Norm of orthogonal projection Consider $\Bbb R^n$ with the standard inner product and let $P$ be an orthogonal projection defined on $\Bbb R^n$. It is known that the operator norm of $P$ induced by the inner product is less than or equal $1$. Let $\|\cdot\|$ be any arbitrary norm on $\Bbb R^n$. Is it true that the operator norm of $P$ induced by the new norm is less than or equal to $1$? 
 A: First, if $P$ is a nontrivial projection ($P\neq 0$), any submultiplicative norm of $P$ is bounded from below by one. This follows simply from the fact that $P$ is idempotent and $\|P\|=\|P^2\|\leq\|P\|^2$ which gives $\|P\|\geq 1$. Consequently, if $\|\cdot\|$ is the matrix norm induced by the vector norm induced by the scalar product w.r.t. which $P$ is orthogonal, then either $\|P\|=0$ or $\|P\|=1$ with the first option possible if and only if $P=0$.
The only way how to make a norm of a nontrivial projection smaller than one is to use a matrix norm which is not submultiplicative. This excludes, e.g., $p$-norms, Frobenius norm, and any matrix norm induced by a vector norm. On the other hand, some matrix norms are not submultiplicative; e.g., the $\max$-norm $\|P\|_\max:=\max\limits_{i,j}|p_{ij}|$.
E.g., with (stealing the $P$ from Omnomnomnom)
$$
P=\frac{1}{5}\pmatrix{1&2\\2&4},
$$
$$
\|P\|_2=1, \quad \|P\|_1=\|P\|_\infty=\frac{6}{5}>1, \quad \|P\|_\max=\frac{4}{5}<1.
$$
A: Not necessarily.  As an example, take
$$
P = \frac{1}{5}  \pmatrix{1&2\\2&4}
$$
which is the projection onto the span of $(1,2)$.  If we take the operator norm derived from $\|\cdot\|_{1}$ (the taxi-cab norm), we find
$$
\|P\|_1 = \frac 65 > 1
$$
Or, barring that computation, it is sufficient to note that
$$
\|e_2\|_1 = 1; \quad \|Pe_2\|_1 = \|(2/5,4/5)^T\| = \frac 65
$$
where $e_2 = (0,1)^T$
