Let $V$ be finite dim. inner product space, and let $\alpha\in\text{End}(V)$ satisfy the condition that

Question

every eigenvector of $\beta=\alpha+\alpha^*$ is also an eigenvector of $\gamma=\alpha-\alpha^*$, where $\alpha^*$ is the Hermitian conjugate. Prove that $\alpha$ is normal.

So we need to show that $\alpha\alpha^*=\alpha^*\alpha$. Let $v$ be an eigenvector for $\beta$, then there exists a scalar $c$ such that $\beta(x)=\alpha(x)+\alpha^*(x)=cx$. Now $x$ is also an eigenvector for $\gamma$, so there is some scalar $d$ such that $\gamma(x)=\alpha(x)-\alpha^*(x)=dx$. I don't know how to proceed from here. My professor did not do much with eigenvectors/spaces/values but I have quals coming up and they definitely require this knowledge. Thanks for any help in advance!

Answer 1

If $\langle \cdot, \cdot \rangle$ denotes the inner product on the complex vector space $V$, then presumably, given $\alpha \in \text{End}(V)$, $\alpha^\ast \in \text{End}(V)$ is that unique element of $\text{End}(V)$ satisfying

$\langle \alpha^\ast x, y \rangle = \langle x, \alpha y \rangle, \; \forall x, y \in V; \tag 1$

that is, $\alpha^\ast$ is the Hermitian adjoint of $\alpha$. We note that this definition yields

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

since

$\langle \alpha^{\ast \ast} x, y \rangle = \langle x, \alpha^\ast y \rangle = \overline{\langle \alpha^\ast y, x \rangle} = \overline{\langle y, \alpha x \rangle} = \langle \alpha x, y \rangle, \; \forall x, y \in V, \tag 3$

whence

$\alpha^{\ast \ast}x = \alpha x, \; \forall x \in V, \tag 4$

whence

$\alpha^{\ast \ast} = \alpha; \tag 5$

we also have, if in addition $\beta \in \text{End}(V)$,

$\langle (\alpha + \beta)^\ast x, y \rangle = \langle x, (\alpha + \beta)y \rangle = \langle x, \alpha y + \beta y \rangle$ $= \langle x, \alpha y \rangle + \langle x, \beta y \rangle$ $= \langle \alpha^\ast x, y \rangle + \langle \beta^\ast x, y \rangle$ $= \langle \alpha^\ast x + \beta^\ast x, y \rangle = \langle (\alpha^\ast + \beta^\ast) x, y \rangle, \; \forall x, y \in V, \tag 6$

so that

$(\alpha + \beta)^\ast = \alpha^\ast + \beta^\ast; \tag 7$

further, for $c \in \Bbb C$,

$\langle (c \alpha)^\ast x, y \rangle = \langle x, c \alpha y \rangle = \bar c \langle x, \alpha y \rangle = \langle \bar c \alpha^\ast x, y \rangle, \tag 8$

and so

$(c \alpha)^\ast = \bar c \alpha^\ast; \tag 9$

from properties (5) and (7) we have

$(\alpha + \alpha^\ast)^\ast = \alpha^\ast + \alpha^{\ast \ast} = \alpha^\ast + \alpha, \tag{10}$

which shows that $\alpha + \alpha^\ast$ is Hermitian. It is well-known that Hermitian operators on finite-dimensional inner product spaces such as $V$ admit a basis of eigenvectors for $V$; that is, there exist vectors $v_i \in V$, $1 \le i \le \dim V$, with

$(\alpha + \alpha^\ast) v_i = \mu_i v_i, \; \mu_i \in \Bbb C; \tag{11}$

by hypothesis, the $v_i$ are also eigenvectors for $\alpha - \alpha^\ast$:

$(\alpha - \alpha^\ast)v_i = \nu_i v_i, \; \nu_i \in \Bbb C; \tag{12}$

therefore,

$(\alpha + \alpha^\ast)(\alpha - \alpha^\ast)v_i = (\alpha + \alpha^\ast) \nu_i v_i = \nu_i (\alpha + \alpha^\ast)v_i = \nu_i \mu_i v_i; \tag{13}$

$(\alpha - \alpha^\ast)(\alpha + \alpha^\ast)v_i = (\alpha - \alpha^\ast) \mu_i v_i = \mu_i (\alpha + \alpha^\ast)v_i = \mu_i \nu_i v_i; \tag{13}$

thus,

$(\alpha + \alpha^\ast)(\alpha - \alpha^\ast)v_i = (\alpha - \alpha^\ast)(\alpha + \alpha^\ast)v_i, \tag{14}$

and we perform a few simple algebraic maneuvers:

$(\alpha + \alpha^\ast)(\alpha - \alpha^\ast)v_i = (\alpha^2 - \alpha \alpha^\ast + \alpha^\ast \alpha - \alpha^{\ast 2}) v_i, \tag{15}$

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

therefore,

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

whence,

$( - \alpha \alpha^\ast + \alpha^\ast \alpha) v_i = ( \alpha \alpha^\ast - \alpha^\ast \alpha) v_i, \tag{18}$

$2(\alpha \alpha^\ast - \alpha^\ast \alpha)v_i = 0, \tag{19}$

$(\alpha \alpha^\ast - \alpha^\ast \alpha)v_i = 0; \tag{20}$

since (20) binds for every vector of the basis $\{v_i, \; 1 \le i \le \dim V \}$ which spans $V$, by linearity we find

$(\alpha \alpha^\ast - \alpha^\ast \alpha) w = 0, \; \forall w \in V; \tag{21}$

thus

$\alpha \alpha^\ast = \alpha^\ast \alpha, \tag{22}$

and we see that $\alpha \in \text{End}(V)$ is normal.

Answer 2

I suppose $ \alpha^*$ is the Hermitian conjugate of $\alpha$?

Then $\alpha+\alpha^*$ is Hermitian, so has a basis of eigenvectors. If $x$ is one of then, then, as you say, $(\alpha+\alpha^*)x=cx$ and $(\alpha-\alpha^*)x=dx$. Then $\alpha x=\frac12(c+d)x$ and $\alpha^* x=\frac12(c-d)x$. So what are $\alpha\alpha^*x$ and $\alpha^*\alpha x$, etc.?