# [SOLVED]Self Adjoint and Unitary Transformations

#### Sudharaka

##### Well-known member
MHB Math Helper
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.

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

##### MHB Oldtimer
Staff member
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").]

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#### Sudharaka

##### Well-known member
MHB Math Helper
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.