MHB Self Adjoint and Unitary Transformations

[Sudharaka]

Hi everyone, :)

Here is a question I encountered recently.

Question:

Let \(V\) be a unitary space. Give the definitions of a self adjoint and unitary linear transformations of \(V\). Prove that \(f_1 g_1=f_2 g_2\), where \(f_1,\,f_2\) are self adjoint positive and \(g_1,\,g_2\) unitary implies \(f_1=f_2,\, g_1=g_2\) as soon as all transformations are non singular.

My Answer:

I know the definitions of the self adjoint and unitary linear transformations. As we have been taught in class they are as follows.

Let \(f:V\rightarrow V\) be a linear transformation and \((.\,,\,.)\) denote the associated Bilinear Form. Then \(f\) is called self adjoint if, \((f(x),\,y)=(x,\,f(y))\) for all \(x,\,y\in V\). Similarly \(f\) is called unitary if \((f(x),\,f(y))=(x,\,y)\) for all \(x,\,y \in V\).

Now the problem I have is how to tackle the second part of the question. If anybody could give me a hint on how to proceed that would be really nice. :)

 

[Opalg]

Let \(V\) be a unitary space. Give the definitions of a self adjoint and unitary linear transformations of \(V\). Prove that \(f_1 g_1=f_2 g_2\), where \(f_1,\,f_2\) are self adjoint positive and \(g_1,\,g_2\) unitary implies \(f_1=f_2,\, g_1=g_2\) as soon as all transformations are non singular.

I think you need to use two facts here:

(1) A positive selfadjoint transformation has a unique positive selfadjoint square root.
(2) If $g$ is unitary then $gg^*$ is the identity transformation. (The star denotes the adjoint.)​

Let $h = f_1 g_1=f_2 g_2$. Then $hh^* = f_1 g_1g_1^*f_1^* = f_1f_1^* = f_1^2$, and similarly $hh^* = f_2^2$. But $hh^*$ is selfadjoint and positive, so by (1) its square roots $f_1$ and $f_2$ must be equal. You can then use the fact that $f_1$ is nonsingular, and therefore has an inverse, to deduce that $g_1=g_2$.

[This result is a generalisation of the fact that a complex number has a unique expression in the form $z = re^{i\theta}$, where $r$ is real (or "selfadjoint") and positive, and $e^{i\theta}$ is on the unit sircle (and thus "unitary").]

 

[Sudharaka]

I think you need to use two facts here:

(1) A positive selfadjoint transformation has a unique positive selfadjoint square root.
(2) If $g$ is unitary then $gg^*$ is the identity transformation. (The star denotes the adjoint.)​

Let $h = f_1 g_1=f_2 g_2$. Then $hh^* = f_1 g_1g_1^*f_1^* = f_1f_1^* = f_1^2$, and similarly $hh^* = f_2^2$. But $hh^*$ is selfadjoint and positive, so by (1) its square roots $f_1$ and $f_2$ must be equal. You can then use the fact that $f_1$ is nonsingular, and therefore has an inverse, to deduce that $g_1=g_2$.

[This result is a generalisation of the fact that a complex number has a unique expression in the form $z = re^{i\theta}$, where $r$ is real (or "selfadjoint") and positive, and $e^{i\theta}$ is on the unit sircle (and thus "unitary").]

Hi Opalg, :)

Thank you so much. I understand everything perfectly now. Thanks again. :)