prove that an endomorphism is normal if and only if

Question

Let $\alpha \in End(V)$ where V is a complex inner product space. Define $$\alpha_1 = \frac{1}{2}(\alpha+\alpha^*)$$

and

$$\alpha_2 = \frac{1}{2i}(\alpha + \alpha^*)$$

Prove that $\alpha$ is normal if and only if $\alpha_1 \alpha_2 = \alpha_2 \alpha_1$

I have already found that $\alpha_1$ and $\alpha_2$ are selfadjoint and that $\alpha = \alpha_1+i\alpha_2$.

I also found that $\beta_1 = \alpha_1$ and $\beta_2 = \alpha_2$.

However, I am having trouble proving normal

I know that an endomorphism is normal if and only if $\alpha^*$ exists and satisfies $\alpha^*$$\alpha$ = $\alpha$$\alpha^*$

Answer 1

This problem has a problem of its own; with

$\alpha_1 = \dfrac{1}{2}(\alpha + \alpha^\ast) \tag 1$

and

$\alpha_2 = \dfrac{1}{2i}(\alpha + \alpha^\ast), \tag 2$

it follows that

$\alpha_1 \alpha_2 = \dfrac{1}{4i}(\alpha + \alpha^\ast)(\alpha + \alpha^\ast) = \dfrac{1}{4i}(\alpha^2 + \alpha \alpha^\ast + \alpha^\ast \alpha+ \alpha^{\ast 2}), \tag 3$

and

$\alpha_2 \alpha_1 = \dfrac{1}{4i}(\alpha^2 + \alpha \alpha^\ast + \alpha^\ast \alpha+ \alpha^{\ast 2}); \tag 4$

we see that

$\alpha_1 \alpha_2 = \alpha_2 \alpha_1 \tag 5$

whether or not we have

$\alpha \alpha^\ast = \alpha^\ast \alpha. \tag 6$

We also note that $\alpha_2$ as defined is not in fact self-adjoint, but skew-adjoint instead:

$\alpha_2^\ast = (\dfrac{1}{2i}(\alpha + \alpha^\ast))^\ast = (\dfrac{1}{2i})^\ast (\alpha + \alpha^\ast)^\ast = -\dfrac{1}{2i}(\alpha + \alpha^\ast) = -\alpha_2; \tag 7$

furthermore,

$\alpha_1 + i \alpha_2 = \dfrac{1}{2}(\alpha + \alpha^\ast) + i \dfrac{1}{2i}(\alpha + \alpha^\ast) = \alpha + \alpha^\ast; \tag 8$

if

$\alpha = \alpha_1 + i \alpha_2 = \alpha + \alpha^\ast, \tag 9$

then

$\alpha^\ast = 0, \tag{10}$

whence

$\alpha = (\alpha^\ast)^\ast = 0^\ast = 0. \tag{11}$

The above discussion indicates that $\alpha_2$ as presented in the text of the question has a few difficulties. If instead we set

$\alpha_2 = \dfrac{1}{2i}(\alpha - \alpha^\ast), \tag{12}$

then

$\alpha_2^\ast = (\dfrac{1}{2i}(\alpha - \alpha^\ast))^\ast = (\dfrac{1}{2i})^\ast(\alpha - \alpha^\ast)^\ast$ $= -\dfrac{1}{2i}(\alpha^\ast - \alpha^{\ast \ast}) = -\dfrac{1}{2i}(\alpha^\ast - \alpha) = \dfrac{1}{2i}(\alpha - \alpha^\ast) = \alpha_2, \tag{13}$

i.e., $\alpha_2$ is self-adjoint. Furthermore, with $\alpha_2$ as (12)

$\alpha_1 + i \alpha_2 = \dfrac{1}{2}(\alpha + \alpha^\ast) + i \dfrac{1}{2i}(\alpha - \alpha^\ast) = \alpha; \tag {14}$

if (6) binds, we clearly have

$\alpha_1 \alpha_2 = \alpha_2 \alpha_1, \tag{15}$

since in that case

$(\alpha + \alpha^\ast)(\alpha - \alpha^\ast) = \alpha^2 - \alpha^{\ast 2} = (\alpha - \alpha^\ast)(\alpha + \alpha^\ast); \tag{16}$

likewise, if (15) holds, then without assuming (6) we find by means of a calculation similar but not identical to (16),

$\alpha^2 - \alpha \alpha^\ast + \alpha^\ast \alpha - \alpha^{\ast 2} = \alpha^2 + \alpha \alpha^\ast - \alpha^\ast \alpha - \alpha^{\ast 2}, \tag{17}$

or

$2 \alpha^\ast \alpha = 2 \alpha \alpha^\ast,\tag{18}$

from which

$\alpha^\ast \alpha = \alpha \alpha^\ast, \tag{19}$

the sought-for result.