Matrix norm and intermediate value property The operator norm of a square matrix gives the maximum stretch of a vector. Does this operator norm satisfy the intermediate value property? More specifically, for every $0<\alpha<\left\vert A\right\vert$ where A is a matrix with entries in $\mathbb{R}$,can we find a vector $x\in\mathbb {R}  n$ such that $\left\vert Ax\right\vert=\alpha?$ I am not clear about application of continuity to a correct function to get the solution. Please help.
 A: Can we find a vector $x$ with $|Ax|=\alpha$? The answer to that is so obviously yes it seems clear you meant to ask a different question; since you're asking about the "stretch" surely you meant to ask whether there exists $x$ with $|x|=1$ and $|Ax|=\alpha$.
The answer is no; consider for example $A=I$ and $\alpha\ne1$.
A: Since $|A| >0$, we have $A \ne 0$, hence there is $z \in \mathbb R^n$ such that $|Az| \ne 0.$ Now put $x= \frac{\alpha}{|Az|} z$. Then we get $|Ax|= \alpha.$
A: Assuming you define the operator norm as the 2-norm of a matrix. If you update the question of finding $\lVert x \rVert_2=1$ such that $\sigma_n \leq \alpha \leq \sigma_1 = \lVert A \rVert_2$ where $\alpha := \lVert Ax \rVert_2$ and $\sigma_1$ and $\sigma_n$ are the maximum and minimum singular values of $A$ respectively, then the answer is yes. You just solve these equations
$$\begin{align}
\alpha_1 \sigma_1 + \alpha_n \sigma_n &= \alpha \\
\alpha_1^2 + \alpha_n^2 = 1
\end{align}$$
to find nonnegative $\alpha_1$ and $\alpha_n$, then $x=\alpha_1 v_1 + \alpha_n v_n$ where $v_i$ is the right singular vectors of $A$.
To see why this is true, note that $v_1^T v_n=0$, $\lVert v_i \rVert_2=1$ and $\lVert Av_i \rVert_2=\sigma_i.$
