How to prove that normal matrix with property $A^2=A$ is Hermitian?

Question

I am given a matrix $A\in M(n\times n, \mathbb{C})$ normal (in matrix form $AA^*=A^*A$) and $A^2=A$. The task is to prove that the matrix is Hermitian.

But when I try something like $A^*=\,\,...$ , then I can't reach $A$, because I can't "get rid of star" in expression. Also it is not enough to show $BA=BA^*$ for some $B$ since matrix don't form a field, and I haven't got any other thoughts.

Thanks in advance!

Showing that $BA=BA^*$ for some $B$ wouldn't suffice even in a field. It would suffice in any case if $B$ were invertible (or at least not a zero divisor; but those two coincide for square matrices). — tomasz, Jul 01 '13 at 23:43
Something looks wrong to my untrained eye. If $A$ is invertible, then $A^2=A$ implies $A=I$. — bob.sacamento, Jul 01 '13 at 23:52

score 6 · Accepted Answer · answered Jul 01 '13 at 23:53

6

Hint: by spectral theorem, a normal matrix is hermitian if and only if all its eigenvalues are real. What complex numbers have the property that they are equal to their squares?

answered Jul 01 '13 at 23:53

tomasz

35,474

Not just $1$, zero too! – tomasz Jul 02 '13 at 00:07
Oh, for sure. But anyway it is real. Thanks again. – nakajuice Jul 02 '13 at 00:12

score 3 · Answer 2 · answered Jul 01 '13 at 23:55

3

Hint By spectral theorem, a normal matrix is diagonalizable by a unitary matrix.

Then what are the eigenvalues of a diagonalizable matrix $A$ which satisfies $$A^2=A?$$

answered Jul 01 '13 at 23:55

Sungjin Kim

20,102

How to prove that normal matrix with property $A^2=A$ is Hermitian?

2 Answers2