a transformation matrix. 
My problem is, given a vector for which I know the maximum and minimum possible values of each element, can I compute the maximum and minimum values of the resultant vector after transforming via $\mathbf{Q}$?
i.e. What are the minimum and maximum values for each element of $\mathbf{w}\in \mathbb{R}^n$ given the real orthogonal matrix $\mathbf{Q}$ and a completely known vector space $\mathbb{V} = \{\mathbf{v}\in\mathbb{R}^n\}$. Where
$$\mathbf{w=Qv}$$
Apologies if my notation is weird, my linear algebra is pretty rusty.
 A: You can determine these by multiplying by $Q$. but I don't think that's what you meant. 
Perhaps what you really meant was "If I consider all vectors $v$ for which each entry $v_i$ lies between $a$ and $b$", what can I say about the coordinates of $\{Qv\}$?" 
In that case, one answer is that the largest entry of $Qv$ will be at most $\sqrt{n}b$, and the least entry will be at least $\frac{1}{\sqrt{n}}a$, at least if $a$ is positive. It's pretty easy to see this in 2-space. For the moment, assume that $a = -b$. And then we might as well assume that $b = 1$, because if you make $b$ be $2$, everything just scales up by a factor of 2, etc. 
So: you transform the square $[-1, 1] \times [-1, 1]$ by an orthogonal matrix, i.e., a rotation/reflection. What can happen? How large can the projection of the result onto the $x$ or $y$ axis be? Answer: rotate by 45 degrees. Then the vector $(1, 1)$ becomes the vector $(\sqrt{2}, 0)$. 
The case for $a > 0$ is similar: spinning around the rectangle will sweep out an annulus, whose inner and outer radii can be similarly computed. The "$\sqrt{2}$" that appeared in my answer did so because there were two entries in the vector; in $n$-space, that becomes a $\sqrt{n}$.
