Prove $\lVert A \rVert _{p} \geq 1$ for any idempotent matrix $A \neq 0$

Question

As the title states,

I'm facing a problem to prove:

• for any idempotent matrix $A \in \mathbb{C}^{n \times n}$ and $A \neq 0$, $\lVert A \rVert_{p} \geq 1$.

Here the p-norm $\lVert A \rVert_{p}$ satisfies the properties:

1. $\lVert A + B \rVert_{p} \leq \lVert A \rVert_{p} + \lVert B \rVert_{p}$
2. $\lVert A B \rVert_{p} \leq \lVert A \rVert_{p} \lVert B \rVert_{p}$
3. $\lVert \alpha A \rVert_{p} = |\alpha|\lVert A \rVert_{p}$
4. $\lVert A \rVert_{p} = 0$ iff $A = 0$.

and is defined as: $\lVert A \rVert_{p} = \max_{\lVert x \rVert_{p} = 1} \lVert Ax \rVert_{p}$

I'm quite confused about how to build a connection between any p-norm of a complex matrix and a real number.

I have a feeling that maybe this connection need to be built by eigenvalue? But haven't figured out yet.

I have proved that

• $I-A$ is idempotent if $A$ is idempotent.
• $0$ and $1$ are the only possible eigenvalues of an idempotent matrix.

Any hints or solutions you could offer? Thank you very much!

Answer 1

Note that for any square matrix, the eigenvalues are bounded by the operator (p-) norm, i.e. $$ \|A\|_p \geq \lambda$$ for any eigenvector $\lambda$. This is a basic fact: Let $x$ be an eigenvector corresponding to $\lambda$, i.e. $Ax = \lambda x$. Then, $$\|Ax\|_p \leq \|A\|_p\|x\|_p$$ but also $$\|Ax\|_p = \|\lambda x\|_p = |\lambda|\|x\|_p.$$

So, if you have shown that an idempotent matrix $A$ only has eigenvalues $1$ and $0$, you already know $\|A\|_p \geq 1$.