$A^2=A $ prove $A$ is hermite matrix

Asked Dec 09 '15 at 13:04

Active Dec 09 '15 at 13:12

Viewed 92 times

Let matrix $A\in M_n(\mathbb{C})$ satisfying $A^2=A$ , for every $n\times 1$ vector $x$ we have $|Ax|\le|x|$ where $| |$ denotes the usual norm of vector , prove $A$ is a hermite matrix.

edited Dec 09 '15 at 13:12

Angelo Mark

6,226

asked Dec 09 '15 at 13:04

Idele

1,968

1

your thoughts/attempts? You are reputed enough to know that questions without personal attempt shown are discouraged here! – Jesse P Francis Dec 09 '15 at 13:07
1

See my answer here – Ben Grossmann Dec 09 '15 at 13:11
@JessePFrancis well , I don't know how the inequality condition show any information of A , and so I really don't have any thoughts , can you help me – Idele Dec 09 '15 at 13:14

$A^2=A $ prove $A$ is hermite matrix

0 Answers0