2
$\begingroup$

and in the other way, orthogonal nullspace and range require symmetry?

Thanks,

R

2 Answers 2

4

Hint: any $v\in V$ can be written uniquely as $v_1+v_2$, where $v_1$ is in the nullspace of $A$ and $v_2$ in the range space. Thus written, $Av=v_2$.

  • 0
    Could you give me another hint? :)2010-10-06
  • 0
    For the first direction, you need to show that $=$. Write each $u$ and $v$ as above.2010-10-06
  • 0
    By the way, to see why any $v$ can be written uniquely like that, put $v=Av+(I-A)v$.2010-10-06
2

Let $f^2=f$. Then $f$ is self-adjoint iff $f = f^* f$ iff $\langle fx,y \rangle=\langle f^* f,y \rangle $ iff $\langle fx,y \rangle =\langle fx,fy \rangle$ iff $\langle fx,fy-y \rangle =0$ for all $x,y$ iff $ran(f)$ and $ker(f)$ are orthogonal.

  • 0
    Thanks Martin. I am trying to get the intuition beyond this. I do not see in your explanation where symmetry comes in.2010-10-06
  • 0
    Well, self-adjoint = symmetric in the real case.2010-10-07