MHB Issue 2 - Tapp - Characterizations of the Orthogonal Groups

[Math Amateur]

I am reading Kristopher Tapp's book: Matrix Groups for Undergraduates.

I am currently focussed on and studying Section 2 in Chapter 3, namely:

"2. Several Characterizations of the Orthogonal Groups".

I need help in fully understanding the proof of Proposition 3.10.

Section 2 in Ch. 3, including Proposition 3.10 and its proof reads as follows:https://www.physicsforums.com/attachments/4002
https://www.physicsforums.com/attachments/4003
https://www.physicsforums.com/attachments/4004

In the proof of Proposition 3.10 (see bottom of above text) we read:

" ... ... If $$ A \in GL_n ( \mathbb{C} ) $$, then

$$ \rho_n (A) \cdot \rho_n (A)^* = \rho_n (A) \cdot \rho_n (A^*) = \rho_n (A \cdot A^* ) $$

which shows that $$A \in U(n)$$ if and only if $$\rho_n (A) \in O(2n) $$. ... ... "

I do not see how or why this follows ... ...My question, then, is as follows:

Can someone show formally and rigorously that

$$ \rho_n (A) \cdot \rho_n (A)^* = \rho_n (A) \cdot \rho_n (A^*) = \rho_n (A \cdot A^* ) $$

implies that

$$A \in U(n)$$ if and only if $$\rho_n (A) \in O(2n) $$?

I wold be grateful for some help in this matter ...

Peter
***NOTE***

Tapp introduces $$\rho_n$$ in Section 1 of Ch. 2 (pages 24-25) ... so I am providing these pages as follows:https://www.physicsforums.com/attachments/4005
https://www.physicsforums.com/attachments/4006

 

[GJA]

Hi Peter,

Given the equation

$$\rho_{n}(A)\rho_{n}(A)^{*}=\rho_{n}(A\cdot A^{*}),$$

we can get the iff as follows:

Suppose $$A\in U(n).$$ By Proposition 3.9.4, this means that $$A\cdot A^{*}=I.$$ From the above equation this gives

$$\rho_{n}(A)\rho_{n}(A)^{*}=\rho_{n}(I)=I$$

Since $$\rho_{n}(A)$$ is a real matrix we have $$\rho_{n}(A)^{*}=\rho_{n}(A)^{T},$$ where $$T$$ denotes transpose. Thus the above becomes

$$\rho_{n}(A)\rho_{n}(A)^{T}=I,$$

which means that $$\rho_{n}(A)$$ is an orthogonal matrix; i.e. $$\rho_{n}(A)\in O(2n).$$

To prove the converse, suppose $$\rho_{n}(A)\in O(2n)$$ and, essentially, reverse the argument outlined above.

Does this help? Let me know if you'd like more details on the converse proof/if anything else is unclear.

 

[Math Amateur]

Hi Peter,

Given the equation

$$\rho_{n}(A)\rho_{n}(A)^{*}=\rho_{n}(A\cdot A^{*}),$$

we can get the iff as follows:

Suppose $$A\in U(n).$$ By Proposition 3.9.4, this means that $$A\cdot A^{*}=I.$$ From the above equation this gives

$$\rho_{n}(A)\rho_{n}(A)^{*}=\rho_{n}(I)=I$$

Since $$\rho_{n}(A)$$ is a real matrix we have $$\rho_{n}(A)^{*}=\rho_{n}(A)^{T},$$ where $$T$$ denotes transpose. Thus the above becomes

$$\rho_{n}(A)\rho_{n}(A)^{T}=I,$$

which means that $$\rho_{n}(A)$$ is an orthogonal matrix; i.e. $$\rho_{n}(A)\in O(2n).$$

To prove the converse, suppose $$\rho_{n}(A)\in O(2n)$$ and, essentially, reverse the argument outlined above.

Does this help? Let me know if you'd like more details on the converse proof/if anything else is unclear.

Thank you GJA ... that seems very clear ... just working through the logic to ensure that I fully understand it ...

Peter***EDIT***

Having shown that:$$\rho_n(A) \rho_n(A)^* = I
$$could we immediately conclude from this (using Proposition 3.9) that:

$$\rho_n (A) \in O(2n)$$ We can do this since ... ... If in Proposition 3.9 we take $$\mathbb{K} = \mathbb{R}$$ and $$A := \rho_n (A)$$ - that is replace $$A$$ with $$\rho_n (A)$$ in the proposition we get:

$$A \in O_{2n} ( \mathbb{R} ) = O(2n) \ \ \ \Longleftrightarrow \ \ \ \rho_n(A) \rho_n(A)^* = I$$Is the above analysis correct? Can someone confirm that the above is correct?

Peter

 

[GJA]

Hi Again Peter,

Applying Proposition 3.9 a second time in the manner you describe would be correct as well.

 

[Math Amateur]

Hi Again Peter,

Applying Proposition 3.9 a second time in the manner you describe would be correct as well.

Thanks to GJA for the help in seeing that $$ \rho_n (A) \cdot \rho_n (A)^* = \rho_n (A) \cdot \rho_n (A^*) = \rho_n (A \cdot A^* ) $$

implies that

$$A \in U(n)$$ if and only if $$\rho_n (A) \in O(2n) $$ ... ... ... BUT ... I now have another problem/issue with Proposition 3.10 (1).
I would have thought that the previous analysis by GJC showed that

$$\rho_n ( U(n) ) = O (2n)
$$

... ... BUT ...

... Proposition 3.10 (1) reads as follows:$$\rho_n ( U(n) ) = O (2n) \cap \rho_n ( GL_n ( \mathbb{C} )
$$My question is as follows:Why do we have the term $$\rho_n ( GL_n ( \mathbb{C} )$$ intersecting with $$O (2n)$$?Can someone please explain what is going on here?

Peter

...

 

[Fallen Angel]

Hi Peter,

From here

$$A \in U(n)$$ if and only if $$\rho_n (A) \in O(2n) $$ ...

You can't conclude this

$$\rho_n ( U(n) ) = O (2n)
$$

They are simply different statements.

As an example, take a matrix $A=\left(\begin{array}{cccc}
1 & 2 & 3 & 4 \\ 5 & 6 & 7 & 8 \\ 9 & 10 & 11 & 12 \\ 13 & 14 & 15 & 16
\end{array}\right)$

And now, computing it's QR factorization we got that

$Q=\left(\begin{array}{cccc}
-0.0602 & -0.8345 & 0.2702 & -0.4765 \\ -0.3010 & -0.4576 & -0.0051 & 0.8366\\-0.5417 & -0.0808 & -0.8003 & -0.2439\\ -0.7825 & 0.2961 &0.5352 & -0.1163
\end{array}\right)$
$Q$ is computed numerically so it got some little error but we can just forgot about that in this precise example.

Then, $Q$ is orthogonal ($QQ^T=I$), so $Q\in \mathcal{O}(2n)$ but it's not the image of any complex matrix via $\rho_{n}$ (it is obvious since $Q_{1,1}\neq Q_{2,2}$).